Answer:
C. two atoms of oxygen.
Explanation:
Step 1: Data given
Silicon has 14 electrons
Silicon is part of Group IV, all the elements there have 4 valence electrons.
It can form a compound when 4 valence electrons bind with the 4 valence elctrons of silicon
A. four atoms of calcium.
Calcium has 2 valence elctrons. 4 atoms of calcium cannot bind on 1 atom of silicon since there are only 4 valence electrons.
B. one atom of chlorine.
1 atom of chlorine has 7 valence electrons. Chlorine can bind with an atom with 1 valence electron. Since silicon has 4 valence electrons, they will not bind.
Silicon can bind with 4 atoms of chlorine to form SiCl4
C. two atoms of oxygen.
Oxygen has 6 valence electrons, this means oxygen can bind with an element with 2 valence electrons.
Since silicon has 4 valence electrons, it can bind with 2 atoms of oxygen to form SiO2 (silicon dioxide).
D. three atoms of hydrogen.
Hydrogen has 1 valence electron. 1 hydrogen atom can bind with an element that has 7 valence electrons.
Three atoms of hydrogen can bind with an element that has 5 valence electrons.
Silicon will not bind with 3 atoms of hydrogen ( but can bind with 4 atoms of hydrogen)
Use the following information to determine the empirical formula if a compound is found to have 18.7% Li, 16.3% C, and 65.5% O Answer questions 4-6 with this data
Answer:
The empirical formula is Li2CO3
Explanation:
Step 1: Data given
Suppose the mass of the compound = 100 grams
The compound contains :
18.7% Li = 18.7 grams of Li
16.3 % C = 16.3 grams of C
65.0% O = 65.0 grams of O
Total = 100%
Molar mass of Li = 6.94 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16 g/mol
Step 2: Calculate moles
Moles = Mass / molar mass
Moles Li = 18.7 grams / 6.94 g/mol = 2.65 moles
Moles C = 16.3 grams / 12.01 = 1.36 moles
Moles O = 65.0 grams / 16.0 g/mol = 4.06 moles
Step 3: Calculate the mol ratio
We divide by the smallest number of moles
Li: 2.65/1.36 ≈ 2
C: 1.36/1.36 = 1
O: 4.06/1.36 ≈ 3
The empirical formula is Li2CO3
A myelin sheath is a______.
A. large band of neural fibers connecting the two hemispheres of the brain.
B. fatty tissue layer encasing the axon of a nerve cell.
C. bushy extension of a neuron that conducts impulses toward the cell body.
D. nerve network within the brainstem that controls physical arousal.
Answer:B. Fatty tissue layer encasing the axon of a nerve cell
Explanation: Myelin is an insulating layer that forms around a nerves including those of brain and spinal cord,its is made up if protein and fatty substances.The myelin sheath aids in the quick and efficient transfer of electrical impulses along a nerve cells. Damage to it can slow down the impulse,for example multiple sclerosis diseases.
Suppose you could watch radioactive atoms decay. It would probably get quite boring as time went by. Why?
a. The initial decay rate is very fast, but the decay rate decreases over time.
b. Due to randomness, the last couple of radioactive atoms may take a long time before they become nonradioactive.
c. The pattern becomes very predictable.
d. Only a few radioactive nuclei are left to decay, so fewer and fewer atoms decay.
e. The rate of decay decreases with each half-life cycle.
f. The probability that an atom will decay is reduced with each half-life cycle.
Answer: the answer should and most definitely be D.
Explanation: I mean think about it after a while only a few radioactive nuclei are left which means it will dye down after a while which also makes it very boring hope this helps :)
Answer:
a,b,c,d,e
Explanation:
just did it on ed 2020
Compared to the energy and charge of the electrons in the first shell of a Be atom, the electrons in the second shell of this atom have
(1) less energy and the same charge(2) less energy and a different charge(3) more energy and the same charge(4) more energy and a different charge
Answer:
4. More energy and a different charge.
Electrons in the second shell of a beryllium atom possess more energy due to being further from the nucleus but have the same negative charge as those in the first shell.
When comparing the energy and charge of electrons in different shells of a beryllium (Be) atom, we find that electrons in the first shell (1s) are closer to the nucleus and thus have lower energy than electrons in the second shell (2s and 2p). The second shell is at a higher energy level due to being further away from the nucleus and its electrons experiencing less attractive force because of the shielding effect of the inner-shell electrons. Electron-electron repulsion also impacts the energies of electrons in the atom. Despite the differences in energy levels, the charge on electrons in any shell of an atom remains the same, negative in nature.
Therefore, compared to the electrons in the first shell of a Be atom, the electrons in the second shell have more energy but the same charge. This leads us to the correct answer to the student's question: (3) more energy and the same charge.
In acetyl CoA formation, the carbon-containing compound from glycolysis is oxidized to produce acetyl CoA. From the following compounds involved in cellular respiration, choose those that are the net inputs and net outputs of acetyl CoA formation.
Answer:
- Net Input: NAD⁺, coenzyme A, pyruvate
- Net Output: NADH, acetyl CoA, CO₂
- Not input or output: O₂, ADP, glucose and ATP
Explanation:
Hello,
In this case, it is important to recall that acetyl-CoA is produced either by oxidative decarboxylation of pyruvate derived from glycolysis, which is carried out into the mitochondrial matrix, by cause of the oxidation of high-order fatty acids, or by oxidative degradation of very specific amino acids. Acetyl-CoA then enters in the citric acid cycle where it is oxidized in the light of energy production.
