Be sure to answer all parts. Which of the following ions possess a dipole moment? (a) ClF2+ (b) ClF2− has a dipole moment has no dipole moment cannot be determined has a dipole moment has no dipole moment cannot be determined (c) IF4+ has a dipole moment has no dipole moment cannot be determined (d) IF4− has a dipole moment has no dipole moment cannot be determined
Answers
Answer:
(A). Has Dipole moment, (B). No dipole moment, (C). Has dipole moment, (D). It has no Dipole moment.
Explanation:
In order to determine if a specie has dipole moment or not there is the need for us to draw the Lewis structure, please check attached file for the Lewis structures of each species.
(A). ClF2+: it HAS dipole moment because of Asymmetry. Note that the Fluorine atoms are more Electronegative that the chlorine atom.
(B). ClF2− : Normally, it should have a dipole moment because Fluorine atoms are more Electronegative that the chlorine atom on each side BUT due to its geometry (according to VSEPR theory) which makes it to have a linear geometry, the dipole moment will cancel out ,hence, NO DIPOLE MOMENT.
(C). IF4+ : it HAS dipole moment. Although, the dipole moment of two out of the four Fluorine will cancel out the other two will not cancel out making it to have Dipole moment.
(D). IF4-: it has no dipole moment because the four bonds are in opposite directions
ClF2+ and IF4+ has dipole moment whereas ClF2− and IF4− has no dipole moment.
ClF2+ has dipole moment because of Asymmetry i.e. the Fluorine atoms has higher electronegativity value that the chlorine atom.
ClF2− has no dipole moment due to its geometry i.e. it has a linear geometry, the dipole moment will cancel out and no dipole moment occur.
IF4+ has dipole moment because the dipole moment of two out of the four Fluorine will cancel out but the other two will not cancel out making it to have Dipole moment.
IF4- has no dipole moment because the four bonds are in opposite directions.
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Related Questions
Consider an electrochemical cell based on the spontaneous reaction 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+. If the zinc ion concentration is kept constant at 1 M, and the chlorine ion concentration is decreased from 1 M to 0.001 M, the cell voltage should A) increase by 0.06 V. D) decrease by 0.18 V. B) increase by 0.18 V. E) increase by 0.35 V. C) decrease by 0.06 V.
Answers
Answer:
B) increase by 0.18 V
Explanation:
The given chemical spontaneous reaction is :
[tex]2 AgCl_{(s)} + Zn _{(s)} \to 2Ag (s) + 2Cl^- _{(aq)} + Zn ^{2+} _{(aq)}[/tex]
By applying Nernst Equation:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
here ;
n = number of electrons transferred in the reaction
n =2
[tex]E^0 = E^0_{cathode } - E^0_{anode}[/tex]
[tex]E^0 = E^0_{Ag^+/Ag } - E^0_{Zn^{2+}/Zn}[/tex]
[tex]E^0 =+0.80 \ V -(-0.76 \ V)[/tex]
[tex]E^0 =1.56 \ V[/tex]
When it happens to occur that the concentration of chlorine (aq) and Zn²⁺ (aq) is 1 M ;
[tex]E^0_{cell} = E^0[/tex] is as follows:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log [\frac{[Zn^{2+}]}{[Cl^-]^2}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log (1) \ \ \ \ \ \ \ ( where \ log (1) = 0)[/tex]
[tex]E_{cell} = 1.56 \ V[/tex]
Now; the [tex]E_{cell}[/tex] value in the decreased concentration of chlorine (aq) ion is calculated as:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log [\frac{[Zn^{2+}]}{[Cl^-]^2}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log (1*0.001^2)[/tex]
[tex]E_{cell} = +1.737 \ V[/tex]
Hence; the change in voltage = [tex]E_{cell} - E^0[/tex]
= 1.737 - 1.56
= 0.177 V
≅ 0.18 V
We therefore conclude that: since the [tex]E^0_{cell}[/tex] value after the decreased concentration of Chlorine is greater than the [tex]E^0[/tex] before the change; then there is increase in the value by 0.18 V
Which of the following is a component of cell theory that is supported by the diagram
Answers
Option 4. All living organisms are composed of cells. The component of cell theory that is supported by the diagram is option 4.
The cell theory, which is a fundamental concept in biology, states the following:
All living organisms are composed of one or more cells: This means that cells are the basic structural and functional units of all living things.
The cell is the basic unit of life: Cells are the smallest units that can carry out all the processes necessary for life, including metabolism and reproduction.
All cells arise from pre-existing cells: New cells are produced by the division of existing cells, and no new cells are spontaneously generated.
From the diagram, the first image is a simple unicellular organism and the others are multicellular organism tissues. These shows that all living things are made up of cell/cells depending on the complexity of the organism.
Complete question
EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pKa3=2.00 , p K a4 = 2.69 pKa4=2.69 , p K a5 = 6.13 pKa5=6.13 , and p K a6 = 10.37 pKa6=10.37 . The distribution of the various protonated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTA EDTA , it is convenient to calculate the fraction of EDTA EDTA that is in the completely unprotonated form, Y 4 − Y4− . This fraction is designated α Y 4 − αY4− . Calculate α Y 4 − αY4− at two pH values.
