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3 years ago

A 35 L sample of N2 is at 65 mmHg. What volume would the gas occupy at 74 mmHg? *

Chemistry

1 answer:

kvv77 [185]3 years ago

8 0

Answer:

About 40 L

Explanation:

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If 100 ml of a 0.5 M HCI Solution is diluted with water to 1000ml, what is the new concetration?

Vesna [10]

Answer:

$0.05\ \text{M HCl}$

Explanation:

$V_1$ = Initial volume = 100 mL

$V_2$ = Final volume = 1000 mL

$M_1$ = Initial concentration = 0.5 M

$M_2$ = Final concentration

We have the relation

$\dfrac{M_1}{M_2}=\dfrac{V_2}{V_1}\\\Rightarrow M_2=M_1\dfrac{V_1}{V_2}\\\Rightarrow M_2=0.5\times \dfrac{100}{1000}\\\Rightarrow M_2=0.05\ \text{M HCl}$

The new concentration is $0.05\ \text{M HCl}$ .

3 0

3 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago

The label on a bottle of medicine reads "Each 5 mL teaspoonful contains glucose, 1.87 g; levulose, 1.87 g; and phosphoric a

Alexandra [31]

(a) 43.6 mg; (b) 520 mg

(a) Mass of phosphoric acid (PA) in a dose

Mass of PA = 2 tsp × (21.8 mg PA/1 tsp) = 43.6 mg PA

(b) Mass of PA in the bottle

Step 1. Convert ounces to millilitres

Volume = 4 oz × (30 mL/1 oz) = 120 mL

Step 2. Calculate the mass of PA

Mass of PA = 120 mL × (21.8 mg PA/5 mL) ≈ 520 mg PA

4 0

3 years ago

A substance with a high pOH would likely have which of the following?

Anna [14]

Answer: a low $OH^-$ and low pH.

Explanation:

pH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pOH=-log[OH^-]$

$pOH=log\frac {1}{OH^-}$

Thus as pOH and $OH^-$ are inversely related, a solution having higher pOH will have less amount of $OH^-$ concentration. And a solution having more pOH will have less pH.

Thus a substance with a high pOH would likely have low $OH^-$ concentration and low pH.

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3 years ago

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White-tailed deer eat green plants, acorns, fruits, nuts, and twigs.

ICE Princess25 [194]

D Hetero-tropic herbivore

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3 years ago

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