In this manner, during such processes, there are some net inputs and outputs, therefore, they are sorted as show below, considering there some of them not classified neither as input nor output:
- Net Input: NAD⁺, coenzyme A, pyruvate
- Net Output: NADH, acetyl CoA, CO₂
- Not input or output: O₂, ADP, glucose and ATP
Best regards.
What is a common property of metals? a) insulates from electricity b) conducts electricity c) has very little reactivity' d) is liquid at room temperature
Answer:
B) conducts electricity
Explanation:
This answer can be answered by the process of elimination. Metals are excellent conductors of electricity, which means A is out, and metals are also known to have medium reactivity and have high melting points, meaning C and D are also out.
According to the concept of conduction, the common property of metals is that they conduct electricity .
What is conduction?Conduction is defined as a process as a means of which heat is transferred from the hotter end of the body to it's cooler end.Heat flows spontaneously from a body which is hot to a body which is cold.
In the process of conduction,heat flow is within the body and through itself.In solids the conduction of heat is due to the vibrations and collisions of molecules while in liquids and gases it is due to the random motion of the molecules .
When conduction takes place, heat is usually transferred from one molecule to another as they are in direct contact with each other.There are 2 types of conduction:1) steady state conduction 2) transient conduction.According to the type of energy conduction is of three types:
1) heat conduction
2) electrical conduction
3)sound conduction
Learn more about conduction,here:
https://brainly.com/question/12136944
#SPJ5
Joseph Proust showed that when elements combine to form new substances, they do so in specific mass ratios.a) trueb) false
Answer: option A) True
Explanation:
Joseph Proust is the founder of the Law of Definite proportion. This law states that two elements will always combine together to form a chemical compound and this will be in the same proportion by mass.
For Example: 2 moles of Hydrogen (H2) ALWAYS combine with 1 atom of Oxygen (O) to yield water (H2O)
So, the answer is True
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Answer:
[tex]Ag_2SO_4[/tex]
Explanation:
Formula for the calculation of no. of Mol is as follows:
[tex]mol=\frac{mass\ (g)}{molecular\ mass}[/tex]
Molecular mass of Ag = 107.87 g/mol
Amount of Ag = 5.723 g
[tex]mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol[/tex]
Molecular mass of S = 32 g/mol
Amount of S = 0.852 g
[tex]mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol[/tex]
Molecular mass of O = 16 g/mol
Amount of O = 1.695 g
[tex]mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol[/tex]
In order to get integer value, divide mol by smallest no.
Therefore, divide by 0.02657
[tex]Ag, \frac{0.05305}{0.02657} \approx 2[/tex]
[tex]S, \frac{0.02657}{0.02657} \approx 1[/tex]
[tex]O, \frac{0.10594}{0.02657} \approx 4[/tex]
Therefore, empirical formula of the compound = [tex]Ag_2SO_4[/tex]
Answer:
The empirical formula is Ag2SO4
Explanation:
Step 1: Data given
A compound contains Ag, S and O
Mass of Ag = 5.723 grams
Mass of S = 0.852 grams
Mass of O = 1.695 grams
Molar mass of Ag = 107.87 g/mol
Molar mass of S = 32.065 g/mol
Molar mass of O = 16 g/mol
Total mass = Mass of Ag + mass of S + mass of O = 5.723g + 0.852g + 1.695g = 8.27 grams
Step 2: Calculate moles
Moles = mass / molar mass
Moles of Ag = 5.723 grams / 107.87 g/mol = 0.05305 moles
Moles S = 0.852 grams / 32.065 g/mol = 0.0266 moles
Moles O = 1.695 grams / 16.00 g/mol = 0.1059 moles
Step 3: Calculate mole ratio
We divide by the smallest amount of moles
Ag: 0.05305/0.0266 = 2
S: 0.0266/0.0266 = 1
O: 0.1059 / 0.0266 = 4
The empirical formula is Ag2SO4
In a combustion reaction, 46.0 of ethanol reacts with 96.0 of oxygen to produce water and carbon dioxide. If 54.0g of water is produced, how much carbon dioxide is produced?
approximately 132.03 grams of carbon dioxide is produced.
To solve this problem, we first need to write the balanced chemical equation for the combustion of ethanol [tex](\(C_2H_5OH\))[/tex]:
[tex]\[ C_2H_5OH + O_2 \rightarrow H_2O + CO_2 \][/tex]
Now, we can balance the equation:
[tex]\[ C_2H_5OH + 3O_2 \rightarrow 2H_2O + 2CO_2 \][/tex]
From the balanced equation, we can see that 1 mole of ethanol [tex](C_2H_5OH\))[/tex] reacts with 3 moles of oxygen [tex](\(O_2\))[/tex] to produce 2 moles of water [tex](\(H_2O\))[/tex] and 2 moles of carbon dioxide [tex](\(CO_2\))[/tex].
