Answer:
.0027 M
Explanation:
We must calculate the threshold concentration of PO3−4 using Ksp and the given concentration of Ca2+:
Ca3(PO4)2(s)⇌3Ca2+(aq)+2PO3−4(aq)
Ksp=8.6×10−19=[Ca2+]3[PO3−4]2=(4.9×10−5M)3[PO3−4]2
[PO3−4]=0.0027 M
How many mL of a 0.130 M aqueous solution of chromium(II) nitrate, Cr(NO3)2, must be taken to obtain 5.08 grams of the salt
Answer:
222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.
Explanation:
Being:
Cr: 52 g/moleN: 14 g/moleO: 16 g/molethe molar mass of chromium (II) nitrate, Cr(NO₃)₂ is:
Cr(NO₃)₂ = 52 g/mole + 2* (14 g/mole + 3* 16 g/mole)= 176 g/mole
So: if 176 grams are present in 1 mole of the compound, 5.08 grams in how many moles of the compound will be present?
[tex]amount of moles=\frac{5.08 grams* 1 mole}{176 grams}[/tex]
amount of moles=0.0289 moles
Molarity (M) is the number of moles of solute that are dissolved in a given volume. It is then calculated by dividing the moles of the solute by the volume of the solution:
[tex]molarity (M)=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in [tex]\frac{moles}{liter}[/tex]
So in this case:
molarity= 0.130 Mnumber of moles of solute= 0.0289 molesvolume= ?Replacing:
[tex]0.130 M=0.130 \frac{moles}{liter} =\frac{0.0289 moles}{volume}[/tex]
Solving:
[tex]volume=\frac{0.0289 moles}{0.130 \frac{moles}{liter} }[/tex]
volume=0.2223 liters
Being 1 L= 1,000 mL:
volume=0.222 liters= 222.3 mL
222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.
which of the following compounds are polar: CH2Cl2, HBr?
Answer : HBr polar
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Consider the following reaction at 298K.
I2 (s) + Pb (s) = 2 I- (aq) + Pb2+ (aq)
Which of the following statements are correct?
Choose all that apply.
ΔGo > 0
The reaction is product-favored.
K < 1
Eocell > 0
n = 2 mol electrons
B-
Answer:
Eªcell > 0; n = 2
Explanation:
The reaction:
I2 (s) + Pb (s) → 2 I- (aq) + Pb2+ (aq)
Is product favored.
A reaction that is product favored has ΔG < 0 (Spontaneous)
K > 1 (Because concentration of products is >>>> concentration reactants).
Eªcell > 0 Because reaction is spontaneous.
And n = 2 electrons because Pb(s) is oxidizing to Pb2+ and I₂ is reducing to I⁻ (2 electrons). Statements that are true are:
Eªcell > 0; n = 2Which of the following is a salt that could be generated by combining a weak acid and a weak base? Select the correct answer below:
a) NaCl
b) Na2SO4
c) NH4NO3
d) NH4F
Answer:
d) NH4F
Explanation:
Hello,
In this case, the base resulting from mixing a weak acid and a weak base is d) NH4F since ammonium hydroxide is a wear base and hydrofluoric acid is a weak acid.
Ammonium hydroxide is a weak base since it is not completely ionized in ammonium and hydroxyl ions:
[tex]NH_4OH\rightarrow NH_4^++OH^-[/tex]
Moreover, hydrofluoric acid is a weak acid since it is not completely ionized in hydrogen and fluoride ions:
[tex]HF\rightleftharpoons H^++F^-[/tex]
For the both of the substances, the limit is established by the basic and the acid dissociation constant respectively.
Regards.
The two common properties of all solids are fixed _____ and _____.
Answer:
shape
volume
Hope this helps! （づ￣3￣）づ╭❤～
Explanation:
Which statements about spontaneous processes are true? Select all that apply: A spontaneous process is one that occurs very quickly.
Answer: Here are the complete options.
A spontaneous process is one that occurs very quickly. A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium. A spontaneous process is one that occurs without continuous input of energy from outside the system. A process is spontaneous if it must be continuously forced or driven.
The correct option is
A spontaneous process is one that occurs without continuous input of energy from outside the system.