Answers
To calculate αY4-, we need to find the fraction of EDTA in its protonated forms at the given pH and divide the concentration of the completely unprotonated form by the total concentration of EDTA.
Explanation:To calculate the fraction of EDTA that is in the completely unprotonated form, Y4- (designated as αY4-), at a given pH value, we need to find the values of αY4- at two different pH values.
To calculate αY4-, we first need to find the fraction of EDTA in its protonated forms at the given pH.
For example, at pH = 2, we can calculate the fractional composition of each protonated form by dividing the concentration of the protonated form by the total concentration of all protonation states of EDTA. We can repeat this calculation for another pH value to find αY4- at that pH.
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The fraction of EDTA in the completely unprotonated form at pH3 is approximately [tex]\( 9.28 \times 10^{-4} \)[/tex], and at pH8 is approximately [tex]\( 8.83 \times 10^{-7} \)[/tex].
To calculate the fraction of EDTA that is in the completely unprotonated form, [tex]\( \alpha_{Y^{4-}} \)[/tex], at a given pH, we use the following equation:
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + \sum_{i=1}^{6} 10^{(pH - pKa_i)}} \][/tex]
Here, [tex]\( pKa_i \)[/tex] represents the [tex]\( i \)-th[/tex] acid dissociation constant for EDTA. Since EDTA is a hexaprotic acid, it has six [tex]\( pKa \)[/tex] values, which are given as:
[tex]\( pKa_1 = 0.00 \), \( pKa_2 = 1.50 \), \( pKa_3 = 2.00 \), \( pKa_4 = 2.69 \), \( pKa_5 = 6.13 \), and \( pKa_6 = 10.37 \).[/tex]
Calculate [tex]\( \alpha_{Y^{4-}} \)[/tex] for two pH values, say pH = 3 and pH = 8.
For pH = 3:
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{(3 - 0.00)} + 10^{(3 - 1.50)} + 10^{(3 - 2.00)} + 10^{(3 - 2.69)} + 10^{(3 - 6.13)} + 10^{(3 - 10.37)}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{3} + 10^{1.5} + 10^{1} + 10^{0.31} + 10^{-3.13} + 10^{-7.37}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 1000 + 31.62 + 10 + 2.08 + 0.00074 + 4.17 \times 10^{-8}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1 + 1000 + 31.62 + 10 + 2.08} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1033.62 + 31.62 + 10 + 2.08} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1077.32} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx 9.28 \times 10^{-4} \][/tex]
For pH = 8:
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{(8 - 0.00)} + 10^{(8 - 1.50)} + 10^{(8 - 2.00)} + 10^{(8 - 2.69)} + 10^{(8 - 6.13)} + 10^{(8 - 10.37)}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{8} + 10^{6.5} + 10^{6} + 10^{5.31} + 10^{1.87} + 10^{-2.37}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10000000 + 316227.76 + 1000000 + 20794.42 + 74.12 + 0.00417} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{10000000 + 316227.76 + 1000000 + 20794.42 + 74.12} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1131992.3} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx 8.83 \times 10^{-7} \][/tex]
Therefore, the fraction of EDTA in the completely unprotonated form at pH 3 is approximately [tex]\( 9.28 \times 10^{-4} \)[/tex], and at pH 8 is approximately [tex]\( 8.83 \times 10^{-7} \)[/tex].
The complete question is:
EDTA is a hexaprotic system with the PK, values:
PKa1 = 0.00, pKa2 = 1.50, pKa3 2.00, pKa4 = 2.69,
pKa5 6.13, and pKa6 10.37.
The distribution of the various protonated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTA, it is convenient to calculate the fraction of EDTA that is in the completely unprotonated form, Y4-. This fraction is designated αY4−. Calculate αY4− at two pH values.
pH = 3.10
pH = 10.55
In addition to displacing halide ions, the acetylide ion also adds to carbonyl groups. 2-Methyl-3-butyn-2-ol (MBI) is an acetylenic alcohol used in the manufacture of products for the agrochemical and specialty chemical industry. It can be synthesized by the addition of acetylene to acetone to form the alkoxide ion and, as a second step, protonation of the alkoxide ion to produce the alcohol. Complete the mechanism for 2-methyl-3-butyn-2-ol production by drawing in the products of each step and the missing curved arrows. Sodium amide deprotonates the terminal alkyne to form sodium ethynide. Draw all missing reactants and/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. Add charges where needed. Electron flow arrows should start on an atom or a bond and should end on an atom, bond, or location where a new bond should be create
Answers
Answer:
Explanation:
check below for explanation.
Question 7 When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of iron(III) chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for iron(III) chloride in . Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.
Answers
Answer:
See explaination
Explanation:
moles of benzamide = mass / Molar mass of it = 70.4g / ( 121.14g/mol) = 0.58 mol
Molality = moles of solute ( benzamide) / ( solvent mass in kg) = 0.58 mol / ( 0.85kg) = 0.6837
we have formula dT = i x Kf x m , where dT = change in freezing point = 2.7C , i = vantoff factor = 1 for non dissociable solutes , Kf = freezing oint constant of solvent , m = 0.6837
hence 2.7C = 1 x Kf x 0.6837m
Kf = 3.949 C/m
we use this Kf value for calculating i for NH4Cl , where moles of NH4Cl = ( 70.4g/53.491g/mol) =1.316 mol
molality = ( 1.316mol) / ( 0.85kg) = 1.5484 , dT = 9.9
hence 9.9 = i x 3.949C/m x 1.5484 m
i = 1.62
Using the appropriate Ksp values, find the concentration of K+ ions in the solution at equilibrium after 600 mL of 0.45 M aqueous Cu(NO3)2 solution has been mixed with 450 mL of 0.25 M aqueous KOH solution. (Enter in M.) (Ksp for Cu(OH)2 is 2.6x10-19).
