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jenyasd209 [6]
3 years ago
11

Assume that 4 molecules of nitrogen and 9 molecules of hydrogen are present? after complete reaction, how many molecules of ammo

nia are proceded?
Chemistry
1 answer:
Anastasy [175]3 years ago
5 0
<span>2<span>N2</span>+3<span>H2</span>→2N<span>H3

That's the balanced equation. As shown, 4 molecules of N2 would react with 12 molecules of H2. Since not that much H2 molecules exist, then it would serve as the limiting reactant. Therefore:

9 mol H2 * (2NH3 / 3H2) = 6 molecules of ammonia
</span></span>
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At 300.0 K and 0.987 atm pressure, what will be the volume of 2.30 mol of Ne?
Nadya [2.5K]

Answer:

V = 57.39 L

Explanation:

Given that,

Temperature, T = 300 K

Pressure, P = 0.987 atm

No. of moles of Ne, n = 2.30 mol

We need to find the volume of Ne. We know that, the ideal gas law is as follows :

PV = nRT

Where

P is pressure and R is gas constant

V=\dfrac{nRT}{P}\\\\V=\dfrac{2.3\times 0.0821\times 300}{0.987 }\\\\V=57.39\ L

So, the volume of the Ne is 57.39 L.

3 0
3 years ago
If an atom has no electric charge, what can be said about the number of protons and electrons it contains?
vitfil [10]

then the electrons and protons would have a even amount of negetive electric charges

8 0
3 years ago
Which reaction has a positive δs°? question 6 options:
Nastasia [14]
B. Because there are 3 molecules in right and 2 molecules in the left, so entropy rises.
3 0
3 years ago
There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
How many grams of argon would it take to fill a large light bulb with a volume of 0.745 L at STP?
EastWind [94]

Answer:

Mass = 1.33 g

Explanation:

Given data:

Mass of argon required = ?

Volume of bulb = 0.745 L

Temperature and pressure = standard

Solution:

We will calculate the number of moles of argon first.

Formula:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

By putting values,

1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K

0.745 atm. L = n × 22.43 atm.L/mol

n = 0.745 atm. L / 22.43 atm.L/mol

n = 0.0332 mol

Mass of argon:

Mass = number of moles × molar mass

Mass = 0.0332 mol × 39.95 g/mol

Mass = 1.33 g

8 0
3 years ago
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