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icang [17]
1 year ago
9

Acetylene gas (ethyne; HC = CH) burns in an oxyacetylene torch to produce carbon dioxide and water vapor. The heat of reaction f

or the combustion of acetylene is 1259 kJ/mol.
(c) How many grams ofCO₂ form?
Chemistry
1 answer:
Yanka [14]1 year ago
6 0

The mass of CO2 produced by 26g of acetylene is 88g.

Given ,

In an oxyacetylene torch, acetylene gas (ethyne; HCCH) burns to produce carbon dioxide and water vapour.

The acetylene combustion reaction is given by,

H2O + HCCH + 5/2 O=O 2CO2

Heat of reaction for acetylene combustion = 1259kj/mol

CO2 has a molecular mass of 44g/mol.

2 moles of CO2 have a molecular mass of 88g.

On combustion, 1 mole of acetylene yields 2 moles of CO2.

Thus, 26g of acetylene produces 88g of CO2 when burned.

As a result, the mass of carbon dioxide produced by 26g of acetylene is 88g.

Learn more about acetylene here :

brainly.com/question/15346128

#SPJ4

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What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent in the overall r
Sergeu [11.5K]
Answer:
             5 moles of electrons

Explanation:

The balance equation is as follow,

<span>                   5 Ag</span>⁺ + Mn⁺²<span> + 4 H</span>₂O    →<span>    5 Ag + MnO</span>₄⁻<span> + 8 H</span>⁺

Reduction of Ag:

                         Ag⁺ + 1 e⁻    →    Ag
Or,
                         5 Ag⁺  +  5 e⁻    →    5 Ag

Oxidation of Mn:

                          Mn⁺²   →    MnO₄⁻  +  5 e⁻

Result:
          Hence 5 moles of Ag⁺ accepts 5 electrons from 1 mole of Mn⁺².
4 0
3 years ago
What is the density of a liquid that has a mass of 27g and a volume of 30mL?
NNADVOKAT [17]
  • Answer: 7 g/cm to the third power
  • Hope this helps
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-Carrie

8 0
3 years ago
A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A
Kryger [21]

Answer:

T_{2}=16,97^{\circ}C

Explanation:

The specific heats of water and steel are  

Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}

Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}

Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium T_{2}_{w}= T_{2}_{s}  

An energy balance can be written as

m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}  

Replacing

1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)

Then, the temperature T_{2}=16,97^{\circ}C

8 0
3 years ago
A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 32.5 kPa. If Po2 =
HACTEHA [7]

Answer:

A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 32.5 kPa.

<u>The pressure for oxygen is 3 kPa</u>

Explanation:

According to Dalton's Law of Partial Pressure total exerted by the mixture of non-reacting gases is equal to sum of the partial pressure of each gas.

P_{total}=P_{1}+P_{2}+P_{3}

So,

For , a gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure:

P_{total}=P_{O_{2}}+P_{N_{2}}+P_{CO_{2}}

P_{total} = 32.5kPa

P_{O_{2}} = 6.5kPa

P_{N_{2}} = 23.0kPa

Insert the values in :

P_{total}=P_{O_{2}}+P_{N_{2}}+P_{CO_{2}}

32.5 kPa = 6.5 kPa + 23.0 kPa +P_{CO_{2}}

32.5 kPa = 29.5 kPa +P_{CO_{2}}

P_{CO_{2}}= 32.5 - 29.5

P_{CO_{2}}= 3kPa

6 0
3 years ago
How well can you apply Charles’s law to this sample of gas that experiences changes in pressure and volume? Assume that pressure
Phoenix [80]

Answer:

As temperature increases the volume of given amount of gas increases while pressure and number of moles remain constant.

Explanation:

According to the charle's law,

The volume of given amount of gas is directly proportional to the temperature at constant pressure and number of moles of gas.

Mathematical expression:

V ∝ T

V = KT

V/T = K

When temperature changes from T₁ to T₂ and volume changes from V₁ to V₂.

V₁/T₁ = K        V₂/T₂ = K

or

V₁/T₁  = V₂/T₂

Thus, the ratio of volume and temperature remain constant for constant amount of gas at constant pressure.

6 0
3 years ago
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