Mass percent is a technique of expressing or defining a concentration or a component in a combination. The mass percent of oxygen in chromium(III) nitrate is approximately 60.5%.
To calculate the mass percent of oxygen in chromium(III) nitrate (Cr(NO₃)₃), we need to determine the molar mass of the compound and the molar mass of the oxygen atoms in the compound.
The molar mass of Cr(NO₃)₃ can be calculated by adding the molar masses of the individual atoms in the compound:
Molar mass of Cr(NO₃)₃ = (1 x molar mass of Cr) + (3 x molar mass of N) + (9 x molar mass of O)
Molar mass of Cr(NO₃)₃ = (1 x 52.00 g/mol) + (3 x 14.01 g/mol) + (9 x 16.00 g/mol)
Molar mass of Cr(NO₃)₃ = 238.03 g/mol
The molar mass of the oxygen atoms in Cr(NO₃)₃ can be calculated by multiplying the molar mass of oxygen by the number of oxygen atoms in the compound:
Molar mass of O in Cr(NO₃)₃ = 16.00 g/mol x 9
Molar mass of O in Cr(NO₃)₃ = 144.00 g/mol
To calculate the mass percent of oxygen in Cr(NO₃)₃, we can use the formula:
The mass percent of oxygen = (mass of O in compound / molar mass of compound) x 100%
Mass percent of oxygen = (144.00 g/mol / 238.03 g/mol) x 100%
The mass percent of oxygen ≈ 60.5%
Therefore, the mass percent of oxygen in chromium(III) nitrate is approximately 60.5%.
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To calculate the mass percent of oxygen in chromium(III) nitrate, we need to determine the molar mass of the compound and the molar mass of oxygen. The molar mass of Cr(NO3)3 is calculated to be 291.97 g/mol. The mass percent of oxygen in chromium(III) nitrate is approximately 55.24%.
Explanation:To calculate the mass percent of oxygen in chromium(III) nitrate, we need to determine the molar mass of the compound and the molar mass of oxygen. The formula for chromium(III) nitrate is Cr(NO3)3.
The molar mass of Cr(NO3)3 can be calculated as follows:
Molar mass of Cr = 51.996 g/mol
Molar mass of N = 14.007 g/mol
Molar mass of O = 15.999 g/mol
Molar mass of Cr(NO3)3 = (51.996 g/mol) + 3*(14.007 g/mol) + 9*(15.999 g/mol) = 291.97 g/mol
The molar mass of oxygen is 15.999 g/mol. The mass percent of oxygen in chromium(III) nitrate can be calculated as:
Mass percent of O = (mass of O / molar mass of Cr(NO3)3) * 100%
Mass percent of O = (9*(15.999 g/mol) / 291.97 g/mol) * 100% = 55.24%
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If you are asked to make 40 mL of a 1.0% (w/v %) agarose gel, how many grams of agarose will you add to the 40ml of buffer
Answer:
0.4g
Explanation:
1.0% (w/v%) = 1 g of agarose 100 ml of Tris-Acetate-EDTA, this is the buffer that agarose is run with
the amount of agarose for 40 ml = 1 /100 × 40 ml = 0.4 g
Final answer:
To prepare 40 mL of a 1.0% (w/v %) agarose gel, measure and mix 0.4 grams of agarose powder with 40 mL of buffer solution in an Erlenmeyer flask.
Explanation:
If you are asked to make 40 mL of a 1.0% (w/v %) agarose gel, you will add 0.4 grams of agarose to the 40 mL of buffer. To made this calculation, you need to understand the meaning of w/v %, which stands for weight/volume percentage. It is defined as the mass of a solute (in this case agarose powder) divided by the volume of the solution, and then multiplied by 100 to get the percentage.
For a 1.0% (w/v %) agarose gel, this means that you need 1 gram of agarose powder for every 100 mL of solution. Therefore, for 40 mL of solution, you simply use the proportion:
1 gram/100 mL = X grams/40 mL
Solving for X gives you:
X = (1 gram/100 mL) * 40 mL = 0.4 grams
Conclusion: To prepare a 40 mL agarose gel at 1.0% w/v concentration, weigh out 0.4 grams of agarose powder using an electronic scale, and then blend it with the appropriate buffer in an Erlenmeyer flask.
Student measured the absorbance of substance X in aqueous solution depending on concentration. He used 0.5 cm cuvette for his measurements; concentration was expressed in moles/L and plotted on x-axis, while absorbance was on y-axis. His data were best fitted by the straight line with equation y = 2120x + 0.0112. The extinction coefficient for substance X in aqueous solution is:
Answer:
The extinction coefficient is 4240 M⁻¹cm⁻¹
Explanation:
The Beer-Lambert law for the absorbance of a monochromatic light by a molecule is given by the following mathematical expression:
[tex]A=EbC[/tex]----------------------(1)
Here, E is the extinction coefficient, b is the length of light pathway through the solution and C is the concentration in mol/L. This is a linear equation and fits well with its formula
y = mx + C-------------------(2)
Here, m is the slope and C is the y intercept. y and x are the two variables. Comparing equation 1 and 2, Absorbance (A) and molar concentration (C) are the two variables, while the product of extinction coefficient (E) and path length (b) makes the slope of the line. The y-intercept in the given equation is due to the solvent interference.