Given:
- Mass of ethanol [tex](\(C_2H_5OH\)):[/tex] 46.0 g
- Mass of water [tex](\(H_2O\))[/tex] produced: 54.0 g
First, we need to calculate the moles of water produced:
[tex]\[ \text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of water}} \][/tex]
[tex]\[ \text{Moles of water} = \frac{54.0 \, \text{g}}{18.015 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of water} \approx 3.0 \, \text{mol} \][/tex]
According to the balanced equation, the ratio of moles of water to moles of carbon dioxide is 2:2. Therefore, the moles of carbon dioxide produced is also approximately 3.0 mol.
Now, to find the mass of carbon dioxide produced:
[tex]\[ \text{Mass of } CO_2 = \text{Moles of } CO_2 \times \text{Molar mass of } CO_2 \][/tex]
[tex]\[ \text{Mass of } CO_2 = 3.0 \, \text{mol} \times 44.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of } CO_2 = 132.03 \, \text{g} \][/tex]
Therefore, approximately 132.03 grams of carbon dioxide is produced.
A food product contains 15 g of carbohydrates, 5 g of protein and 4 g of fat. How many calories are in one serving of this product?
Answer: 116 calories of energy
Explanation:
A calorie is a non-standard unit of energy.
On combustion,
1 gram Carbohydrates = 4 calories,
1 gram protein = 4 calories,
1gram fat = 9 calories.
Therefore,
15 grams of carbohydrates = (15*4) = 60
5 grams of protein = (5*4) = 20
4 grams of fats = (4*9) = 36
Then add up: 60 + 20 + 36 = 116 calories of energy
A 300-g aluminum cup contains 700 g of water in thermal equilibrium with the cup at 60°C. The combination of cup and water is cooled uniformly so that the temperature decreases by 2.2°C per minute. At what rate is energy being removed? Express your answer in watts.
Answer:
117.3 W is being removed.
Explanation:
The heat removed can be calculated as:
Q = m*c*ΔT
Where m is the mass, c is the specific heat and ΔT is the temperature variation. Because there're two components:
Q = mwater*cwater*ΔT + maluminum*caluminum*ΔT
Q = (mwater*cwater + maluminum*caluminum)*ΔT
Searching in a thermodynamic table:
cwater = 4.184 J/g°C
caluminium = 0.9 J/g°C
In 1 minute, the temperature decreases 2.2°C, so ΔT = -2.2°C
Q = (700*4.184 + 300*0.9) * (-2.2)
Q = -7037.36 J
The rate of energy is the potency (P), which is the heat divided by the time. So, for 1 minute (60 s):
P = -7037.36/60
P = -117.3 J/s
P = -117.3 W
The minus signal indicates that the energy is being removed.
Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine, CO(g) Cl2(g) <-------> COCl2(g) The value of Kc for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which PCO?
The question is incomplete, here is the complete question:
Phosgene, [tex]COCl_2[/tex], gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:
[tex]CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)[/tex]
The value of [tex]K_c[/tex] for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which [tex]p_{CO}=p_{Cl_2}=0.265atm[/tex] and [tex]p_{COCl_2}=0.000atm[/tex] ?
Answer: The equilibrium partial pressure of CO, [tex]Cl_2\text{ and }COCl_2[/tex] is 0.257 atm, 0.257 atm and 0.008 atm respectively.
Explanation:
The relation of [tex]K_c\text{ and }K_p[/tex] is given by:
[tex]K_p=K_c(RT)^{\Delta n_g}[/tex]
[tex]K_p[/tex] = Equilibrium constant in terms of partial pressure
[tex]K_c[/tex] = Equilibrium constant in terms of concentration = 5.79
[tex]\Delta n_g[/tex] = Difference between gaseous moles on product side and reactant side = [tex]n_{g,p}-n_{g,r}=1-2=-1[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
T = Temperature = 570 K
Putting values in above equation, we get:
[tex]K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124[/tex]
We are given:
Initial partial pressure of CO = 0.265 atm
Initial partial pressure of chlorine gas = 0.265 atm
Initial partial pressure of phosgene = 0.00 atm
The given chemical equation follows:
[tex]CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)[/tex]
Initial: 0.265 0.265
At eqllm: 0.265-x 0.265-x x
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}[/tex]
Putting values in above equation, we get:
[tex]0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59[/tex]
Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure
So, the equilibrium partial pressure of CO = [tex](0.265-x)=(0.265-0.008)=0.257atm[/tex]
The equilibrium partial pressure of [tex]Cl_2=(0.265-x)=(0.265-0.008)=0.257atm[/tex]
The equilibrium partial pressure of [tex]COCl_2=x=0.008atm[/tex]
Hence, the equilibrium partial pressure of CO, [tex]Cl_2\text{ and }COCl_2[/tex] is 0.257 atm, 0.257 atm and 0.008 atm respectively.
The equilibrium partial pressures of CO, Cl2, and COCl2 are 5.79 atm, 5.79 atm, and 0.10 atm, respectively.
Explanation:The equilibrium partial pressures of the three gases can be determined using the equilibrium expression and the given information. The equilibrium expression for the phosgene-forming reaction is:
PCOCl2 = (PCO * PCl2) / PCOCl2,eq
Given that the equilibrium constant (Kc) is 5.79, the equilibrium partial pressures can be calculated as follows:
PCO = (PCOCl2,eq / PCl2) * Kc
PCl2 = (PCOCl2,eq / PCO) * Kc
Substituting the given information:
PCO = (0.10 atm / 0.10 atm) * 5.79 = 5.79 atm
PCl2 = (0.10 atm / 0.10 atm) * 5.79 = 5.79 atm
PCOCl2,eq = 0.10 atm
The equilibrium partial pressures of CO, Cl2, and COCl2 are 5.79 atm, 5.79 atm, and 0.10 atm, respectively.
