Chemistry, 11.07.2019 15:00 joelpimentel

Calculate the mass of ammonia that contains a trillion hydrogen atoms.

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Answer from: DaisyHargrove23

9.4 × 10⁻¹² g of NH₃

Solution:

Step 1: Calculate Number of NH₃ Molecules containing 1 Trillion Hydrogen Atoms:

As,

3 Hydrogen Atoms are present in = 1 Molecule of NH₃

So,

1 × 10¹² Hydrogen Atoms will be present in = X Molecules of NH₃

Solving for X,

X = (1 × 10¹² × 1) ÷ 3

X = 3.33 × 10¹¹ Molecules of NH₃

Step 2: Calculate Moles of NH₃:

As,

Moles = Number of Molecules ÷ 6.022 × 10²³

Putting value,

Moles = 3.33 × 10¹¹ Molecules ÷ 6.022 × 10²³

Moles = 5.53 × 10⁻¹³ mol

Step 3: Calculate Mass of NH₃:

As,

Moles = Mass ÷ M.Mass

Solving for Mass,

Mass = Moles × M.Mass

Putting values,

Mass = 5.53 × 10⁻¹³ mol × 17 g.mol⁻¹

Mass = 9.4 × 10⁻¹² g of NH₃

Answer from: Quest

when [oh-] is 10-5 m, and [h3o+] is 10-9 m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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