A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium
Explanation:
spontaneous process is one that occurs without continuous input of energy from outside the system and occur on its own because spontaneous processes are thermodynamically favorable characterized by a decrease in the system's free energy, they do not need to be driven by an outside source of energy. Which means that the initial energy is higher than the final energy.
A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium which will result to The sign of ΔG will change from positive to negative (or vice versa) where T = ΔH/ΔS. In cases where ΔG is: negative
What are periodic trends if ionic radii
Answer:
Explan ionization energy, atomic radius, and electron affinityation:
This question most likely has answer choices. The possible answer choices are as followed:
Ionic radii tend to increase down a group.Ionic radii tend to decrease across a period.Anionic radii tend to increase across a period.Cationic radii tend to decrease across a period.Ionic radii increase when switching from cations to anions in a period.The answers are Ionic radii tend to increase down a group, Cationic radii tend to decrease across a period, and Ionic radii increase when switching from cations to anions in a period (1st, 4th, and 5th options).
the following glassware was found in a lab drawer: 12 beakers, 10 flasks and 60 test tubes. what percent of the glassware are test tubes?
Answer:
73.1707317073%
=> Approximately 73.2%
Explanation:
Total = 60 + 10 + 12
=> 82
Test tubes are 60/82
=> 30/41
=> 73.1707317073%
=> Approximately 73.2%
Answer:
73.17%
Explanation:
To find the percentage of test tubes to the overall glassware, we need to get the number of test tubes divided by the total number of glassware.
12 beakers + 10 flasks + 60 test tubes = 82 glassware
% test tube = 60 / 82 = .7317 ==> 73.17 %
So 73.17 % of the glassware was test tubes.
Cheers.
A chemist prepares a solution of sodium chloride by measuring out 25.4 grams of sodium chloride into a 100. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's sodium chloride solution. Be sure your answer has the correct number of significant digits.
Answer:
The concentration in mol/L = 4.342 mol/L
Explanation:
Given that :
mass of sodium chloride = 25.4 grams
Volume of the volumetric flask = 100 mL
We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol
and number of moles = mass/molar mass
The number of moles of sodium chloride = 25.4 g/58.5 g/mol
The number of moles of sodium chloride = 0.434188 mol
The concentration in mol/L = number of mol/ volume of the solution
The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L
The concentration in mol/L = 4.34188 mol/L
The concentration in mol/L = 4.342 mol/L
Using the standard reduction potentials Ni2+(aq) + 2 e‑Ni(s) ‑0.25 volt Fe3+(aq) + e‑Fe2+(aq) +0.77 volt Calculate the value of E°cell for the cell with the following reaction. Ni2+(aq) + 2 Fe2+(aq) →Ni(s) + 2 Fe3+(aq)
Answer:
The correct answer is - 1.02 V
Explanation:
From the reduction-oxidation reaction:
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:
Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s) Eº= ‑0.25 V
Oxidation (anode) : 2 x (Fe²⁺ → Fe³⁺ + e-)(aq) Eº= -0.77 V
-------------------------------------
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):
Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V
Calculate the concentration of H3O+ in a solution that contains 5.5 × 10-5 M OH- at 25°C. Identify the solution as acidic, basic, or neutral.
Explanation:
To calculate [H3O+] in the solution we must first find the pH from the [ OH-]
That's
pH + pOH = 14
pH = 14 - pOH
To calculate the pOH we use the formula
pOH = - log [OH-]
And [OH-] = 5.5 × 10^-5 M
So we have
pOH = - log 5.5 × 10^ - 5
pOH = 4.26
Since we've found the pOH we can now find the pH
That's
pH = 14 - 4.26
pH = 9.74
Now we can find the concentration of H3O+ in the solution using the formula
pH = - log H3O+
9.74 = - log H3O+
Find the antilog of both sides
H3O+ = 1.8 × 10^ - 10 MThe solution is basic since it's pH lies in the basic region.
Hope this helps you
Because of movements at the Mid-Atlantic Ridge, the Atlantic Ocean widens by about 2.5 centimeters each year. Explain which type of plate boundary causes this motion.
Answer:
A divergent plate boundary
Explanation:
At a divergent boundary, the plates pull away from each other and generate new crust.
Answer:
Because the ocean becomes larger, this is a divergent plate boundary. Divergent plates cause the ocean floor to expand, making the ocean larger.