Bicarbonate concentrate mixers may have a which are replaced on a routine basic.
A) True
B) False
Answer: True the bicarbonate mixture can help save time and few routine.
Explanation:
For the purpose of making dialysate for hemodialysis patient therapies a bicarbonate mixing and delivering systems designed to prepare a liquid sodium bicarbonate formulation comes in handy.
Certain systems like the SDS unit also allow for the transfer and distribution of acid concentrate solutions. We also provide stand-alone acid concentrate delivery systems using a variety of holding tanks and delivery methods.
A challenge for hemodialysis providers is to properly provide bicarbonate solution in a cost effective manner. Preparation and disinfection can be time-consuming and labor intensive.
Bicarbonate however can corrode certain metals and painted surfaces leaving your preparation area encrusted and grimy.
Furthermore, if not mixed properly, bicarbonate can negatively affect the dialysate solution.
The answer to the above is true the bicarbonate mixture can help save time and few routine.
How many liters of a 1.5 M solution can you make if you have .50 mol of KCl?
Answer:
The answer to your question is 0.33 liters
Explanation:
Data
Volume = ?
Molarity = 1.5 M
number of moles = 0.5
Formula
[tex]Molarity = \frac{number of moles }{volume}[/tex]
Solve for V
[tex]Volume = \frac{number of moles}{molarity}[/tex]
Substitution
[tex]Volume = \frac{0.5}{1.5}[/tex]
Simplification and result
Volume = 0.33 l
Which statement regarding the gold foil experiment is FALSE? 1. The α particles were repelled by electrons. 2. It suggested that atoms are mostly empty space. 3. It suggested the nuclear model of the atom. 4. Most of the α particles passed through the foil undeflected. 5. It was performed by Rutherford and his research group early in the 20th century.
Answer:
1. The α particles were repelled by electrons.
Explanation:
The gold foil experiment was performed by Rutherford and his research group in 1911 (at the beginning of the 20th century). In this experiment, α particles were bombed to gold foils, and films were placed surround it to collect the particles.
It was observed that most of the particles passed through of the foil undeflected, and for that, Rutherford stated that the atom was a "huge empty". Some particles were deflected, because they're attracted to the electrons at the electrosphere, and a small number of particles were complete deflected to the origin because they chocked with the small positive nuclei.
Thus, the experiment suggested the nuclear model of the atom, called the planetary model, that was improved after by Bohr and other scientists in the quantum model.
The false statement about the gold foil experiment is that α particles were repelled by electrons. In reality, their deflection was due to the encounter with a small, dense, positively charged nucleus, leading to the nuclear model of the atom. Therefore, the correct option is 1.
The α particles were repelled by electrons.
It suggested that atoms are mostly empty space.
It suggested the nuclear model of the atom.
Most of the α particles passed through the foil undeflected.
It was performed by Rutherford and his research group early in the 20th century.
The false statement here is statement 1: The α particles were repelled by electrons. This is inaccurate because α particles are positively charged and were repelled when they encountered the dense, positively charged nucleus, not electrons. The gold foil experiment led to the discovery of the atomic nucleus and overturned the plum pudding model of the atom, which assumed a uniform distribution of mass and charge. The experiment showed that atoms are mostly empty space, with most α particles passing through the foil undeflected, a small number deflected at large angles, and a very few reversing direction, indicating the presence of a small, dense, positively charged nucleus.
A student assisting with the experiment would observe all of the following about the electron transport chain EXCEPT:A. Electrons are passed from carriers with lower reduction potential to those with higher reduction potential.B. The first electron carrier is also a proton pump.C. All electron carriers are mobile and hydrophobic.D. The electron carriers can transport a maximum of 2 electrons.
Answer:
C. All electron carriers are mobile and hydrophobic
Explanation:
Hello,
In this case, it is widely known that the electron carriers move inside the inner mitochondrial membrane and consequently move electrons from one to another. In such a way, they are mobile, therefore they are largely hydrophobic as long as they are inside the membrane.
For instance, the cytochrome c is a water-soluble protein in a large range, therefore, the answer is: C. All electron carriers are mobile and hydrophobic.
Best regards.
A sled slides along a horizontal surface for which the coefficient of kinetic friction is 0.25. Its velocity at point a is 8.0m/s and at point b is 5.0m/s.
The question is incomplete, here is a complete question.
A sled slides along a horizontal surface for which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.0m/s and at point B is 5.0m/s.
Use the impulse-momentum theorem to find how long the sled takes to travel from A to B.
Answer : The sled takes time to travel from A to B is, 1.2 seconds.