Now find the concentration of OH? ions in this solution at equilibrium.
Answers
Answer:
[K⁺] = 0.107 M
[OH⁻] = 1.13 × 10⁻⁹ M
Explanation:
600 mL of 0.45 M Cu(NO3)2 gives equal mole of Cu²⁺ and (NO₃)²⁻
⇒ 0.45 × 600 × 10⁻³
= 0.27 moles of Cu²⁺ and (NO₃)²⁻
450 mL of 0.25 M KOH gives equal moles of K⁺ and OH⁻
⇒ 0.25 × 450 × 10⁻³
= 0.1125 moles of K⁺ and OH⁻
Now after mixing 0.1125 moles of OH⁻ precipitates 0.05625 moles of Cu²⁺ (because 1 Cu²⁺ needs 2 OH⁻)
Therefore , moles of remaining Cu²⁺ = 0.27 - 0.05625
=0.21375 moles which is equal to :
⇒ 0.21375/(( 600+450))× 10⁻³
= 0.21375/1050 × 10⁻³
= 0.20357 M
Given that :
(Ksp for Cu(OH)2 is 2.6 × 10⁻¹⁹)
We know that , Ksp = [Cu²⁺][OH⁻]²
2.6 × 10⁻¹⁹ = 0.20357 × [OH⁻]²
[OH⁻]² = 2.6 × 10⁻¹⁹/0.20357
[OH⁻] = 1.13 × 10⁻⁹ M
[K⁺] = moles of K⁺ /total volume
[K⁺] = 0.1125 / 1050 × 10⁻³
[K⁺] = 0.107 M
_ H₂ + O2
H20
balance equation
Answers
Answer:
2H2 + O2 = 2H2O
Explanation:
From the original equation, you first need to write each component separately.
Left side: H = 2 ; O = 2 (number based on the subscript)
Right side: H = 2 ; O = 1 (number based on the subscripts; no subscript means that the element is just 1)
Notice that the number of H is already balance but the number of O is not. In order to balance the O, you need to multiply the element by 2, but you CANNOT do this by simply changing the subscript.
Left side: H = 2 ; O = 2
Right side: H = 2 x 2 = 4 ; O = 1 x 2 = 2
Now, notice that the number of O is now balance (both are 2) but the number of H is not (since
You also multiply the left side H by 2. Hence,
Left side: H = 2 x 2 = 4 ; O = 2
Right side: H = 2 x 2 = 4 ; O = 1 x 2 = 2
The equation is now balanced.
What is the molarity of a hydrochloric acid solution, HCl (aq), if 30.00 mL of the solution is required to completely react with 25.00 mL of a 0.200 M magnesium hydroxide solution, Mg(OH)2 (aq)? Be sure to write out the balanced chemical reaction between these two compounds.
Answers
Answer:
0.333 M
Explanation:
Step 1: Write the balanced equation
2 HCl(aq) + Mg(OH)₂(aq) = MgCl₂(aq) + 2 H₂O(l)
Step 2: Calculate the reacting moles of Mg(OH)₂
25.00 mL of a 0.200 M magnesium hydroxide react. The reacting moles of Mg(OH)₂ are:
[tex]25.00 \times 10^{-3} L \times \frac{0.200mol}{L} = 5.00 \times 10^{-3} mol[/tex]
Step 3: Calculate the reacting moles of HCl
The molar ratio of HCl(aq) to Mg(OH)₂ is 2:1. The reacting moles of HCl are:
[tex]5.00 \times 10^{-3} molMg(OH)_2 \times \frac{2mol}{1molMg(OH)_2} =1.00 \times 10^{-2}molHCl[/tex]
Step 4: Calculate the molarity of HCl
[tex]\frac{1.00 \times 10^{-2}mol}{30.00 \times 10^{-3}L} = 0.333 M[/tex]
A galvanic cell is powered by the following redox reaction: br2 (l) 2NO(g) (l)(aq) (aq) (aq) Answer the following questions about this cell. If you need any electrochemical data, be sure you get it from the ALEKS Data tab. Write a balanced equation for the half-reaction that takes place at the cathode. Write a balanced equation for the half-reaction that takes place at the anode. Calculate the cell voltage under standard conditions. Round your answer to decimal places.
Answers
Final answer:
In the galvanic cell, the reduction of Br₂ to 2Br⁻ occurs at the cathode, while the oxidation of 2NO to N₂O₄ occurs at the anode. The cathode is the silver electrode and the anode is the copper electrode. The silver electrode is the positive electrode and the copper electrode is the negative electrode.
Explanation:
In the galvanic cell powered by the redox reaction between Br₂ and 2NO, the half-reaction that takes place at the cathode is the reduction of Br₂ to 2Br⁻. The half-reaction that takes place at the anode is the oxidation of 2NO to N₂O₄.
