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alexgriva [62]
3 years ago
14

In reality, reactants don't have to react in perfect whole-numbers of moles. In a two-reactant synthesis reaction, usually one

Chemistry
1 answer:
Fofino [41]3 years ago
5 0

Answer:

srry i cant help u with this im so confuse like u just put all of it on there how would anybody answer tht!

?!Explanation:

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3 H2 (g) + N2 (g) 2 NH3 (g)
ale4655 [162]

Answer:

Mass = 0.697 g

Explanation:

Given data:

Volume of hydrogen = 1.36 L

Mass of ammonia produced = ?

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Solution:

Chemical equation:

3H₂ + N₂       →      2NH₃

First of all we will calculate the number of moles of hydrogen:

PV  = nRT

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

1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K

1.36 atm.L = n × 22.43 atm.L/mol

n = 1.36 atm.L / 22.43 atm.L/mol

n = 0.061 mol

Now we will compare the moles of hydrogen and ammonia:

                 H₂         :          NH₃

                  3          :            2

                0.061     :         2/3×0.061 = 0.041

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.041 mol × 17 g/mol

Mass = 0.697 g

4 0
3 years ago
Where was all the hydrogen in the universe formed?
Maslowich
The low-mass elements, hydrogen and helium, were produced in the hot, dense conditions of the birth of the universe itself. The birth, life, and death of a star is described in terms of nuclear reactions. The chemical elements that make up the matter we observe throughout the universe were created in these reactions.
5 0
3 years ago
in example 5.11 of the text the molar volume of n2 at STP is given as 22.42 L/mol how is this number calculatd how does the mola
Valentin [98]

Answer:

V = 22.42 L/mol

N₂ and H₂ Same molar Volume at STP

Explanation:

Data Given:

molar volume of N₂ at STP = 22.42 L/mol

Calculation of molar volume of N₂ at STP  = ?

Comparison of molar volume of H₂ and N₂ = ?

Solution:

Molar Volume of Gas:

The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol

Molar volume can be calculated by using ideal gas formula  

                               PV = nRT

Rearrange the equation for Volume

                            V = nRT / P . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

Standard values

P = 1 atm

T = 273 K

n = 1 mole

R = 0.08206 L.atm / mol. K

Now put the value in formula (1) to calculate volume for 1 mole of N₂

                   V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm

                   V = 22.42 L/mol

Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.

6 0
3 years ago
The heat of fusion for water is 80. cal/g. How many calories of heat are needed to melt a 35 g ice cube that has a temperature o
HACTEHA [7]

Answer: 2800 calories

Explanation:

Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.

Amount of heat required to fuse 1 gram of water = 80 cal

Mass of ice given = 35 gram

Heat required to fuse 1 g of ice at 0^0C = 80 cal

Thus Heat required to fuse 35 g of ice =\frac{80}{1}\times 35=2800cal

Thus 2800 calories of energy is required to melt 35 g ice cube

6 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
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