Explanation :
Impulse-momentum theorem:
Impulse = Change in momentum
[tex]\Delta P=F\times \Delta t\\\\m\nu_f-m\nu_i=F\times \Delta t\\\\m(\nu_f-\nu_i)=\mu \times m\times g\times \Delta t\\\\(\nu_f-\nu_i)=\mu \times g\times \Delta t[/tex]
where,
m = mass
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
[tex]\mu[/tex] = coefficient of kinetic friction = 0.25
[tex]\nu_f[/tex] = final velocity = 8.0 m/s
[tex]\nu_i[/tex] = initial velocity = 5.0 m/s
t = time
Now put all the given values in the above formula, we get:
[tex](\nu_f-\nu_i)=\mu \times g\times \Delta t[/tex]
[tex](8.0-5.0)=0.25\times 9.8\times \Delta t[/tex]
[tex]\Delta t=1.2s[/tex]
Therefore, the sled takes time to travel from A to B is, 1.2 seconds.
The sled is decelerating due to kinetic friction, and we can calculate this deceleration using the equations of motion and the given coefficient of kinetic friction. The normal force on the sled can also be found by dividing the frictional force by the coefficient of kinetic friction.
Explanation:This problem pertains to the concept of kinetic friction and motion in Physics. We know that the sled's velocity decreases from 8.0m/s to 5.0m/s over some distance, which implies that it's decelerating due to kinetic friction. We can make use of the equation of motion v^2 = u^2 - 2*a*d where v is final velocity, u is initial velocity, a is deceleration (friction) and d is distance.
First, let's find the deceleration caused by the friction force using this equation. As we know that the frictional force can be expressed as F_kinetic = μk*m*g, where m is the mass of the sled (not given) and g is the acceleration due to gravity (9.8 m/s^2). From here, we can also find the normal force on the sled by knowing that the normal force is equal to kinetic friction divided by the coefficient of kinetic friction.
Learn more about Kinetic Friction here:
https://brainly.com/question/36370309
#SPJ3
The following list contains some common radicals. Using the charges on these ions and the idea of valence, predict the formulas for the following:_________.
1. nitrate NO3-1 phosphate PO4-3
2. sulfate SO4-2 acetate C2H3O2-1
3. ammonium NH4+ chromate CrO4-2
4. carbonate CO3-2 dichromate Cr2O7-2
5. permanganate MnO4-1 sulfide S-2
6. sodium carbonate
7. potassium permanganate
1. nitrate [tex]NO_{3} ^{-1}[/tex] : The valency of nitrogen is 5 and oxygen is 2. Therefore, the formula for nitrate is [tex]NO_{3} ^{-1}[/tex]
2. sulfate [tex]SO_{4}^{ -2}[/tex] : The valency of sulfur is 6 and oxygen is 2. Therefore, the formula for sulfate is [tex]SO_{4}^{ -2}[/tex].
3. ammonium [tex]NH_{4} ^+[/tex] : The valency of nitrogen is 5 and hydrogen is 1. Therefore, the formula for ammonium is [tex]NH_{4}^ +.[/tex]
4. carbonate [tex]CO_{3}^{ -2}[/tex] : The valency of carbon is 4 and oxygen is 2. Therefore, the formula for carbonate is [tex]CO_{3}^{ -2}[/tex].
5. permanganate [tex]MnO_{4}^{ -1}[/tex] : The valency of manganese is 7 and oxygen is 2. Therefore, the formula for permanganate is [tex]MnO_{4}^{ -1}.[/tex]
6. sodium carbonate : The formula for sodium carbonate is [tex]Na_{2} CO_{3}[/tex]. Sodium has a valency of 1 and carbonate has a valency of 2.
7. potassium permanganate : The formula for potassium permanganate is [tex]KMnO_{4}[/tex]. Potassium has a valency of 1, manganese has a valency of 7, and oxygen has a valency of 2.
1. For nitrate [tex](NO_{3} ^{-1})[/tex], the charge on the nitrate ion is -1. To determine the formula, we need to balance the charges. Since the nitrate ion has a charge of -1, it will combine with a cation that has a charge of +1. The cation that can balance the charge is sodium [tex](Na^{+1})[/tex]. Therefore, the formula for nitrate is [tex]NaNO_{3}[/tex].
2. For phosphate [tex](PO_{4}^{ -3})[/tex], the charge on the phosphate ion is -3. Again, we need to balance the charges. The cation that can balance the charge is calcium [tex](Ca^{+2})[/tex]. Therefore, the formula for phosphate is [tex]Ca_{3} (PO_{4} )_{2}[/tex].
3. For ammonium [tex]NH_{4} ^+.[/tex], the charge on the ammonium ion is +1. This is a polyatomic ion, meaning it is a group of atoms with an overall charge. The formula for ammonium is [tex]NH_{4} ^+.[/tex]
4. For carbonate [tex](CO_{3} ^{-2})[/tex], the charge on the carbonate ion is -2. The cation that can balance the charge is calcium (Ca+2). Therefore, the formula for carbonate is [tex]CaCO_{3}[/tex].
5. For sulfide [tex](S^{-2})[/tex], the charge on the sulfide ion is -2. The cation that can balance the charge is magnesium [tex](Mg^{+2})[/tex]. Therefore, the formula for sulfide is MgS.
6. Sodium carbonate is a compound formed by the combination of sodium [tex](Na^+)[/tex] and carbonate [tex]((CO_{3}) ^{-2})[/tex] ions. The formula for sodium carbonate is [tex]Na_{2} CO_{3}[/tex].