Explanation:
PLATO ANSWER
The absorption spectrum of argon has a line at 515 nm. What is the energy of
this line? (The speed of light in a vacuum is 3.00 x 108 m/s, and Planck's
constant is 6.626 x 10-34 Jos.)
O A. 2.59 x 1027j
O B. 3.86 x 10-28 J
O C. 3.86 x 10-19 J
O D. 2.59 x 1018 J
Answer:
OPTION C is correct
3.86 x 10-19 J
Explanation:
Energy of the line can be calculated using below formula
E= h ν.................(1)
Where E= energy
h= plank constant= 6.626 10-34 J s
c=speed of light=3 x 108 m/s
But we know that Velocity V= = c / λ
Then substitute into equation (1) we have
E = h c / λ.............(2)
We can calculate our( hc ) in nm for unit consistency
h c =( 6.626 ×10^-34)x(3×108)
h c = (1.986 x 10-16 )
hc = 1.986 x 10-16 J nm then since our (hc) and λ are in the same unit , were good to go then substitute into equation(2)
E = h c / λ = (1.986 x 10-16) / 515
E = 3.86 x 10-19 J
Therefore, the Energy is 3.86 x 10-19 J
A chemist prepares a solution of silver(I) nitrate(AgNO3) by measuring out 269. mu mol of silver(I) nitrate into a 300. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mmol/L of the chemist's silver(I) nitrate solution.
Answer:
concentration in mmol/L = 8.97 × 10⁻¹ mmol/L
Explanation:
Given that:
the number of moles of silver(I) nitrate(AgNO3) the chemist used in preparing a solution = 269 mmol = 269 × 10⁻³ mmol
The volume of the volumetric flask = 300 mL = 300 × 10⁻³ L
In order to calculate the concentration in mmol/L of the chemist's silver(I) nitrate (AgNO3) solution , we used the formula which can be expressed as;
[tex]concentration \ in \ mmol/L = \dfrac{ number \ of \ mmol}{vol. \ of \ the \ solution}[/tex]
[tex]concentration \ in \ mmol/L = \dfrac{ 269 * 10^{-3 } \ mmol }{300 * 10^{-3} \ L }[/tex]
concentration in mmol/L = 0.8966 mmol/L
concentration in mmol/L = 8.97 × 10⁻¹ mmol/L
A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup calorimeter fitted with a lid through which a thermometer passes. The acid-base reaction is as follows:
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
The temperature of each solution before mixing is 22.3 °C. After mixing, the temperature of the solution mixture reaches a maximum temperature of 31.4 °C. Assume the density of the solution mixture is 1.00 g/mL, its specific heat is 4.18 J/g.°C, and no heat is lost to the surroundings. Calculate the enthalpy change, in kj, per mole of H2SO4 in the reaction.
a. +85.6 kJ/mol.
b. -85.6 kJ/mol.
c. +5.71 kJ/mol.
d. -5.71 kJ/mol.
e. -114 kJ/mol.
Answer:
THE ENTHALPY CHANGE IN KJ/MOLE IS +114 KJ/MOLE.
Explanation:
Heat = mass * specific heat capacity * temperature rise
Total volume = 100 + 50 = 150 mL
Total mass = density * volume
Total mass = 1 * 150 mL = 150 g
So therefore, the heat evolved during the reaction is:
Heat = 150 * 4.18 * ( 31.4 - 22.3)
Heat = 150 * 4.18 * 9.1
Heat = 5705.7 J
Equation for the reaction:
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
From the equation, 2 moles of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water
50 mL of 1 M of H2SO4 contains
50 * 1 / 1000 mole of acid
= 0.05 mole of acid
The production of 1 mole of water evolved 5705.7 J of heat and hence the enthalpy changein kJ per mole will be:
0.05 mole of H2SO4 produces 5705.7 J of heat
1 mole of H2SO4 will produce 5705.7 / 0.05 J
= 114,114 J / mole
In kj/mole = 114 kJ/mole.
Hence, the enthalpy change of the reaction in kJ /mole is +114 kJ/mole.