From this information, we can determine the extinction coefficient by the slope of the line with the following formula.
[tex]m=Eb[/tex]
[tex]E=\frac{2120 (M^{-1} )}{0.5(cm)}[/tex]
[tex]E= 4240 M^{-1}cm^{-1}[/tex]
A chemist prepares a solution of silver perchlorate by measuring out of silver perchlorate into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's silver perchlorate solution. Round your answer to significant digits.
To calculate the concentration of a solution, moles of solute are divided by the final volume in liters. Without the specific mass of silver perchlorate, we cannot determine the concentration of the solution.
Explanation:To calculate the concentration of the chemist's silver perchlorate solution, we need to know the mass of silver perchlorate used and the final volume of the solution. The details of the mass have been omitted from the question, but typically, you would use the molar mass of silver perchlorate to determine the number of moles. To prepare a solution of desired concentration, the chemist would dissolve silver perchlorate in a volumetric flask and add water to the mark, ensuring the final volume accounts for the space occupied by the dissolved solute.
Once a known mass of silver perchlorate is dissolved to make a known final volume, the molarity (M) can be calculated by dividing the number of moles of silver perchlorate by the final volume of the solution in liters. For example, if we dissolved 50 grams of silver perchlorate and the molar mass is X g/mol, we would first calculate the moles as 50 g / X g/mol. Then if the final volume of the solution is 1 L, the concentration would be 50 g / X g/mol / 1 L = Y M.
If the question had provided specific values, we would use them in the calculation and round the answer to the appropriate number of significant digits. However, without the exact mass of silver perchlorate used, we cannot complete this calculation.
A student is carrying out a biological preparation that requires 1 M NaCl to maintain an ionic strength of 1.0. The student chooses to use 1.0 M ammonium sulfate instead. Why is this a serious error?
Explanation:
It is known that the ionic strength of a solution, I, is represented as follows.
[tex]I = \frac{1}{2} \sum m_{i}z^{2}_{i}[/tex]
where, [tex]m_{i}[/tex] = the concentration of the ion,
[tex]z_{i}[/tex] = the charge of the ion
Now, for 1 M NaCl solution, the ionic strength will be calculated as follows.
I = [tex]\frac{1}{2} [(1 \times 1^{2}) + (1 \times 1^{2})][/tex]
= 1.00
And, for 1 M [tex](NH_{4})_{2}SO_{4}[/tex], concentration of [tex][NH_{4}][/tex] = 2 M and [tex][SO_{4}][/tex] = 1 M.
I = [tex]\frac{1}{2} [(2 \times 1^{2}) + ((1 \times 2^{2})][/tex]
= 3.00
As the ionic strength of 1 M [tex](NH_{4})_{2}SO_{4}[/tex], is different therefore it is a serious error.
Using 1.0 M ammonium sulfate instead of 1 M NaCl is a serious error in the biological preparation due to the differences in dissociation and specific properties of the compounds.
Explanation:The student's choice to use 1.0 M ammonium sulfate instead of 1 M NaCl is a serious error because they are not equivalent compounds. Ammonium sulfate dissociates into two ions (NH4+ and SO42-), while NaCl only dissociates into two ions (Na+ and Cl-). Therefore, the ionic strength of the solution will not be the same.
In addition, the presence of different ions in ammonium sulfate will affect the specific properties and behavior of the solution compared to NaCl. For example, the presence of ammonium ions can affect the pH and the precipitation of certain compounds.
It is important to use the correct reagents in a biological preparation to maintain the desired conditions and ensure accurate results.
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For a certain chemical reaction, ΔH∘=−156kJ. Assuming the reaction is at equilibrium, classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the net reaction.
a. Increase the temperature.
b. Decrease the temperature.
Explanation:
As per Le Chatelier's principle, any disturbance caused in an equilibrium reaction will tend to shift the equilibrium in a direction away from the disturbance.
Since, the given reaction is exothermic in nature so, when we increase the temperature then reaction will shift in the direction opposing the increase in temperature. Hence, the equilibrium will shift on left side when we increase the temperature.
On the other hand, when we decrease the temperature then the equilibrium will shift on the right side.
The correct classifications for the actions based on the given are as follows:
a. Increase the temperature: This causes a rightward shift in the direction of the net reaction.
b. Decrease the temperature: This causes a leftward shift in the direction of the net reaction.
For a chemical reaction at equilibrium, Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
The reaction in question has a standard enthalpy change of 156 kJ, which indicates that the reaction is exothermic (releases heat).
a. Increase the temperature:
When the temperature is increased for an exothermic reaction, the system tries to absorb the excess heat to restore equilibrium. Since the forward reaction releases heat, the system will favor the reverse reaction to absorb heat. However, because the forward reaction is exothermic and already releases heat, increasing the temperature actually provides energy to the system, which drives the reaction forward to consume the added energy. Therefore, increasing the temperature causes a rightward shift in the direction of the net reaction.
b. Decrease the temperature:
Conversely, when the temperature is decreased, the system tries to release heat to restore equilibrium. For an exothermic reaction, the forward reaction releases heat, so decreasing the temperature will favor the forward reaction to release heat and restore equilibrium. Therefore, decreasing the temperature causes a leftward shift in the direction of the net reaction, favoring the products since the forward reaction is exothermic and will release heat.