7. Potassium permanganate is a compound formed by the combination of potassium [tex](K^+)[/tex] and permanganate [tex]MnO_{4} ^{-1}[/tex] ions. The formula for potassium permanganate is [tex]KMnO_{4}[/tex].
An odorless, colorless, poisonous gas is produced by the incomplete burning of fossil fuels. This gas also interferes with the blood's ability to carry oxygen. What is the name of this air pollutant?
A. asbestos
B. carbon monoxide
C. radon
D. sulfur dioxide
Answer:
B
Explanation:
All the properties listed above is for carbon ii oxide.
It is odorless, colorless and poisonous
It is formed from the incomplete combustion of fossil fuels
And it can actually interfere with the blood’s ability to carry oxygen
Yes bottom line it is Carbon Monoxide!!! mine was D though (different verion)
for any substance solid liquid or gas mass increases as volume increases because density represnets mass per volume
Answer:
A) False
B) True
Explanation:
The question is incomplete. The complete question is as follows.
There are two statements. Identify which statement is true or false.
A) For any substance, solid, liquid or gas, mass increases as volume increases.
B) Density represents mass per volume.
Statement A
Mass is defined as the amount of matter in an object. Its fixed and can't be changed. Volume on the other hand is the amount of space an object takes up.Since the amount of matter is fixed, an increase in volume doesn't cause an increase in mass. For example, a test tube, with a volume of 10ml, contains liquid fragrance. When heated, the fragrance vaporizes and fill the entire room, until you can smell it in every corner. Note that previously the fragrance of same mass was occupying a 10ml space in the test-tube. After heating, the same mass of fragrance occupied the entire room (volume increased but mass remained constant).
It happened because as the fragrance is heated, the space between molecules increases causing them to occupy more space BUT in the whole scenarios, the number of molecules stayed unaffected. Hence statement A is false.
Statement B
Density measures the compactness of the substance. It's defined as the amount of mass per unit volume. It increases as the atoms/molecules in a substance comes closer. Density of solids are greatest as the atoms/molecules are tightly packed together. It is least in gases as the molecules are far apart from each other. Hence the statement is true.
Be sure to answer all parts. this is a two-part question. first, draw the minor alkene product that should be formed in the reaction. second, draw a stepwise mechanism that shows the formation of the major product: part 1: 2xsafari + view structure major product minor product part 2: view structure h5mech30504 view structure + br+ br− ch3obr+ ch3oh2+ part 3 out of 3 edit structure ... arr edit structure ... + br+ ch3oh2+ ch3obr+ br−
Answer:
See explanation below
Explanation:
The question is incomplete, however, I found a question very similar to this, and I'm assuming this is the question you are asking to answer. If it's not, please tell me which one it is. Here's a tip for you to get an idea of how to solve.
Picture 1, would be the original question. Picture 2 is the answer of it.
Now, This is a E1 reaction where this type of reactions are taking place in two steps. The first step is the formation of the carbon cation, this step is always slow. The secon step is the addition of a nucleophyle, or, in this case, formation of a pi bond, and we get a alkene.
Hope this can help you
The minor alkene product in the reaction of 2-bromobutane with methanol in the presence of HBr is trans-2-butene. The major product is cis-2-butene. The stepwise mechanism for the formation of the major product is shown above.
The reaction of 2-bromobutane with methanol in the presence of HBr is a Markovnikov addition reaction. This means that the bromide ion will add to the carbon of the 2-bromobutane molecule that is more substituted with alkyl groups. In this case, the carbon that is more substituted with alkyl groups is the secondary carbon.
The carbocation intermediate that is formed in step 1 of the mechanism is a secondary carbocation. Secondary carbocations are more stable than primary carbocations, so they are more likely to form.
In step 2 of the mechanism, the methanol molecule can attack the carbocation intermediate from either side. However, the methanol molecule is more likely to attack the carbocation intermediate from the side that is less hindered by the other alkyl groups. In this case, the side of the carbocation intermediate that is less hindered by the other alkyl groups is the side that is facing the hydrogen atom of the methanol molecule.
In step 3 of the mechanism, the proton from the HBr molecule is transferred to the oxygen atom of the methyl ether intermediate. This is a simple acid-base reaction.
The overall mechanism of the reaction is shown below:
2-bromobutane + methanol + HBr → cis-2-butene + methyl bromide + water
For more such information on: alkene
https://brainly.com/question/29120960
#SPJ6
Which of the following are molecular solids and which are covalent solids?
1. Se8
2. Hbr
3. Si
4. Co2
5. C
6. P4o6
7. Sih4
Answer:
Covalent solids: Se8, Si, C
Molecular solids: HBr, CO2, P4O6, SiH4
Explanation:
A covalent solid consists only of one type of atom infinitely linked by covalent bonds to form a three dimensional rigid solid while a molecular solid consists of discreet molecules held together by intermolecular forces.
Molecular Solid are Se8 and So.
Covalent Solid are Hbr ,Sih4, Co2, P4O6 and C.
What is Molecular Solid?molecular solid refer to solid that consist of discrete molecules. The corces that bind the molecules together are van der Waals forces, dipole-dipole interactions, quadrupole and so on.
What is Covalent Solid?Covalent Solid refer to solid that the atoms are bonded together by Covalent bonds.
Therefore,
Molecular Solid are Se8 and So.
Covalent Solid are Hbr ,Sih4, Co2, P4O6 and C.