For the electrode assignment, the cathode is the electrode where reduction occurs, so it is the silver electrode. The anode is the electrode where oxidation occurs, so it is the copper electrode.
The positive electrode in this galvanic cell is the cathode, which is the silver electrode, and the negative electrode is the anode, which is the copper electrode.
Retigeranic Acid II: The Endgame. This question deals with transformations that were employed in the last few steps of a total synthesis of Retigeranic Acid. (a) Provide reagents that will accomplish the transformation of molecule 6 to molecule 7. (b) Draw the mechanism of the transformation of molecule 7 to molecule 8 when treated with sodium hydroxide in water. (It is ok to abbreviate parts of the molecule not involved in the reaction to make drawing a little easier).
Answers
Answer:
see explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
is 8 inches enough to satisfy a girl
Answers
Answer:
Yes
Explanation:
of course
explanation:
yes
Acid chlorides can be converted into primary amines with the loss of a single carbon atom via the Curtius rearrangement. The reaction involves treating an acid chloride with sodium azide to form an acyl azide. The acyl azide then loses N2 to give an isocyanate, which is hydrolyzed to release CO2 and the primary amine. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answers
Answer:
Find the given attachment.
Footballs used by the NFL are inflated to 12.5 psi. 11 of 12 footballs measured 10.5 psi after the playoff game when the temperature was 5°C. At what temperature were the balls inflated to explain this pressure drop?
Answers
Final answer:
After calculating T1, we will have our answer in Kelvin, which can be converted back to degrees Celsius by subtracting 273.15.
Explanation:
To determine at what temperature the NFL footballs were inflated before the pressure dropped to 10.5 psi from 12.5 psi, we can use the ideal gas law and assume that the volume of the footballs remains constant (since they're made of materials that don't expand or contract much with temperature). The ideal gas law implies a direct relationship between pressure and temperature, meaning that if the temperature drops, the pressure will also drop, provided the number of moles of gas and the volume remain constant.
Starting with the initial condition, where P1 = 12.5 psi and the final condition where P2 = 10.5 psi at T2 = 5°C, we can find the initial temperature (T1) using the following relation from Charles's Law:
P1/T1 = P2/T2
However, to use this formula, we need to convert the temperatures into Kelvin:
T2 (in Kelvin) = 5°C + 273.15 = 278.15 K
T1 = (P1 * T2) / P2
T1 = (12.5 psi × 278.15 K) / 10.5 psi
After calculating T1, we will have our answer in Kelvin, which can be converted back to degrees Celsius by subtracting 273.15.
How many molecules are there in 0.250 moles of sodium hydroxide
Answers
Answer: The mass of sodium hydroxide is 10 grams.
Explanation: the atomic weight: 39.997 g/mol
A sample of hexane (C6H14) has a mass of 0.580 g. The sample is burned in a bomb calorimeter that has a mass of 1.900 kg and a specific heat of 3.21 J/giK. What amount of heat is produced during the combustion of hexane if the temperature of the calorimeter increases by 4.542 K?
A. 8.46 kJ
B. 16.1 kJ
C. 27.7 kJ
D. 47.8 kJ
Answers
Answer:
27.7 KJ
Explanation:
Q= mC dT
m= 1900 g+0.580 g= 1900.58 g
Q= (1900.58 g * 3.21 KJ/ gK* 4.542 K)
Q=27710 J= 27.7 KJ
Answer:
C) 27.7 kJ
Explanation:
20.0 g of Nitrogen is produced when oxygen gas reacts with NO gas. 29.8 L of oxygen is required at STP to produce this 20.0 g of NO. A.yes B.false
Answers
Answer:
There will be produced 30.64 grams of nitrogen dioxide
The statement is false
Explanation:
Step 1: Data given
MAss of nitrogen produced = 20.0 grams
Molar mass of N2 = 28.0 g/mol
Mass of NO = 20.0 grams
Molar mass of NO = 30.01 g/mol
Volume of O2 at STP = 29.8 L
Step 2: the balanced equation
2NO + O2 → 2N02
Step 3: Calculate moles NO
Moles NO = mass NO / molar mas NO
Moles NO = 20.0 grams / 30.01 g/mol
Moles NO = 0.666 moles
Step 4: Calculate moles O2
1 mol O2 at STP = 22.4 L
29.8 L = 29.8/22.4 = 1.33 moles
Step 5: Calculate the limiting reactant
For 2 moles NO we need 1 mol O2 to produce 2 moles NO2
NO is the limiting reactant. IT will completely be consumed (0.666 moles).