A 1 liter solution contains 0.436 M hypochlorous acid and 0.581 M potassium hypochlorite. Addition of 0.479 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
Answer:
Exceed the buffer capacity and Raise the pH by several units
Explanation:
Options are:
Raise the pH slightly
Lower the pH slightly
Raise the pH by several units
Lower the pH by several units
Not change the pH
Exceed the buffer capacity
The hypochlorous acid, HClO, is in equilibrium with Hypochlorite ion (From potassium hypochlorite, ClO⁻) producing a buffer. Using H-H equation, pH of initial buffer is:
pH = pKa + log [ClO⁻] / [HClO]
pKa for hypochlorous acid is 7.53
pH = 7.53 + log [0.581M] / [0.436M]
pH = 7.65
Barium hydroxide reacts with HClO producing more ClO⁻, thus:
Ba(OH)₂ + 2HClO → 2ClO⁻ + 2H₂O
As 0.479 moles of Barium hdroxide are added. For a complete reaction you require 0.479mol * 2 = 0.958 moles of HClO
As you have just 0.436 moles (Volume = 1L),
The addition will:
Exceed the buffer capacityThe Ba(OH)₂ that reacts is:
0.436 moles HClO * (1mole (Ba(OH)₂ / 2 mol HClO) = 0.218 moles Ba(OH)₂ and will remain:
0.479 mol - 0.218 mol = 0.261 moles Ba(OH)₂
As 1 mole of Ba(OH)₂ contains 2 moles of OH⁻, moles of OH⁻ and molarity is:
0.261 moles* 2 = 0.522 moles OH⁻ = [OH⁻]
pOH = -log [OH⁻]
pOH = 0.28
And pH = 14 - pOH:
pH = 13.72
Thus, after the addition the pH change from 7.65 to 13.62:
Raise the pH by several units
Which part of an atom is mostly empty space?
A. nucleus
B. proton cloud
C. electron cloud
D. neutron
Answer:
C. Electron cloud
the electron is around 1/2000 times the size of the proton.
If you imagine the proton a a marble in the middle of a football field, the electrons will revolve around the last row
There are approximately 2 × 1022 molecules and atoms in each breath we take and the concentration of CO in the air is approximately 9 ppm. How many CO molecules are in each breath we take? solution
Answer:
1.8x10¹⁷ molecules of CO are in each breath we take
Explanation:
Parts per million, ppm, is an unit of concentration in chemistry used for very diluted solutions.
A 9ppm of X in a solution means in 1 million of molecules (1x10⁶) you have only 9 molecules of X.
In a breath we have 2x10²² molecules and 9 ppm are CO. Thus, CO molecules in each breath are:
2x10²² molecules × (9 molecules CO / 1x10⁶ molecules) =
1.8x10¹⁷ molecules of CO are in each breath we take[tex]1.8\times 10^{17}[/tex] molecules of CO are in each breath we take
The calculation is as follows:A 9ppm of X in a solution represent in 1 million of molecules[tex](1\times10^6)[/tex]you have only 9 molecules of X.
Now CO molecules in each breath is
[tex]= 2\times 10^{22}\ vmolecules \times (9\ molecules\ CO \div 1\times 10^6 molecules) \\\\= 1.8\times 10^{17}[/tex]
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A piece of solid Fe metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement redox reaction that may be predicted. Assume that the oxidation state of in the resulted solution is 2 . (Use the lowest possible coefficients for the reaction. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If no reaction occurs, leave all boxes blank and click on Submit.)
Answer:
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
Explanation:
The ionic equation shows the actual reaction that took place. It excludes the spectator ions. Spectator ions are ions that do not really participate in the reaction even though they are present in the system.
For the reaction between iron and copper II nitrate, the molecular reaction equation is;
Fe(s) + Cu(NO3)2(aq)----> Fe(NO3)2(aq) +Cu(s)
Ionically;
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
Calculate the energy required to heat 1.30kg of water from 22.4°C to 34.2°C . Assume the specific heat capacity of water under these conditions is 4.18·J·g−1K−1 . Round your answer to 3 significant digits.
Answer:
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
Explanation:
Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.
The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:
Q = c * m * ΔT
Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).