In summary, for an exothermic reaction, increasing the temperature will shift the equilibrium to the right (towards the products), and decreasing the temperature will shift the equilibrium to the left (towards the reactants).
If you boil a sample of water, the steam that results will have: Select the correct answer below: less mass than the water more mass than the water the same mass as the water impossible to predict
Answer:
it same mass as water
Explanation:
Final answer:
The steam resulting from boiled water will have the same mass as the water from which it was boiled, adhering to the law of conservation of mass, as long as the system is closed and no mass is lost due to external factors.
Explanation:
When you boil a sample of water, the steam that results will have the same mass as the water if you ignore any minor mass loss due to gases dissolved in water escaping or small amounts of water that may stick to the container. This is because the mass of a substance remains constant during a phase change; only the state of the substance changes from liquid to gas. This is in accordance with the law of conservation of mass.
The mass of a substance does not change as it undergoes physical transformations; thus, when you boil water, all of the mass is converted to steam provided the system is closed and no mass escapes. In an open system, there might be a loss of mass due to external factors, but the boiling process itself does not add or remove mass from the water.
Under what conditions does an increase in temperature turn a nonspontaneous process into a spontaneous process?
Choose one or more:
a. ΔH < 0, ΔS > 0
b. ΔH > 0, ΔS > 0
c. ΔH > 0, ΔS < 0
d. ΔH < 0, ΔS < 0
(What I'm trying to figure out is if you need to consider T being negative OR positive to start)
Answer:ΔH < 0, ΔS > 0
Explanation:
Consider ∆G= ∆H-T∆S
If ∆H is negative and and ∆S is positive, when T is increased, T∆S becomes more positive until the result of the difference between ∆H-T∆S becomes negative. Remember that a reaction is only spontaneous when ∆G is negative. Hence under these conditions, ∆G becomes negative and the reaction becomes spontaneous.
) A 0.907 M lead II nitrate solution has a density of 1.252 g/cm3. Find the molality of the solution.
Answer: The molality of the solution is 0.953 m
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute
[tex]V_s[/tex] = volume of solution in ml
Given : 0.907 moles of lead(ii)nitrate is dissolved in 1000 ml of the solution.
density of solution= 1.252 g/ml
Thus mass of solution = [tex]Density\times volume=1.252\times 1000 ml=1252g[/tex]
mass of solute =[tex]moles\times {\text {Molar mass}}=0.907mol\times 331g/mol=300g[/tex]
mass of solvent =mass of solution - mass of solute = (1252-300)g= 952 g
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.
[tex]Molality=\frac{n\times 1000}{W_s}[/tex]
where,
n = moles of solute = 0.907 moles
[tex]W_s[/tex]= weight of solvent in g = 952 g
[tex]Molality = \frac{0.907\times 1000}{952g}=0.953m[/tex]
Thus molality of the solution is 0.953 m
Final answer:
To find the molality of the 0.907 M lead(II) nitrate solution, we first calculate the mass of the solvent in kilograms and then divide the amount in moles of the solute by this mass. The final molality is approximately 0.953 mol/kg.
Explanation:
The question is asking to find the molality of a 0.907 M lead(II) nitrate solution with a density of 1.252 g/cm3. To calculate the molality, we need the number of moles of solute and the mass of the solvent in kilograms.
First, calculate the mass of 1 litre (1000 cm3) of solution using the density:
Mass of solution = Density imes Volume = 1.252 g/cm3 imes 1000 cm3 = 1252 g
Next, the mass of the solute in 1 litre of solution:
Mass of solute = Molarity imes Molar Mass imes Volume = 0.907 mol/L imes 331.2 g/mol imes 1 L = 300.2794 g
Now, find the mass of the solvent (water) by subtracting the mass of solute from the total mass of the solution:
Mass of solvent = Mass of solution - Mass of solute = 1252 g - 300.2794 g = 951.7206 g
Convert mass of solvent to kilograms:
Mass of solvent (kg) = 951.7206 g imes (1 kg / 1000 g) = 0.9517206 kg
Finally, calculate the molality (m):
Molality (m) = Moles of solute / Mass of solvent (kg) = 0.907 mol / 0.9517206 kg = 0.953 m
Therefore, the molality of the lead(II) nitrate solution is approximately 0.953 mol/kg.
A graduated cylinder contains 20.8 mL of water. What is the new water level, in milliliters, after 35.2 g of silver metal is submerged in the water?
