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What is the formula for the compound nitrogen monoxide? express your answer as a chemical formula?

Lina20 [59]

<span>Formula for the compound nitrogen monoxide is NO.
Below is the structure:</span>

5 0

2 years ago

A gas mixture contains 3.00 atm of H2 and 1.00 atm of O2 in a 1.00 L vessel at 400K. If the mixture burns to form water while th

sleet_krkn [62]

Answer:

$p_{H_2O}=2.00atm$

Explanation:

Hello!

In this case, according to the following chemical reaction:

$2H_2+O_2\rightarrow 2H_2O$

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

$n_{H_2}=\frac{3.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0914molH_2 \\\\n_{O_2}=\frac{1.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0305molH_2$

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

$n_{H_2O}^{by\ H_2} = 0.0914molH_2*\frac{2molH_2O}{2molH_2} =0.0914molH_2O\\\\n_{H_2O}^{by\ O_2} = 0.0305molO_2*\frac{2molH_2O}{1molO_2} =0.0609molH_2O$

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

$p_{H_2O}=\frac{0.0609molH_2O*0.08206\frac{atm*L}{mol*K}*400K}{1.00L}\\\\ p_{H_2O}=2.00atm$

Best regards!

7 0

2 years ago

The carbohydrates seen here contains 3 common elements.They are....

Llana [10]

The answer is A),<span>carbon,hydrogen,and oxygen

Hope this helps
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6 0

2 years ago

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How many grams of oxygen gas occupy 12.3 L of space at 109.4 kPa and 15.4oC?

I am Lyosha [343]

17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C. Details about how to calculate mass can be found below.

<h3>How to calculate mass?</h3>

The mass of a given gas can be calculated by multiplying the number of moles of the substance by its molar mass.

However, the number of moles of the gas must be calculated first as follows:

PV = nRT

Where;

P = pressure = 1.0796941atm
V = volume = 12.3L
n = number of moles
T = temperature = 288.4K
R = gas law constant = 0.0821 Latm/molK

1.079 × 12.3 = n × 0.0821 × 288.4

13.27 = 23.68n

n = 13.27/23.68

n = 0.56mol

Mass = 0.56 × 32

mass of oxygen gas = 17.93g

Therefore, 17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C.

Learn more about mass at: brainly.com/question/19694949

3 0

1 year ago

Please hurry! classify the following as a

Talja [164]

Answer:

pure substance.

Explanation:

i hope that helps

5 0

1 year ago

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