In this case:
[tex]c=4.18 \frac{J}{g*K}[/tex]m= 1.30 kg= 1,300 g (1 kg=1,000 g)ΔT= 34.2 °C - 22.4 °C= 11.8 °C= 11.8 °K Being a temperature difference, it is independent if they are degrees Celsius or degrees Kelvin. That is, the temperature difference is the same in degrees Celsius or degrees Kelvin.Replacing:
[tex]Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K[/tex]
Q= 64,121.2 J
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
The number of moles of H2O which contains 4g of oxygen?
Answer:
16G = 1 mole ; then 4G = how many moles? 4/16 = 0.25 mole; That means 4 grams of oxygen is 0.25 moles.
Explanation:
A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.
One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose an EPA chemist tests a sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
Answer:
6.5 mg/L.
Explanation:
Step one: write out and Balance the chemical reaction in the Question above:
NiCl2 + 2AgNO3 =====> 2AgCl + Ni(NO3)2.
Step two: Calculate or determine the number of moles of AgCl.
So, we are given that the mass of AgCl = 3.6 mg = 3.6 × 10^-3 g. Therefore, the number of moles of AgCl can be calculated as below:
Number of moles AgCl = mass/molar mass = 3.6 × 10^-3 g / 143.32. = 2.5118 × 10^-5 moles.
Step three: Calculate or determine the number of moles of NiCl2.
Thus, the number of moles of NiCl2 = 2.5118 × 10^-5/ 2 = 1.2559 × 10^-5 moles.
Step four: detemine the mass of NiCl2.
Therefore, the mass of NiCl2 = number of moles × molar mass = 1.2559 × 10^-5 moles × 129.6 = 1.6 × 10^-3 g.
Step five: finally, determine the concentration of NiCl2.
1000/ 250 × 1.6 × 10^-3 g. = 6.5 mg/L.
Calculate LaTeX: \DeltaΔGº for a voltaic cell with Eºcell = +0.24 V if the overall reaction involves a 3 electron reduction.
Answer:
-69 kJ
Explanation:
Step 1: Given data
Standard cell potential (E°cell): +0.24 V
Electrons involved (n): 3 mol
Step 2: Calculate the standard Gibbs free energy change (ΔG°) for the voltaic cell
We will use the following expression.
ΔG° = -n × F × E°cell
where,
F is Faraday's constant (96,485 C/mol e⁻)
ΔG° = -n × F × E°cell
ΔG° = -3 mol e⁻ × 96,485 C/mol e⁻ × 0.24 V
ΔG° = -69 kJ
What is the pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid with 15.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. K a = 1.8 ×× 10-5 for CH3CO2H.
Answer:
pH = 8.72
Explanation:
This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
As this compound acts like a base, we propose this equilibrium:
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰
Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M
So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
We can avoid the quadractic equation because Kb is so small
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH = 8.72
The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
Calculation of the pH of the solution:Since the following equation should be used.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
Now
(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
So,
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
Now
Kw = Ka. Kb
Kb = Kw/Ka
And,
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵
= 5.55×10⁻¹⁰
Now
[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
Now
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH
= 8.72
Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
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How many valence electrons must two atoms share to form a single covalent bond? answers A.2 B.4 C.3 D.1
Answer:
2
Explanation:
A single covalent bond is formed when two electrons are shared between the same two atoms, one electron from each atom.
Answer:
the answer is 2
Explanation:
A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an average volume of the titrant was 38.2 mL. The starting volume of the HCl solution was 20 mL. What's the concentration of the HCl? answer options: A) 0.788 M B) 0.284 M C) 3.34 M D) 0.191 M
Answer: it is A
Explanation: I am sure
Answer:
0.191 M
Explanation:
i took the test.
A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?
Question is incomplete, the complete question is as follows:
A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?
A. Toxicity, because it can be observed by altering the state of the substance
B. Boiling point, because it can be observed by altering the state of the substance
C. Toxicity, because it can be observed by replacing the atoms of the substance with new atoms
D. Boiling point, because it can be observed by replacing the atoms of the substance with new atoms
Answer:
B.
Explanation:
A student can examine a substance without altering the bonds within the molecules by examining its boiling point.
The boiling point is the property of a substance, at which the substance changes its state, which is from solid to liquid, liquid to gas and others. So, examining the boiling point will alter the bonds within the molecules as the state of substance will change.
Hence, the correct answer is "B".