Answer: The new water level of the cylinder is 24.16 mL
Explanation:
To calculate the volume of water displaced by silver, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of silver = 10.49 g/mL
Mass of silver = 35.2 g
Putting values in above equation, we get:
[tex]10.49g/mL=\frac{35.2g}{\text{Volume of silver}}\\\\\text{Volume of silver}=\frac{35.2g}{10.49g/mL}=3.36mL[/tex]
We are given:
Volume of graduated cylinder = 20.8 mL
New water level of the cylinder = Volume of graduated cylinder + Volume of water displaced by silver
New water level of the cylinder = (20.8 + 3.36) mL = 24.16 mL
Hence, the new water level of the cylinder is 24.16 mL
A flash distillation chamber operating at 101.3 kpa is separating an ethanol water mixture the feed mixture contains z weight ethanol and a molar flow rate of a feed is
Answer:
The answer is [tex][\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454[/tex] mol/hr
[tex][\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454[/tex]mol/hr
Explanation:
For flash distillation
F = V+L
[tex]\frac{V}{F} + \frac{L}{F} = 1[/tex]
[tex]\frac{F}{V} -\frac{L}{V} = 1[/tex]
Fz = Vy+Lx
Y = [tex]\frac{F}{V}\times Z - \frac{L}{V}\times X[/tex] let, [tex]\frac{V}{F} = F[/tex]
[tex]y = \frac{Z}{F} -[ \frac{1}{F} -1]\times X[/tex]
Highlighted reading
F = 299; [tex]\frac{V}{F}[/tex] = 0.85 ; z = 0.36
y = [tex]\frac{0.36}{0.85} - (-0.15)\times X[/tex]
= 0.423 + 0.15x ------------(i)
[tex]y^{*}[/tex] = -43.99713[tex]x^{6}[/tex] + 148.27274[tex]x^{5}[/tex] - 195.46[tex]x^{4}[/tex]+127.99[tex]x^{3}[/tex]-43.3[tex]x^{2}[/tex]+ 7.469[tex]x^{}[/tex]+ 0.02011
At equilibrium, [tex]y^{*}[/tex] = y
0.423+0.15[tex]x^{}[/tex] = [tex]y^{*}[/tex]
-43.99713[tex]x^{6}[/tex]+ 148.27274[tex]x^{5}[/tex] - 195.46[tex]x^{4}[/tex]+127.99[tex]x^{3}[/tex]-43.3[tex]x^{2}[/tex]+ 7.319[tex]x^{}[/tex]-0.403
F(x) for Newton's Law
Let [tex]x_{0} = 0[/tex]
[tex]x_{1}[/tex] = [tex]\frac{0-[{-0.403}]}{7.319}[/tex]
= 0.055
[tex]x_{2}[/tex] = [tex]\frac{{0.055}-{f(0.055)} {{{{{{{}}}}}}}}{f^{'} (0.055)}[/tex]
= [tex]\frac{{(0.055)}-(-0.11)}{3.59}[/tex]
= 0.085
[tex]x^{3}[/tex] = [tex]\frac{{0.085}-(0.024)}{2.289}[/tex]
= 0.095
[tex]x^{4}[/tex] = [tex]\frac{{0.095}-(-0.0353)}{-1.410}[/tex]
= 0.07
From This x and y are found from equation (i) and L and V are obtained from [tex]\frac{V}{F}[/tex] and F values
[tex][\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454[/tex] mol/hr
[tex][\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454[/tex]mol/hr
The question relates to flash distillation with a focus on an ethanol-water azeotropic mixture. It addresses the limitation of distillation due to the azeotrope formation where the maximum ethanol concentration is about 95.6%. Understanding vapor-liquid equilibrium and distillation curves is essential for this process.
Explanation:The question is concerned with the principles of flash distillation and how it relates to azeotropic mixtures, specifically an ethanol-water mixture. When dealing with flash distillation, we are interested in how the composition of the mixture changes as it goes through vaporization and condensation cycles. Flash distillation is often used for the separation of components in a liquid mixture that have significantly different volatilities. The crux of this process is that the vapor formed will be richer in the more volatile components, in this case, ethanol.
Regarding the ethanol-water azeotrope, the mixture cannot be distilled beyond a certain concentration of ethanol due to the constant boiling nature at that specific composition. An azeotrope is a mixture of two or more liquids whose proportions cannot be altered by simple distillation. In the case of ethanol and water, the maximum concentration of ethanol that can be obtained by simple distillation is approximately 95.6%. This is because, at this concentration, the mixture behaves as if it were a single component with a unique boiling point, and further separation by boiling alone is not possible without additional chemical agents or advanced separation techniques like extractive distillation.
In chemical engineering, it's also important to consider the vapor-liquid equilibrium and how it is represented by phase diagrams, such as the distillation curve given in this problem. Understanding these diagrams is critical for designing and operating distillation processes, whether in the context of a laboratory setup or an industrial scale operation.
By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monopro-tic
Answer: The molar mass of monoprotic acid is 135.9 g/mol
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of NaOH solution = 0.1008 M
Volume of solution = 15.0 mL
Putting values in above equation, we get:
[tex]0.1008M=\frac{\text{Moles of NaOH}\times 1000}{15.0}\\\\\text{Moles of NaOH}=\frac{(0.1008\times 15.0)}{1000}=0.00151mol[/tex]
As, the acid is monoprotic, it contains 1 hydrogen ion
1 mole of [tex]OH^-[/tex] ion of NaOH neutralizes 1 mole of [tex]H^+[/tex] ion of monoprotic acid
So, 0.00151 moles of [tex]OH^-[/tex] ion of NaOH will neutralize [tex]\frac{1}{1}\times 0.00151=0.00151mol[/tex] of [tex]H^+[/tex] ion of monoprotic acid
Moles of monoprotic acid = 0.00151 moles
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of monoprotic acid = 0.00151 mole
Given mass of monoprotic acid = 0.2053 g
Putting values in above equation, we get:
[tex]0.00151mol=\frac{0.2053g}{\text{Molar mass of monoprotic acid}}\\\\\text{Molar mass of monoprotic acid}=\frac{0.2053g}{0.00151mol}=135.9g/mol[/tex]
Hence, the molar mass of monoprotic acid is 135.9 g/mol
The molar mass of the monoprotic acid, which can be calculated using the titration method with sodium hydroxide, was found to be approximately 135.7 g/mol.