O2 is in excess. There will react 0.666/2 = 0.333 moles O2
There will remain 1.333 - 0.333 = 0.997 moles O2
Step 6: Calculate moles NO2
For 2 moles NO we need 1 mol O2 to produce 2 moles NO2
For 0.666 moles NO we'll have 0.666 moles NO2
Step 7: Calculate mass NO2
Mass NO2 = moles NO2 * molar mass NO2
Mass NO2 = 0.666 moles * 46.0 g/mol
Mass NO2 = 30.64 grams
There will be produced 30.64 grams of nitrogen dioxide
The statement is false
Electrophilic addition to an alkene proceeds via Markovnikov regiochemistry due to the formation of the more stable carbocation intermediate. In the case of conjugated dienes, that is dienes that are separated by one sigma bond, the carbocation that is formed is stabilized additionally by resonance. Addition of the nucleophile to the carbocation intermediate can therefore give two types of products: direct addition to the double bond, also called 1,2-addition, and conjugate addition to the resonance stabilized carbocation, also called 1,4-addition. Allylic carbocation stability is affected by both the nature of the carbocation (primary allylic, secondary allylic, or tertiary allylic) and by the degree of substitution of the double bond. The latter is typically the dominant effect and so a primary allylic carbocation with a trisubstituted double bond is more stable than a tertiary allylic carbocation with a monosubstituted double bond. Electrophilic addition to a conjugated diene is temperature dependent where reaction at or below room temperature typically leads to a mixture of products in which the 1,2 adduct (or direct addition product) predominates, this is termed kinetic control. At elevated temperatures the reactions have time to come to equilibrium and typically the 1,4 adduct (or conjugate addition product) will predominate, this is termed thermodynamic control. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below for the step by step explanation to the question above.
To how much water should 50mL of 12M hydrochloric acid be added to produce a 4M solution?
Answers
Answer:
50 mL of 12M HCl should be added into 100 mL of water to produce a 4M solution.
Explanation:
According to law of dilution-
[tex]C_{1}V_{1}=C_{2}V_{2}[/tex]
where [tex]C_{1},C_{2},V_{1}[/tex] and [tex]V_{2}[/tex] are initial concentration, final concentration, initial volume and final volume respectively.
Here, [tex]C_{1}=12M[/tex] , [tex]C_{2}=4M[/tex] , [tex]V_{1}=50mL[/tex]
So, [tex]V_{2}=\frac{C_{1}V_{1}}{C_{2}}[/tex] = [tex]\frac{(12M)\times (50mL)}{(4M)}[/tex] = 150 mL
Hence, final volume of HCl solution should be 150 mL.
So, volume of water needed = (150-50) mL = 100 mL
Therefore 50 mL of 12M HCl should be added into 100 mL of water to produce a 4M solution.
You need to add [tex]\( {150} \)[/tex] mL of water to 50 mL of 12 M hydrochloric acid to produce a 4 M solution.
To determine how much water should be added to 50 mL of 12 M hydrochloric acid (HCl) to produce a 4 M solution, we can use the dilution formula which states:
[tex]\[ C_1V_1 = C_2V_2 \][/tex]
where:
- [tex]\( C_1 \)[/tex] and [tex]\( V_1 \)[/tex] are the concentration and volume of the initial solution (12 M and 50 mL, respectively),
- [tex]\( C_2 \)[/tex] and [tex]\( V_2 \)[/tex] are the concentration and volume of the final solution (4 M and the unknown volume of water, respectively).
Let's solve for [tex]\( V_2 \)[/tex]:
[tex]\[ C_1V_1 = C_2V_2 \][/tex]
Substitute the given values:
[tex]\[ 12 \times 50 = 4 \times V_2 \][/tex]
Solve for [tex]\( V_2 \)[/tex]:
[tex]\[ 600 = 4 \times V_2 \][/tex]
[tex]\[ V_2 = \frac{600}{4} \][/tex]
[tex]\[ V_2 = 150 \text{ mL} \][/tex]
Two radioactive nuclides X and Y both decay to stable products. The half-life of X is about a day, while that of Y is about a week. Suppose a radioactive sample consists of a mixture of these two nuclides. If the mixture is such that the activities arising from X and Y are initially equal, then a few days later the activity of the sample will be due ______________.
Answers
Answer:Then a few days later the activity of the sample will be due to have more of Nuclides Y than X
Explanation:
This is because half life of nuclide X is about a day which is less than Y having half life of about a week, After a few days, we would observe that X would have disintegrated more while Y will still be predominant since it disintegrate slower than X. The time it takes for X to disintegrate will always be faster than Y.
How is energy or particles released from a nucleus?
Answers
Answer: The decay energy is the energy released by a radioactive decay. Radioactive decay is the process in which an unstable atomic nucleus loses energy by emitting ionizing particles and radiation.
Be sure to answer all parts. Calculate ΔH o for the following reaction in two ways, using the data given below. H2(g) + I2(g) → 2HI(g) (a) Using the equation ΔH o rxn = ∑(BE(reactants)) − ∑(BE(products)) kJ/mol (b) Using the equation ΔH o rxn = ∑(nΔH o f (products)) − ∑(mΔH o f (reactants)) kJ/mol Bond BE kJ/mol Substance ΔH o f kJ/mol H―H 436.4 H2(g) 0 I―I 151 I2(g) 61.0 H―I 298.3 HI(g) 25.9
Answers
Answer:
a) ΔH°rxn = -9.2kJ/mol
b) ΔH°rxn = -9.2kJ/mol
Explanation:
Using Hess's law, you can find ΔH of a reaction from ΔH of formation of the substances involved in the reaction, thus:
ΔH°rxn = ∑(BE(reactants)) − ∑(BE(products))
Or:
ΔH°rxn = ∑(nΔH°f (products)) − ∑(mΔH°f (reactants))
For the reaction:
H₂(g) + I₂(g) → 2HI(g)
a) Using the first equation:
ΔH°rxn = ΔH (H-H) + ΔH (I-I) - 2ΔHBE (H-I)
ΔH°rxn = 436.4kJ + 151kJ - 2×298.3kJ
ΔH°rxn = -9.2kJ/mol
b) Using the second equation:
ΔH°rxn = 2Δ°f (HI) − ΔH°f (H₂) - ΔH°f (I₂)
ΔH°rxn = 2×25.9kJ - 0kJ - 61.0kJ
ΔH°rxn = -9.2kJ/mol
The reaction H₂(g) + I₂(g) → 2HI(g) has a ΔHo value of -9.2 kJ/mol, which is calculated using both bond dissociation energies and enthalpies of formation. This indicates the reaction is exothermic.
