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3 years ago

If the density of nitrogen gas at a certain pressure and temperature is 1.20 g/l, how many moles of nitrogen gas are in 15.0 l

1 answer:

8 0

Density of nitrogen= 1.20 g/l
no. of grams of nitrogen=m= (1.20g/l)(15.0l)= 18 g
molar mass of nitrogen=M= (14*2)= 28 g/mol

no. of moles of nitrogen=m/M=18/28= 0.643 moles

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Answer:

Mass = 0.158 g

Explanation:

Formula used,

P V = n R T

Or,

n = P V / R T

Putting values,

n = 0.948 atm . 0.025 L / 0.0821 L.atm.K⁻¹.mol⁻¹ . 291.45

n = 0.00099 mol

Note: we have changed pressure from mmHg to atm, volume from mL to L and temperature from C to K)

Also,

Mass = n . Molecular Mass

Mass = 0.00099 mol × 159.808 g/mol

Mass = 0.158 g

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What is the pH of a KOH solution that has [H ] = 1. 87 × 10–13 M? What is the pOH of a KOH solution that has [OH− ] = 5. 81 × 10

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pH is the negative log of the hydrogen ion concentration and pOH is the negative log of the hydroxide ion concentration.

The relation between the pH and pOH can be given as, $\rm pOH = 14 - pH$

The pH of KOH can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the first case, the concentration of the KOH is $1. 87 \times 10^{-13}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times 10^{-13}\;\rm M ]\\\\&= 12.73\end{aligned}$

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pOH of KOH can be calculated by the formula,

$\rm pOH = \rm -log [OH^{-}]$

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Substituting value in the equation:

$\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm M ]\\\\&= 2.24 \end{aligned}$

Hence, the pOH of KOH is 2.24

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The pH of NaCl can be calculated by the formula,

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Substituting values in the equation:

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Therefore, KOH is basic and NaCl is approximately neutral.

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2 years ago

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<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
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In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

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3 years ago

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