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bearhunter [10]
4 years ago
15

How many moles of n are in 0.185 g of n2o?

Chemistry
2 answers:
pickupchik [31]4 years ago
8 0

The mass of N_{2}O is 0.185 g.

To calculate the number of moles first calculate the molar mass of N_{2}O .

M_{N_{2}O}=2\times M_{N}+M_{O}

Molar mass of nitrogen and oxygen is 14 g/mol and 16 g/mol respectively

Thus, M_{N_{2}O}=2\times 14 g/mol+18 g/mol=44 g/mol

Now, number of moles can be calculated from mass and molar mass as follows:

n=\frac{m}{M}=\frac{0.185 g}{44 g/mol}=0.0042 mol

Therefore, number of moles of 0.185 g N_{2}O is 0.0042 mol.

solong [7]4 years ago
3 0

Answer : The number of moles of nitrogen are 0.00840 moles.

Explanation : Given,

Mass of N_2O = 0.185 g

Moles of N_2O = 44 g/mole

First we have to calculate the moles of N_2O.

\text{ Moles of }N_2O=\frac{\text{ Mass of }N_2O}{\text{ Molar mass of }N_2O}=\frac{0.185g}{44g/mole}=0.00420moles

Now we have to calculate the moles of nitrogen (N).

In N_2O, there are 2 moles of nitrogen atom and 1 mole of oxygen atom.

As, 1 mole of N_2O contains 2 moles of nitrogen

So, 0.00420 mole of N_2O contains 2 × 0.00420 = 0.00840 moles of nitrogen

Therefore, the number of moles of nitrogen are 0.00840 moles.

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The answer is 1.15m.

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We can find the number of H2SO4 moles by using its molarity

C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288

Since water has a density of 1.00kgL, the mass of solvent is

m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg

Therefore, molality is

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Choose the FALSE statement.
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C. The half-life of C-14 is about 40,000 years.

Explanation:

The only false statement from the options is that the half-life of C-14 is 40,000yrs.

The half-life of an isotope is the time it takes for half of a radioactive material to decay to half of its original amount. C-14 has an half-life of 5730yrs. This implies that during every 5730yrs, C-14 will reduce to half of its initial amount.

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Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?
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Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

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  • O₂: 13 moles
  • CO₂: 8 moles
  • H₂O: 10 moles

Being:

  • C: 12 g/mole
  • H: 1 g/mole
  • O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

  • C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
  • O₂: 2*16 g/mole= 32 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
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Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

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If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10}   }{116 grams of C_{4}H_{10}}

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