Explanation:To calculate ΔHo for the reaction H₂(g) + I₂(g) → 2HI(g),first we will use the bond dissociation energies (BE), and second the enthalpies of formation (ΔHof).
(a) Using the equation ΔHorxn = ∑BE(reactants) − ∑BE(products), we get ΔHorxn = [1(436.4 kJ/mol) + 1(151 kJ/mol)] - 2(298.3 kJ/mol) = -9.2 kJ/mol. This asserts that the reaction is exothermic and energy is released in the process. The sum of bond energies of reactants is smaller than that of the products.
(b) Using the equation ΔHorxn = ∑nΔHof(products) − ∑mΔHof(reactants), we get ΔHorxn = 2(25.9 kJ/mol) - [1(0 kJ/mol) + 1(61.0 kJ/mol)] = -9.2 kJ/mol. This value is same as obtained by the first method and thus validates it.
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4) What is the difference between tetrahedral bent and trigonal planar bent?
Answers
Answer:
A tetrahedral bent molecular geometry involves four electron pairs including two lone pairs and two bonding groups while a trigonal planar bent molecular geometry involves three electron pairs two bonding groups and one lone pair.
Explanation:
According to valence shell electron pair repulsion theory, the number of electron pairs on the central atom of a molecule influences its shape. It follows that, the shape of a molecule is a consequence of arrangement of valence shell electron pairs. Electron pairs must be positioned as far apart in space as possible.
Electron pairs may be bonding pairs or lone pairs. The repulsion between two lone pairs is greater than the repulsion between a lone pair and a bond pair which in turn is greater than the repulsion between two bond pairs. Hence the presence of lone pairs causes much distortion of the expected molecular geometry of the molecule as predicted by VSEPR theory. This distortion usually leads to the assumption of a bent geometry, depending on the expected geometry of the molecule based on VSEPR theory.
A tetrahedral bent molecular geometry involves four electron pairs including two lone pairs and two bonding groups while a trigonal planar bent molecular geometry involves three electron pairs; two bonding groups and one lone pair. For a bent tetrahedral geometry, the bond angle is much less than 109° while for a bent trigonal planar geometry, the bond angle is much less than 120°.
The difference between tetrahedral bent and trigonal planar bent molecular geometries lies in the number of electron pairs around the central atom and the resulting shape.
In a tetrahedral bent geometry, the central atom is surrounded by four electron pairs, which are arranged around the atom in a tetrahedral shape. This arrangement leads to bond angles of approximately 109.5 degrees.
An example of a molecule with a tetrahedral bent shape is water (H2O), where the central oxygen atom has two bonded pairs and two lone pairs of electrons. The lone pairs exert greater repulsion than the bonded pairs, causing the bond angles to be slightly less than the ideal tetrahedral angle, resulting in a bent shape.
On the other hand, in a trigonal planar bent geometry, the central atom is surrounded by three electron pairs, which are arranged in a plane around the atom in a trigonal planar shape.
This arrangement leads to bond angles of approximately 120 degrees. However, when one or more of these electron pairs are lone pairs, the geometry can become bent due to the lone pair-bond pair repulsion being greater than bond pair-bond pair repulsion.
An example of a molecule with a trigonal planar bent shape is sulfur dioxide (SO2), where the central sulfur atom has two bonded pairs and one lone pair of electrons. The presence of the lone pair causes the molecule to have a bent shape with bond angles less than 120 degrees.
In summary, the key differences are:
- Tetrahedral bent involves four electron pairs with a tetrahedral arrangement, while trigonal planar bent involves three electron pairs with a trigonal planar arrangement.
- Tetrahedral bent typically results from two bonded pairs and two lone pairs, leading to a bond angle of slightly less than 109.5 degrees.
- Trigonal planar bent typically results from two bonded pairs and one lone pair, leading to a bond angle of less than 120 degrees.
Therefore, the main difference is the number of electron pairs and the resulting molecular shape and bond angles.
In an isothermal gas chromatography experiment using an ECD detector, 1.69 nmols of nchlorohexane, C6H13Cl, was added as an internal standard to an unknown amount of nchlorodecane, C10H21Cl. The area of the first peak to elute was 32434 units and the area of the second peak to elute was 2022 units. Calculate the amount of chlorodecane in the unknown.
Answers
Answer:
The amount of Chlorodecane in the unknown is 0.105nmols
Explanation:
a) Since the GC is in an isothermal state, Chlorohexane C6H13Cl (1.69 nmols) because of its lower boiling point will elute first and Chlorodecane C12H21Cl will elute second.
