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3 years ago

Can you please help?

Chemistry

1 answer:

KiRa [710]3 years ago

6 0

Answer:

The answer to your question is below

Explanation:

3.-

[HF] = 6.9 N

Volume = ?

[H₃PO₄] = 3.2 N

Volume = 22ml

[HF] x Volume a = [H₃PO₄] x Volume b

Volume a = [H₃PO₄] x Volume b / [HF]

Volume a = 3.2 x 22 / 6.9

= 10.2 ml

4.-

[Al(OH)₃] = 4 N

Volume b = 72 ml

[HBr] = ?

Volume a = 21 ml

[Al(OH)₃] x Volume b = [HBr] x Volume a

[HBr] = [Al(OH)₃] x Volume b / Volume a

[HBr] = 4 x 72 / 21

[HBr] = 13.7 N

5.-

[HC₂H₃O₂] = 0.769 N

Volume a = ?

[Mg(OH)₂] = 1.15 N

Volume b = 31 ml

[HC₂H₃O₂] x Volume a = [Mg(OH)₂] x Volume b

Volume a = [Mg(OH)₂] x Volume b / [HC₂H₃O₂]

Volume a = 1.15 x 31 / 0.769

Volume a = 46.4 ml

Volume a = ?

[HCl] = 4.7 N

Volume b = 15.78 ml

[Ba(OH)₂] = 5.67 N

[HCl] x volume a = [Ba(OH)₂] x volume b

Volume a = [Ba(OH)₂] x volume b / [HCl]

Volume a = 5.67 x 15.78 / 4.7

Volume a = 19 ml

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Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride

IrinaK [193]

The question is incomplete, here is the complete question:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed? Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of hydrogen chloride produced in the reaction will be 4.4 mL

<u>Explanation:</u>

We are given:

Volume of methane gas = 1.1 mL

The chemical equation for the reaction of methane gas and chlorine gas follows:

$CH_4(g)+4Cl_2(g)\rightarrow 4HCl(g)+CCl_4(g)$

Moles of methane gas = 1 mole

Moles of hydrogen chloride gas = 4 moles

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

where,

$V_1\text{ and }n_1$ are the volume and number of moles of methane gas

$V_2\text{ and }n_2$ are the volume and number of moles of hydrogen chloride

We are given:

$V_1=1.1mL\\n_1=1mol\\V_2=?L\\n_2=4mol$

Putting values in above equation, we get:

$\frac{1.1}{1}=\frac{V_2}{4}\\\\V_2=\frac{1.1\times 4}{1}=4.4mL$

Hence, the volume of hydrogen chloride produced in the reaction will be 4.4 mL

5 0

3 years ago

How much heat is evolved in converting 1.00 mol of steam at 155.0 ∘c to ice at -50.0 ∘c? the heat capacity of steam is 2.01 j/(g

Ne4ueva [31]

When the specific heat capacity of the water is 4.18 J/g.°C so, we are going to use this formula to get the heat for cooling three phases changes from steam to liquid and from liquid to ice (solid) :

when Q = M*C*ΔT

Q is the heat in J

and M is the mass in gram = 1 mol H2O * 18 g/mol(molar mass) = 18 g

C is the specific heat J/g.°C

ΔT is the change in temperature

Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice * ΔT ice)]

= 18 g * [(2.01 * (155-100°C)) + (4.18 * (100-0°C)) + (2.09 * (0 - 55 °C))]

∴Q = 7444.8 J

and when we know that the heat of fusion for water = 334J/g

and heat of vaporization for water = 2260J/g

∴Q for the two phases changes = M * (2260+334)

= 18 * (2260+334)

= 46692 J

∴ Q total = 7444.8 + 46692 = 54136.8 J

5 0

3 years ago

Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2

baherus [9]

Answer:

1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) $Cu_{(s)}$ → $Cu^{2+}_{(aq)} +2e$ E°=0.337 v

(reduction) $MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}$ → $MnO_{2 (aq)}+2H_{2}O$ E°=1.679 v

(overall) $2MnO_{4 (aq)}+3Cu_{(s)}$ +8H^{+}_{(aq)}→ $3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O$ E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

$E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}$

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

$E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346$