Learn more about solids from the link below.
https://brainly.com/question/13878434
what is the total number of grams of NaOH needed to react completely with 196 grams of H2 SO4? regents
Answer:
160 g
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
For [tex]H_2SO_4[/tex]:-
Mass of [tex]H_2SO_4[/tex] = 196 g
Molar mass of [tex]H_2SO_4[/tex] = 98 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{196\ g}{98\ g/mol}[/tex]
[tex]Moles\ of\ Sulfuric\ acid= 2\ mol[/tex]
According to the given reaction:
[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O[/tex]
1 mole of sulfuric acid reacts with 2 moles of NaOH
So,
2 moles of sulfuric acid reacts with 2*2 moles of NaOH
Moles of NaOH must react = 4 moles
Molar mass of NaOH = 40 g/mol
Mass = Moles*molar mass = [tex]4\times 40\ g[/tex] = 160 g
Final answer:
To react completely with 196 grams of H₂SO₄, you would need 80 grams of NaOH. This calculation involves balancing the chemical equation and using stoichiometry.
Explanation:
Step 1: Write and balance the chemical equation:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Step 2: Calculate the molar mass of NaOH (40 g/mol) and H₂SO₄ (98 g/mol).
Step 3: Use stoichiometry to find the moles of NaOH needed to react completely with 196 grams of H₂SO₄. From the balanced equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH.
Step 4: Calculate the total number of grams of NaOH needed:
(196 g H₂SO₄) x (1 mol NaOH / 98 g H₂SO₄) x (40 g NaOH / 1 mol NaOH) = 80 grams of NaOH
Identify the type of reaction and then write the formulas for the word equations and then balance each equation.
________ 6. aluminum + chlorine -> aluminum chloride
________ 7. sodium hydroxide + sulfuric acid -> sodium sulfate + water
________ 8. nickel II chlorate -> nickel II chloride + oxygen
________ 9. butane (C4H10) + oxygen -> carbon dioxide + water
________ 10. zinc + hydrochloric acid -> zinc chloride + hydrogen
Answer:
The answer to your question is below
Explanation:
There are 4 types of chemical reactions:
- Synthesis is when two elements or compounds form only one compound.
- Decomposition is when 1 compound is broken into 2 or more products.
- Single replacement is when one element is replaced by another element.
- Double replacement is when the cations of two compounds are interchanged.
6.- Synthesis Al + 3Cl ⇒ AlCl₃
7.- Double replacement 2NaOH + H₂SO₄ ⇒ Na₂SO₄ + 2H₂O
8.- Decomposition Ni(ClO₃)₂ ⇒ NiCl₂ + 3O₂
9.- Combustion 2C₄H₁₀ + 13O₂ ⇒ 8CO₂ + 10H₂O
10.- Single replacement Zn + 2HCl ⇒ ZnCl₂ + H₂
When calcium metal is placed in water, hydrogen gas bubbles out, leaving a highly alkaline (basic) solution. Write the balanced chemical equation for this reaction. Phases are optional. Do not write an ionic equation (i.e., the answer should not show any charges).
Answer:
Ca (s) + 2H₂O (l) → Ca(OH)₂ (aq) + H₂ (g)
Explanation:
When solid calcium reacts with water, it produces the correspondent hydroxid and hydrogen gas.
The hydroxid which is produced, is the calcium hydroxid which is a strong base, that's why you talk about a highly alkaline solution.
Ca (s) + 2H₂O (l) → Ca(OH)₂ (aq) + H₂ (g)
Calcium hydroxide is a strong base, that dissociates in water, as this:
Ca(OH)₂ → Ca²⁺ (aq) + 2OH⁻ (aq)
It's a basic solution, is providing hydroxyl ions to the medium
The balanced chemical equation for the reaction between calcium and water is: Ca (s) + 2H2O (l) --> Ca(OH)2 (s) + H2 (g). Calcium metal reacts with water to form calcium hydroxide and hydrogen gas.
Explanation:The reaction between calcium and water can be represented as Ca (s) + 2H2O (l) --> Ca(OH)2 (s) + H2 (g). This balanced equation represents the reaction of calcium metal with water to form calcium hydroxide and hydrogen gas. It is a single replacement reaction where calcium displaces the hydrogen in water to form the compound calcium hydroxide and release hydrogen gas. It results in an alkaline solution because calcium hydroxide is a strong base.
Learn more about Balanced Chemical Reaction here:https://brainly.com/question/32834309
#SPJ3
What molecule is the ultimate source of hydrogen ions that are secreted into the gastric juice?
Answer: carbonic acid
Explanation:
A student preparing for the experiments inadvertently adds an additional 400 mL of the same acid solution to the dissolution vessel. What will be the new pOH of this solution?
Answer:
POH= 13
Explanation:
A student preparing for the experiments inadvertently adds an additional 400 mL of the same acid solution to the dissolution vessel. What will be the new pOH of this solution?
PH is the measure of the degree of acidity of a solution.
POH is the measure of the degree of alkalinity of a solution
Note that pH + pOH = 14
if concentration remains the same, then volume changes will not affect pH.
The pH of the solution is given as
PH= -log[H+].