The area of the first peak corresponding to Chlorohexane is 32434 units.
The area of the second peak corresponding to chlorodecane is 2022 units.
Since the response factor of the compound is not given in question and considering the response factor is same for both the compounds, the answer will be as follow:
1.69 nmols of Chlorohexane gives 32434 units
How much of chlorodecane gives 2022 units
By cross multiplication;
Moles of Chlorodecane = 2022*1.69/32434
=0.105nmols
Write the chemical reaction for nitrous acid in water, whose equilibrium constant is K a . Include the physical states for each species. K a reaction: Write the chemical reaction for the nitrite ion in water, whose equilibrium constant is K b . Include the physical states for each species.
Answers
Answer:
HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺
NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
Explanation:
According to Brönsted-Lowry acid-base theory, nitrous acid is an acid because it transfers an H⁺ to another compound. The corresponding reaction is:
HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺
According to Brönsted-Lowry acid-base theory, nitrite ion is a base because it accepts an H⁺ from another compound. The corresponding reaction is:
NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
The chemical reaction for nitrous acid in water is HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺
And, the chemical reaction for the nitrite ion in water, is
NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
Brönsted-Lowry acid-base theory:
Since nitrous acid should be an acid due to this it transferred an H⁺ to another compound so here the reaction should be HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺
Here nitrite ion should have a base due to this, it accepts an H⁺ from another compound. So here the reaction is NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
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A container of gas is heated from 250 K to 303 K. What is the new pressure if the initial pressure is 880 kPa? (Assume constant volume).
Answers
Answer:
The new pressure at constant volume is 1066.56 kPa
Explanation:
Assuming constant volume, the pressure is diectly proportional to the temperature of a gas.
Mathematically, P1/T1 = P2 /T2
P1 = 880 kPA= 880 *10^3 Pa
T1 = 250 K
T2 = 303 K
P2 =?
Substituting for P2
P2 = P1 T2/ T1
P2 = 880 kPa * 303 / 250
P2 = 266,640 kPa/ 250
P2 = 1066.56 kPa.
The new pressure of the gas is 1066.56 kPa
At constant volume, if the temperature of the container is heated to the given value, the pressure of the gas increases to 1066.5kPa.
What is Gay-Lussac's law?Gay-Lussac's law states that the pressure exerted by a given quantity of gas varies directly with the absolute temperature of the gas.
It is expressed as;
P₁/T₁ = P₂/T₂
Given the data in the question;
Initial pressure P₁ = 880kPa = ( 880 / 101.325)atm = 8.68492atmInitial temperature T₁ = 250KInitial temperature T₂ = 303KFinal pressure P₂ = ?P₁/T₁ = P₂/T₂
P₁T₂ = P₂T₁
P₂ = P₁T₂ / T₁
P₂ = (8.68492atm × 303K) / 250K
P₂ = (2631.53atmK) / 250K
P₂ = 10.526atm
P₂ = (10.526 × 101.325)kPa
P₂ = 1066.5kPa
Therefore, at constant volume, if the temperature of the container is heated to the given value, the pressure of the gas increases to 1066.5kPa.
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Consider this chemical reaction, where moving from left to right represents moving forward in time. A five panel comic strip. In the first panel, there are ten large red spheres. In the second panel, there are 8 large red spheres and two small blue spheres. In the third panel, there are six large red spheres and four small blue spheres. In the fourth panel, there are four large red spheres and six small blue spheres. In the fifth panel, there are four large red spheres and six small blue spheres. At what point does the reaction first reach equilibrium
Answers
Answer:
The reaction reaches equilibrium at the fourth panel.
Explanation:
Chemical Equilibrium is achieved when the overall properties the system seem to be constant, that is, stop changing.
Although, for chemical equilibrium, the right term for this equilibrium is dynamic equilibrium; the rate of forward reaction balances the rate of backward reaction, but concentrations can keep changing.
The point where equilibrium is achieved is when exactly when we reach the panel where the spheres that make up this panel is the same as the next panel and the next, that is, the specific colour and number of spheres start to become unchanged.
And from the description given in the question,
- In the first panel, there are ten large red spheres.
- In the second panel, there are 8 large red spheres and two small blue spheres.
- In the third panel, there are six large red spheres and four small blue spheres.
- In the fourth panel, there are four large red spheres and six small blue spheres.
- In the fifth panel, there are four large red spheres and six small blue spheres.
It is evident that the make-up of the spheres have become the same as at the fourth and fifth panel. This means that the first point where this final configuration of spheres first appeared is the fourth panel.
Hence, equilibrium is first reached at the fourth panel.
Hope this Helps!!!
The reaction reaches equilibrium at the fourth panel. A reaction is at equilibrium when the amounts of reactants or products no longer change. Chemical equilibrium is a dynamic process
What is Chemical Equilibrium?It is the rate of formation of products by the forward reaction is equal to the rate at which the products re-form reactants by the reverse reaction.
The point where equilibrium is achieved is when exactly when we reach the panel where the spheres that make up this panel is the same as the next panel and the next, that is, the specific color and number of spheres start to become unchanged.