Explanation:The subject of your query concerns on calculating the molar mass of an organic acid using the titration method with known information about sodium hydroxide. First, we need to look into the moles of sodium hydroxide (NaOH) used. Since we know the volume V=15.0 mL and concentration C=0.1008 M of NaOH used, moles (n) can be calculated using the equation n=CV, converting volume to liters, gives n = 0.1008 mol/L * 0.015 L = 0.001512 mol.
Because the sodium hydroxide and monoprotic acid react in a 1:1 ratio (as given by the neutralization equation NaOH + HA -> NaA + H2O), the moles of acid are also 0.001512.
The molar mass of the substance can then be determined by dividing the given mass of the sample by the number of moles. The mass of the organic acid is 0.2053g, so the molar mass (M) is M = mass / n = 0.2053g / 0.001512 mol = 135.7 g/mol.
This means that the molar mass of the monoprotic organic acid is approximately 135.7 g/mol.
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Sometimes, as we learned in this lesson, a single Lewis structure does not describe a molecule, and it instead resonates between two or more structures. What symbol do we place between these structures to show this?
A. ←
B. ≡
C. ↔
D. →
Answer:
C. ↔
Explanation:
Resonating structure -
These are the set of lewis structures of the same compound .
Where the structure helps to show the delocalization of the electrons over the structure .
The charge on all the resonating structure remains the same .
All the structures can be interconverted to each other , and are shown by the arrow ↔ .
Hence , from the given information of the question,
The correct option is C. ↔
If you are trying to pipet an unknown liquid with a pipetman and the liquid keeps running out of the tip before you can transfer it, what are two possible reasons for this? What can you do to remedy the situation?
Answer:
This is due to the physical properties of the sample, since it affects the volume dispensed.
Explanation:
For example, in the case of very dense samples, selected samples to adhere to the surface of the tip, dispensing more slowly. In contrast, ethanol samples are less viscous and more volatile and are dispensed more rapidly. Some of the ways to minimize these inconveniences are the use of ultra low retention pipette tips, since they have a hydrophobic plastic additive that prevents the liquid from adhering to the inside of the tip.
Another way is to use the reverse pipetting.
The dissolving of ammonium nitrate is often used in an instant cold pack used to relieve swelling.
If this process is an endothermic reaction, i.e. absorbs heat, then why does it feel cold to the touch instead of hot?
Answer:
Here's what I get
Explanation:
Does an ice cube feel cold to the hand?
The melting of ice is an endothermic process.
If you hold an ice cube, heat flows from your hand to the ice. When your hand loses heat, it feels cold.
In the same way, the endothermic dissolving of ammonium nitrate removes the heat from your hand, so the pack feels cold.
Write the following numbers in standard notation, maintaining the same number of significant figures.
6.104 x10^2
9.5 x 10^-3
Move the decimal by a certain number of spaces, according to the exponent of 10
Answer:
6.104x10^2=610.4
9.5x10^3=9500
Explanation:
To convert from scientific notation (6.104 x10^2, 9.5 x 10^-3) to standard notation, move the decimal point in accordance with the power of 10 (right for positive, left for negative). The converted numbers become 610.4 and 0.0095, respectively.
Explanation:In scientific notation, the base number is expressed as a decimal between 1 and 10, and tens are represented by powers of 10. In standard notation, we write the number in usual decimal representation.
To convert 6.104 x10^2 to standard notation, move the decimal two places to the right (because the exponent is positive), giving you 610.4.
For 9.5 x 10^-3, move the decimal three places to the left (because the exponent is negative). So, it becomes 0.0095.
Remember that the number of significant figures must remain the same when converting to standard notation.
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Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product?A. 2C(s)+H2(g)→C2H2(g); ΔG∘f=209.2 kJ/molB. N2(g)+3H2(g)→2NH3(g); ΔG∘f=−33.30 kJ/molC. 2C(s)+2H2(g)→C2H4(g); ΔG∘f=68.20 kJ/molD. 2SO(g)+O2(g)→2SO2(g); ΔG∘f=−600.4 kJ/mol
Answer:
Option B and D
Explanation:
The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system.
A feasible way to synthesize the product entails the spontaneity of the reaction.
The spontaneity of a reaction is explained by the standard gibb's free energy.
If Delta-G = -ve ( the reaction is spontaneous)
if Delta -G = +ve ( the reaction is non-spontaneous)
if Delta-G = 0 ( the reaction is at equilibrium)
Hence the option (B and D) with negative Gibb's free energy are the reaction that are spontaneous.
Calculate the molecular weight when a gas at 25.0 ∘C and 752 mmHg has a density of 1.053 g/L . Express your answer using three significant figures.