For this experiment, the dissolution vessel contains 0.1 M HCl, no matter the initial volume of the acid solution
For the molar concentration of the cation, we can propose that a strong acid will dissociate completely,
[H+] = 0.1 = 1 x 10-1 M
. substituting the concentration of the cation
PH=-log[1 x 10-1] = 1.
Note that pH + pOH = 14 for any aqueous solution.
we say that
the pOH = 14 - pH.
pOH = 14-1 = 13.
Two isotopes of carbon, carbon-12 and carbon-13, have masses of 1.993 10-26 kg and 2.159 10-26 kg, respectively. These two isotopes are singly ionized (+e) and each is given a speed of 7.50 105 m/s. The ions then enter the bending region of a mass spectrometer where the magnetic field is 0.6500 T. Determine the spatial separation between the two isotopes after they have traveled through a half-circle.
Answer:
The spatial separation between the two isotopes after they have traveled through a half cycle is [tex]1.65\times 10^{2}m[/tex]
Explanation:
The moving of the isotopes must excert centripetal force which is equals to the magnetic force on the ions due to the magnetic.
The centripetal force of the ions can be calculated by the following formula.
[tex]F_{c}=\frac{mv^{2}}{r}.............(1)[/tex]
Magnetic force on this ions calculated by the following formula.
[tex]F_{m}=qvB.............(2)[/tex]
Equate the equations (1) and (2)
[tex]F_{c}=F_{m}[/tex]
[tex]\frac{mv^{2}}{r}=qvB[/tex]
[tex]v=\frac{mv}{qB}[/tex]
Substitute the values of the both isotopes.
[tex]r_{12}=\frac{1.993\times 10^{-26}\times 6.13105}{1.6\times 10^{-19}\times 0.7700}=9.9041\times 10^{-2}m[/tex]
[tex]r_{13}=\frac{2.159\times 10^{-26}\times 6.13105}{1.6\times 10^{-19}\times 0.7700}=1.0729\times 10^{-1}m[/tex]
Now the distance traveled by both isotopes by half circle.
Therefore, the distance between the two isotopes is the diameter of the circle which is equal to the twice the radius.
[tex]Seperation=2\times(r_{13}-r_{12})[/tex]
[tex]Seperation=2\times(1.0729\times 10^{-1}-9.9401\times 10^{-2})=1.650\times 10^{-2}m[/tex]
Therefore, The spatial separation between the two isotopes after they have traveled through a half cycle is [tex]1.65\times 10^{2}m[/tex]
True or false: (a) For molecules with similar molecular weights, the dispersion forces become stronger as the molecules become more polarizable. (b) For the noble gases the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table. (c) In terms of the total attractive forces for a given substance, dipole–dipole interactions, when present, are always greater than dispersion forces. (d) All other factors being the same, dispersion forces between linear molecules are greater than those between molecules whose shapes are nearly spherical.
(a) True - Dispersion forces strengthen with increased molecular polarizability.
(b) False - Dispersion forces in noble gases increase down the periodic table.
(c) False - Dispersion forces can be significant compared to dipole–dipole interactions.
(d) False - Dispersion forces depend on molecular size and electron cloud, not just molecular shape.
**(a) True:**
For molecules with similar molecular weights, dispersion forces (Van der Waals forces) become stronger as the molecules become more polarizable. Polarizability is a measure of how easily the electron cloud in a molecule can be distorted, and larger, more complex molecules exhibit greater polarizability, leading to stronger dispersion forces.
**(b) False:**
For noble gases, dispersion forces increase as you go down the column in the periodic table, not decrease. This is because larger atoms have more electrons and a greater electron cloud, resulting in stronger dispersion forces. As a consequence, boiling points generally increase down the noble gas column.
**(c) False:**
In terms of total attractive forces, dipole–dipole interactions are not always greater than dispersion forces. Dispersion forces can be significant, especially in large, nonpolar molecules where temporary dipoles can induce other molecules to develop temporary dipoles.
**(d) False:**
Dispersion forces between linear molecules are not always greater than those between nearly spherical molecules. The shape of the molecule influences the surface area of contact, affecting dispersion forces. Spherical molecules can have significant dispersion forces, especially if they are large and have a high number of electrons.
The same amount of energy is required to remove each electron sequentially from an atom.
A. TrueB. False
Ionization energy, also called ionization potential, in chemistry, the amount of energy required to remove an electron from an isolated atom or molecule.
The statement is false because the energy required to remove each subsequent electron from an atom increases due to the increasing electrostatic attraction between the positively charged nucleus and the remaining electrons, making successive ionization energies higher.
The statement that "the same amount of energy is required to remove each electron sequentially from an atom" is false. The process of removing electrons from an atom is called ionization, and the energy required to do so is termed ionization energy. Initially, removing an electron from a neutral atom requires a certain amount of energy.
However, as electrons are removed, the atom becomes more positively charged, increasing the electrostatic attraction between the remaining electrons and the nucleus. This makes each successive electron harder to remove, and hence, requires more energy. Such an increase in ionization energy is evident when comparing the first, second, and subsequent ionization energies of an atom.
Newly formed cations (positively charged ions) hold onto their remaining electrons more tightly than the neutral atom did, which clearly illustrates why successive ionization energies increase. This is further emphasized when attempting to remove core electrons after the valence electrons have been removed, which requires significantly more energy due to the stronger electrostatic attraction to the nucleus.