As per the descriptions of panel given in question:
It is evident that the make-up of the spheres have become the same as at the fourth and fifth panel. This means that the first point where this final configuration of spheres first appeared is the fourth panel.
Hence, equilibrium is first reached at the fourth panel.
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Make the appropriate conclusion. Choose the correct answer below. A. Reject Upper H 0. There is insufficient evidence to indicate blood lactate level is linearly related to perceived recovery at alphaequals0.10. B. Reject Upper H 0. There is sufficient evidence to indicate blood lactate level is linearly related to perceived recovery at alphaequals0.10. C. Do not reject Upper H 0. There is sufficient evidence to indicate blood lactate level is linearly related to perceived recovery at alphaequals0.10. D. Do not reject Upper H 0. There is insufficient evidence to indicate blood lactate level is linearly related to perceived recovery at alphaequals0.10.
Answers
Answer:
A. Reject Upper H 0. There is sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the alphaequals0.10 level of significance.
Explanation:
H0: p1 = p2
Ha: p1 > p2
pcap = (101 + 72)/(489 + 625) = 0.1553
SE = sqrt(0.1553*(1-0.1553)*(1/489 + 1/625)) = 0.0219
Test statistic,
z = (101/489 - 72/625)/0.0219 = 4.1710
p-value = 0.0000
Reject H0
A. Reject Upper H 0. There is sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the alphaequals0.10 level of significance.
Consider the following reaction.
CO2(g) + H2(9)
-
CO(g) + H2O(1)
What is being oxidized?
O carbon
carbon dioxide
oxygen
o hydrogen
Answers
Answer:
Hydrogen
Explanation:
It gains oxygen from carbon dioxide to form water
Oxidation is the addition of oxygen to an element in a chemical reaction
Answer:
Its D
Explanation:
Trust me just get ronas and cheat
a student prepares a dilute solution of sodium hydroxidem, NaOH (aq), starting with 6 M sodium hydroxide. She then titrates a 1.372 g samle of KHP with the dilute sodium hydroxide solution, NaOH (aq), calculate the molar concentration of the sodium hydroxide solution, NAoH (aq)
Answers
Answer:
M = 0.3077 M
Explanation:
As I said in the comments, you are missing the required volume of the base to react with the KHP. I found this on another site, and the volume it used was 21.84 mL.
Now, KHP is a compound often used to standarize NaOH or KOH solutions. This is because it contains a mole ratio of 1:1 with the base, so it's pretty easy to use and standarize any base.
Now, as we are using an acid base titration, the general expression to use when a acid base titration reach the equivalence point would be:
n₁ = n₂ (1)
This, of course, if the mole ratio is 1:1. In the case of KHP and NaOH it is.
Now, we also know that moles can be expressed like this:
n = M * V (2)
And according to this, we are given the volume of base and the required mass of KHP. So, if we want to know the concentration of the base, we need to get the moles of the KHP, because in the equivalence point, these moles are the same moles of base.
The reported molar mass of KHP is 204.22 g/mol, so the moles are:
n = 1.372 / 204.22 = 6.72x10⁻³ moles
Now, we will use expression (2) to get the concentration of the diluted base:
n = M * V
M = n / V
M = 6.72x10⁻³ / 0.02184
M NaOH = 0.3077 M
This is the concentration of the dilute solution of NaOH
The molar concentration of sodium hydroxide ( NaOH ) is ; 0.3077 M
Given data :
Volume of base = 21.84 mL = 0.02184 L ( missing data )
Molar mass of KHP = 204.22 g/mol
mass of KHP = 1.372 g
At equivalence point
n1 = n2 ( ratio of KHP to NaOH )
note : n = M * V ---- ( 1 )
First step : calculate the number of moles of KHP
n = mass / molar mass
n = 1.372 / 204.22 = 6.72 * 10⁻³ moles
Determine the molar concentration of NaOH
From equation ( 2 )
M ( molar concentration ) = n / V
= 6.72 * 10⁻³ moles / 0.02184 L
= 0.3077 M
Hence we can conclude that The molar concentration of sodium hydroxide ( NaOH ) is 0.3077 M
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HURRY PLS HELP PLEASE: A 40.0-L sample of fluorine is heated from 363 Kelvin to 459 K. What volume will the sample occupy at the higher temperature?
19.3 L
31.6 L
50.5 L
82.6 L
Answers
Final answer:
The volume of a 40.0-L sample of fluorine gas heated from 363 K to 459 K can be found using Charles's Law. After setting up the equation from Charles's Law and solving for the new volume (V2), we find that the volume at the higher temperature is 50.5 L.
Explanation:
To determine the new volume of fluorine gas when heated from 363 K to 459 K, we can use Charles's Law, which states that for a given mass of an ideal gas at constant pressure, the volume is directly proportional to its absolute temperature. Specifically, the formula is V1/T1 = V2/T2, where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature.
In this case, the initial volume (V1) is 40.0 L and the initial temperature (T1) is 363 K. The final temperature (T2) is 459 K. Plugging these values into the formula, we have:
40.0 L / 363 K = V2 / 459 K
Multiplying both sides by 459 K to solve for V2 gives us:
V2 = (40.0 L x 459 K) / 363 K
Upon calculation, V2 = 50.5 L. Therefore, the new volume occupied by the sample at 459 K is 50.5 liters.