Answer:
26.0 g/mol is the molar mass of the gas
Explanation:
We have to combine density data with the Ideal Gases Law equation to solve this:
P . V = n . R .T
Let's convert the pressure mmHg to atm by a rule of three:
760 mmHg ____ 1 atm
752 mmHg ____ (752 . 1)/760 = 0.989 atm
In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.
Moles = Mass / molar mass.
We can replace density data as this in the equation:
0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K
(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x
0.0405 mol = 1.053 g / x
x = 1.053 g / 0.0405 mol = 26 g/mol
What are the four important steps needed to obtain accurate meting points on a Mel-Temp? Select four options:
1. Grind the sample into a fine powder.
2. Put the smallest amount you can see in a capillary tube.
3. Fill the capillary tube with about 3mm of powder.
4. Set the voltage to rise quickly to 20 degrees below the expected temperature, then re-adjust so the temp rises at 2 degrees per minute.
5. Set the voltage to rise quickly to 20 degrees below the expected temperature, then re-adjust so the temp rises at 10 degrees per minute.
6. Record the temp at which liquid first appears and the temp at which the last crystal disappears.
7. Record the temp at which the sample completely melts and the temp when it boils.
To obtain accurate melting points on a Mel-Temp, grind the sample into a fine powder, put the smallest visible amount in a capillary tube, fill the tube with about 3mm of powder, set the voltage to rise quickly to 20 degrees below expected temperature and slow down to 2 degrees per minute, and record the temperature at which liquid first appears and the last crystal disappears.
Explanation:The four important steps needed to obtain accurate melting points on a Mel-Temp are as follows:
Grind the sample into a fine powder.Put the smallest amount you can see in a capillary tube. Fill the capillary tube with about 3mm of powder. Set the voltage to rise quickly to 20 degrees below the expected temperature, then re-adjust so the temperature rises at about 2 degrees per minute.Record the temperature at which liquid first appears and the temperature at which the last crystal disappears. This gives the range within which the melting point falls.By following these steps, you can ensure that you are determining the melting point accurately.
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Calculate the volume in liters of a silver(II) oxide solution that contains mg of silver(II) oxide . Round your answer to significant digits
The question is incomplete, here is the complete question:
Calculate the volume in liters of a [tex]8.75\times 10^{-5}M[/tex] silver (II) oxide solution that contains 200.g of silver (II) oxide . Round your answer to 3 significant digits.
Answer: The volume of solution is [tex]1.84\times 10^4L[/tex]
Explanation:
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
We are given:
Molarity of solution = [tex]8.75\times 10^{-5}M[/tex]
Given mass of silver (II) oxide = 200. g
Molar mass of silver (II) oxide = 124 g/mol
Putting values in above equation, we get:
[tex]8.75\times 10^{-5}M=\frac{200}{124\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{200}{124\times 8.75\times 10^{-5}}=1.84\times 10^4L[/tex]
Hence, the volume of solution is [tex]1.84\times 10^4L[/tex]
To calculate the volume in liters of a silver(II) oxide solution, you need to know the amount of the solution in milligrams and the density of the solution. Use the formula Volume (in liters) = x mg / (density in mg/L) and round the final answer to the appropriate number of significant digits.
Explanation:To calculate the volume in liters of a silver(II) oxide solution, we need to know the amount of the solution in milligrams. Let's say the solution contains x mg of silver(II) oxide.
To convert mg to liters, we need to know the density of the solution. Once we have the density, we can use the formula:
Volume (in liters) = x mg / (density in mg/L)
Round the final answer to the appropriate number of significant digits.
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Fracking involves the ______drilling to extract oil and natural gas from the rock. Select one: a. vertical b. lateral c. horizonta
Answer:
Fracking involves the ______drilling to extract oil and natural gas from the rock. Select one: a. vertical b. lateral c. horizontal
The answer is option C (horizontal)
Explanation:
Fracking commonly known as Hydraulic fracturing is an oil and gas production technique that has enabled the extraction of oil and gas from rock (shale rock).
A fracturing fluid typically water with additives is pumped under high pressure into the drilling pipe to widen fractures in the rock or to create new ones and and allow trapped gas or crude oil to flow through a pipe to a wellhead at the surface, deep holes are drilled down into the shale rock. These fractures are held open by an injected proppant, typically sand and release hydrocarbons. Fracking takes about three to five days. Since shale reserves are typically distributed horizontally rather than vertically, horizontal directional drilling method is employed.
Some of the impacts of Fracking on the environment are:
It consume large quantities of fresh waterThe injection of recovered fracking water into underground disposal wells and the high pressure used to extract oil and gas from rock may cause earthquakes.Fracking causes air and water pollution.Deforestation caused by fracking has deep effects on the environment.Fracking, or hydraulic fracturing, primarily involves horizontal drilling. Starting vertically, the drill transitions to a horizontal path to enhance contact with gas or oil-bearing rocks, thereby improving extraction efficiency.
Explanation:Fracking, also known as hydraulic fracturing, involves the horizontal drilling to extract oil and natural gas from the rock. The process begins with a vertical hole that reaches the desired depth and then transitions into lateral or horizontal drilling. This allows for increased contact with the oil or gas-bearing rock, improving the extraction process and its efficiency.
