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Ulleksa [173]
2 years ago
9

The mass ratio of hydrogen to oxygen in water (formula H20) is 0.126 g hydrogen for every 1 g of oxygen. In another compound of

hydrogen and oxygen called hydrogen peroxide, the mass ratio of hydrogen to oxygen is 0.063 g hydrogen for every 1 g of oxygen. What is a possible formula of hydrogen peroxide?​
Chemistry
1 answer:
poizon [28]2 years ago
5 0

Answer: H2O2

Explanation:

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A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h
NikAS [45]

Nitrogen combine with hydrogen to produce ammonia \text{NH}_3 at a 1:3:2 ratio:

\text{N}_2 \; (g) + 3 \;  \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. 3 \; \text{mol} of hydrogen gas would have been consumed while 2 \; \text{mol} of ammonia would have been produced. The final mixture would therefore contain

  • 17 \; \text{mol} of \text{H}_2 \; (g) and
  • 2 \; \text{mol} of \text{NH}_3 \; (g)

Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:

  • \begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  1.926 \times 10^{3} \; \text{kPa} \end{array}
  • \begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  1.723 \times 10^{3} \; \text{kPa} \end{array}
  • \begin{array}{lll} P(\text{NH}_3) &= & n \cdot R \cdot T / V \\ & = & (2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  2.037 \times 10^{2} \; \text{kPa} \end{array}
6 0
3 years ago
How much heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminu
Rzqust [24]
295k=22°c
1050k=777°

total heat needed=
(0.475)(777-22)(specific heat capacity of aluminium)+(0.475)(specific latent heat of aluminium)
7 0
3 years ago
Before the work of Albert Einstein and other prominent scientists of the early 1900's, the predominant view among
Ivanshal [37]
A wave.

Scientists now recognize that light can behave as both a particle and a wave.
3 0
2 years ago
What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?
kari74 [83]
The heat of reaction (i.e. combustion) of butane (C_{4} H_{10}) when reacted with oxygen (O_{2})  is -2658 kJ/mol butane, and the chemical reaction is given by: 

C_{4} H_{10} + \frac{13}{2} O_{2} ---> 4 CO_{2}  + 5 H_{2}O

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants. 

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required. 

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane  = 32.73 grams butane

The mass of carbon dioxide (CO_{2}) can be determined by multiplying the moles of butane (C_{4} H_{10}) with the mole ratio of (CO_{2}) produced to the (C_{4} H_{10}) reacted, and then with the molar mass of (CO_{2}), which is 44 grams/mole. 

Mass of carbon dioxide produced 
    = 0.5643 moles butane * [4 moles CO_{2}/ 1 mole C_{4} H_{10}] * 44 grams/mole CO_{2}

Mass of carbon dioxide produced  
    = 99.32 grams CO_{2}

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams. 
                
4 0
3 years ago
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What is the mass in grams of 7.23 miles of dinitrogen trioxide?
nevsk [136]

Answer:

549.563868

Explanation:

1 mole is equal to 1 moles N2O3, or 76.0116 grams. so 76.0116 x 7.23 = 549.563868

3 0
3 years ago
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