Zinc is used as a coating for steel to protect the steel from environmental corrosion. If a piece of steel is submerged in an electrolysis bath for 24 minutes with a current of 6.5 Amps, how many grams of zinc will be plated out? The molecular weight of Zn is 65.38, and Zn+2 + 2e– → Zn. Question 7 options: A) 3.17 g of Zn B) 1.09 g of Zn C) 6.34 g of Zn D) 12.68 g of Zn
Answer:
A) 3.17 g of Zn
Explanation:
Let's consider the reduction of Zn(II) that occurs in an electrolysis bath.
Zn⁺²(aq) + 2e⁻ → Zn(s)
We can establish the following relations:
1 min = 60 s1 A = 1 C/sThe charge of 1 mole of electrons is 96,468 C (Faraday's constant).When 2 moles of electrons circulate, 1 mole of Zn is deposited.The molar mass of Zn is 65.38 g/molThe mass of Zn deposited under these conditions is:
[tex]24min \times \frac{60s}{1min} \times \frac{6.5C}{s} \times \frac{1mol\ e^{-} }{96,468C} \times \frac{1molZn}{2mol\ e^{-}} \times \frac{65.38g}{1molZn} = 3.17 g[/tex]
Answer:
A.) 3.17
Explanation:
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A mixture of 50ml of 0.1M HCOOH and 50ml of 0.05M NaOH is equivalent to
Answer:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa
Explanation:
The reaction of a weak acid (HOOH) with NaOH is as follows:
HCOOH + NaOH → HCOONa + H₂O
Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).
The initial moles of both species are:
HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH
NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH
After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).
Final moles:
HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles
HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles
As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:
0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa
Thus, the initial mixture is equivalent to:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONaFrom the following balanced equation, CH4(g)+2O2(g)⟶CO2(g)+2H2O(g) how many grams of H2O can be formed when 1.25g CH4 are combined with 1.25×10^23 molecules O2? Use 6.022×10^23 mol−1 for Avogadro's number.
Answer:
2.81 g of H2O.
Explanation:
We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.
This can be obtained as follow:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.
1 mole of O2 = 16x2 = 32 g.
Thus 6.022×10²³ molecules is present in 32 g of O2,
Therefore, 1.25×10²³ molecules will be present in =
(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.
Therefore, 1.25×10²³ molecules present in 6.64 g of O2.
Next, the balanced equation for the reaction. This is given below:
CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)
Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.
This can be obtained as follow:
Molar mass of CH4 = 12 + (4x1) = 16 g/mol.
Mass of CH4 from the balanced equation = 1 x 16 = 16 g
Molar mass of O2 = 16x2 = 32 g/mol.
Mass of O2 from the balanced equation = 2 x 32 = 64 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol.
Mass of H2O from the balanced equation = 2 x 18 = 36 g
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.
Next, we shall determine the limiting reactant.
This can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2.
Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.
From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.
Therefore, CH4 is the limiting reactant.
Finally, we shall determine the mass of H2O produced from the reaction.
In this case, the limiting reactant will be used because it will give the maximum yield of H2O.
The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted to produce produce 36 g if H2O.
Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.
Therefore, 2.81 g of H2O were obtained from the reaction.
The mass in grams of H₂O which can be formed when 1.25g CH₄ are combined with 1.25×10²³ molecules O₂ is 2.8 grams.
What is stoichiometry?Stoichiometry of any reaction tells about the amount of species present before and after the completion of the reaction.
Given chemical reaction is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Moles of CH₄ will b calculate as:
n = W/M, where
W = given mass = 1.25g
M = molar mass = 16g/mol
n = 1.25/16 = 0.078 moles
Molecues of CH₄ in 0.078 moles = 0.078×6.022×10²³ = 0.46×10²³
Given molecules of O₂ = 1.25×10²³
Required molecules of CH₄ is less as compared to the molecules of O₂, so here CH₄ is the limiting reagent and formation of water is depends on it only.
From the stoichiometry of the reaction it is clear that:
1 mole of CH₄ = will produce 2 moles of H₂O
0.078 moles of CH₄ = will produce 2×0.078=0.156 moles of H₂O
Mass of H₂O will be calculated by using its moles as:
W = (0.156)(18) = 2.8g
Hence required mass of H₂O is 2.8g.
To know more about limiting reagent, visit the below link:
https://brainly.com/question/1163339