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When heating a flammable or volatile solvent for a recrystallization, which of these statements are correct? More than one answer may be correct.
1)You should not use an open flame to heat the solvent.
2)You should heat the solvent in a stoppered flask to keep vapor away from any open flames.
3)You should ensure that no one else is using an open flame near your experiment.
Explanation:
A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.
Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.
Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.
And, you should ensure that no one else is using an open flame near your experiment.
Thus, we can conclude that following statements are correct:
You should heat the solvent in a stoppered flask to keep vapor away from any open flames. You should ensure that no one else is using an open flame near your experiment.When heating a flammable or volatile solvent for recrystallization, one should not do the heating using an open flame or heat near any open flame in the laboratory.
What are flammable or volatile liquids?They are liquids that can catch flame easily either directly or through their vapors.
This means that such liquids should not be heated using an open flame or near any open flame to prevent fire accidents.
Heating near an open flame may cause the vapor to catch fire. Heating in a stoppered flask is not ideal because the main aim of heating is to allow the vapor to escape.
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Given the elements Cl, 784 kJ-mol"', match the atoms with their first ionization energies Ge, and K and three values of possible first ionization energies, 418, 1255, a. CI (418), Ge (784), and K (1255 kJ-mol) b. CI (1255), Ge (784), and K (418 kJ-mol) c. CI (784), Ge (1255), and K (418 kJ.mol) d. CI (1255), Ge (418), and K (784 kJ-mol"') e. CI (418), G (1255), and K (784 kJ-mol-)
Answer:
b. CI (1255 kJ/mol), Ge (784 kJ/mol), and K (418 kJ-mol)
Explanation:
In general, the first ionization energy for a given period increases as we go from left to right in the periodic table (there are some exeptions as with every rule), and the first ionization energy decreases a we go down in the periodic table. The reason for this are:
1. As we move from left to right in a group, the effective nuclear charge increases which makes it harder to remove the electron.
2. As we increase the period in going top to bottom in the periodic table we are adding another shell farther away from the nucleus, making it easier to remove the electron.
Given the three values for the first ionization energy in kJ/mol : 418, 784 and 1255, we expect the highest value to correspond to Cl which belongs to period 3 (K and Ge belong to period 4).
Now comparing K and Ge which belong to period 4, Ge will have a higher effective nuclear charge than K .
So the match will be Cl (1255 kJ/mol), Ge (784) and K(418)
Correct answer is b.
Note: there could be some confusion since the value of 784 was misplaced in the question statement, but we can deduce that in this question we are asked to match the values for the atoms.
A block of substance has a width of 3.8 cm, a length of 8.2 cm, and height of 7 cm . Its mass is 0.68 kg. Calculate the density of the substance.
Answer:
d = 3.11 g /cm³
Explanation:
Volume -
The volume of a cuboidal object can be calculated by multiplying the length , width and height ,
Hence , Volume ( V ) = l * w * h ,
From the question ,
l = 8.2 cm
w = 3.8 cm
h = 7 cm
Hence, the volume is ,
V = l * w * h
V = 8.2 cm * 3.8 cm * 7 cm
V = 218.12 cm³
Density -
Density of a substance is given as the mass divided by the volume ,
Hence,
d = m / V
From the question,
m = 0.68 kg
since ,
1 kg = 1000g
m = 0.68 * 1000 = 680 g
V = 218.12 cm³ ( calculated above )
Hence , the value of density can be calculated by using the above equation,
d = m / V
d = 680 g / 218.12 cm³
d = 3.11 g /cm³
A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the volume of the solvent does not change when the aniline dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits. X10 How do I enter a number in scientific notation? molarity = x10 molarity=
Answer:
2.1 × 10⁻¹ M
2.0 × 10⁻¹ m
Explanation:
Molarity
The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:
3.9 g × (1 mol/93.13 g) = 0.042 mol
The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:
M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M
Molality
The moles of solute are 0.042 mol.
The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:
200 mL × 1.05 g/mL = 210 g = 0.210 kg
The molality of aniline is:
m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m
A chemist adds 380.0mL of a ×9.7510−4mM magnesium fluoride MgF2 solution to a reaction flask. Calculate the micromoles of magnesium fluoride the chemist has added to the flask. Round your answer to 3 significant digits.
Answer:
0.370 micromoles of magnesium fluoride the chemist has added to the flask.
Explanation:
[tex]Molarity=\frac{moles}{\text{Volume of solution(L)}}[/tex]
Moles of magnesium fluoride = n
Volume of the solution = 380.0 mL = 0.380 L (1 mL = 0.001 L)
Molarity of the solution = [tex]9.75\times 10^{-4} mM=9.75\times 10^{-7} M[/tex]
(1 mM = 0.001 M)
[tex]9.75\times 10^{-7} M=\frac{n}{0.380 L}[/tex]
[tex]n=3.705\times 10^{-7} mol[/tex]
1 mole = [tex]10^6[/tex] micro mole
[tex]n=3.705\times 10^{-7} \times 10^6 \mu mol=0.3705 \mu mol[/tex]
0.370 micromoles of magnesium fluoride the chemist has added to the flask.
g An example of an element is ____.
a) chicken noodle soup
b) powerade
c) air inside a balloon
d) lead pipe
e) baking soda (NaHCO)
Consider the reaction data. A ⟶ products A⟶products T ( K ) T (K) k ( s − 1 ) k (s−1) 225 225 0.391 0.391 525 525 0.700 0.700 What two points should be plotted to graphically determine the activation energy of this reaction? To avoid rounding errors, use at least three significant figures in all values.
Answer:
The activation energy for the reaction is, 1.90682 KJ/mol.
Explanation:
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
or,
[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_1[/tex] = rate constant at 225 K = [tex]0.391 s^{-1}[/tex]
[tex]K_2[/tex] = rate constant at 525 K = [tex]0.700 s^{-1}[/tex]
[tex]Ea[/tex] = activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
[tex]T_1=225 K, T_2=525 K[/tex]
Now put all the given values in this formula, we get
[tex]\log (\frac{0.700 s^{-1}}{0.391 s^{-1}})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{225 K}-\frac{1}{525 K}][/tex]
[tex]Ea=1,906.82 J/mole=1.90682 KJ/mol[/tex]
Therefore, the activation energy for the reaction is, 1.90682 KJ/mol.
Consider that you have a 100 mM stock solution and you need to prepare 10 mL of a 30 mM solution. How many milliliters of the stock solution do you need? mL How many milliliters of deionized water do you need? mL
Answer:
V₁ = 3.0 mL
V(H₂O) = 7 mL
Explanation:
Given data
Initial concentration (C₁): 100 mMInitial volume (V₁): ?Final concentration (C₂): 30 mMFinal volume (V₂): 10 mLIn order to determine the volume of the stock solution, we will use the dilution rule.
C₁ × V₁ = C₂ × V₂
100 mM × V₁ = 30 mM × 10 mL
V₁ = 3.0 mL
The volume of deionized water required is:
V(H₂O) = 10 mL - 3.0 mL = 7 mL
The volume of the stock solution required is 3 mL.
The stock solution is defined as the solution from which other solutions are prepared. We have the following information;
concentration of stock solution = 100 mMConcentration of diluted solution = 30 mM Volume of diluted solution = 10 mLUsing the formula;
M1V1 = M2V2
V1 = M2V2/M1
V1 = 30mM × 10 mL/100 mL
V1 = 3 mL
The volume of the stock solution required is 3 mL.
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How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00×10−15m (a typical nuclear distance)?
Answer: 7.68×10^14 J
Explanation:
The magnitude of the charge q1=q2 = 1.6×10^-19 C
Proportionality constant K = 9×10^9 Nm²/C²
Da= 2×10^-10m
Db= 3 ×10^-15m
W= k q1q2 ( 1/Db -1/Da)
W= 9×10^9 (1.69×10^-19)²(1/3.0×10^-15 - 1/2×10^-10)
W= 7.68 ×10^-14J
Therefore the amount of work requires is 7.68 ×10^-14J
Calculate the theoretical oxygen demand (mg/L). of a solution containing 450mg of glucose (C6H12O6). in 2 L of distilled water.
Answer:
ThOD =239.792 mg/L
Explanation:
Theorical Oxigen demand (ThOD):
is the theoretical amount of oxygen
required to oxidize the organic fraction of a
waste up to carbon dioxide and water.
⇒ C sln = 450 mg C6H12O6 / 2 L H2O = 225 mg/L sln
∴ mm C6H12O6 = 180.156 g/mol
balanced reaction:
C6H12O6 + 6O2 → 6CO2 + 6H2O∴ mol C6H12O6 = 1 mol
⇒ mass C6H12O6 = (180.156 g/mol)( 1 mol) = 180.156 g
∴ the value of ThOD is determined when 180.156 g C6H12O6 consume mass O2 = 6(32) = 192 g Oxygen; then in a solution of 225 mg/L, you have:
⇒ ThOD = (192/180.156)×225 mg/L
⇒ ThOD = 239.792 mg/L
The theoretical oxygen demand is 0.24 g/L or 240mg/L.
To calculate the theoretical oxygen demand of a solution containing 450mg of glucose (C6H12O6) in 2 L of distilled water, we need to start by determining the stoichiometry of the complete combustion of glucose, which can be represented by the balanced chemical equation:
C6H12O6 + 6O2 → 6CO2 + 6H2O
From the equation, we can see that 1 mole of glucose requires 6 moles of oxygen to completely react. Therefore, to calculate the oxygen demand, we first need to find the number of moles of glucose present in the solution:
The molar mass of glucose (C6H12O6) is approximately 180 g/mol. So, 450 mg of glucose is equal to 0.450 g or 0.450/180 = 0.0025 moles of glucose.
Using the stoichiometry from the equation, the moles of oxygen required is 0.0025 moles of glucose imes 6 moles of oxygen/mole of glucose = 0.015 moles of oxygen.
Now, the molar mass of oxygen (O2) is approximately 32 g/mol, hence 0.015 moles imes 32 g/mol = 0.48 grams of oxygen. Since the solution volume is 2 L, we need to find the amount per liter, which gives us 0.48 g / 2 L = 0.24 g/L or 240 mg/L of theoretical oxygen demand.