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3 years ago

A reasonable pka for a weak base is: 10.3 0.9 5.8 7.4 13.1

Chemistry

1 answer:

andriy [413]3 years ago

7 0

Answer:

0.9

Explanation:

The pka represents the force by which the molecules need to dissociate for the acids ,

Hence , lower the pka stronger will be the acid and that therefore will dissociate completely and vice versa , for a weak acid higher the pka .

And in case of a base , it will be completely reversed , lower pKa , weaker base ,

and higher pKa , stronger base .

From the data of the question ,

0.9 is the lowest value of the pKa , hence , weakest base .

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bulgar [2K]

Answer:

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Explanation:

The density formula is:

$d=\frac{m}{v}$

Rearrange the formula for m, the mass. Multiply both sides of the equation by v.

$d*v=\frac{m}{v}*v$

$d*v=m$

The mass of the gold nugget can be found by multiplying the density and volume. The density is 19.3 grams per cubic centimeter and the volume is 0.87 cubic centimeters.

$d= 19.3 g/cm^3\\v-0.87 cm^3$

Substitute the values into the formula.

$m=d*v$

$m= 19.3 g/cm^3*0.87 cm^3$

Multiply. Note that the cubic centimeters, or cm³ will cancel each other out.

$m=19.3 g*0.87$

$m=16.791 g$

The mass of the gold nugget is 16.791 grams.

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3 years ago

A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h

NikAS [45]

Nitrogen combine with hydrogen to produce ammonia $\text{NH}_3$ at a $1:3:2$ ratio:

$\text{N}_2 \; (g) + 3 \; \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)$

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. $3 \; \text{mol}$ of hydrogen gas would have been consumed while $2 \; \text{mol}$ of ammonia would have been produced. The final mixture would therefore contain

$17 \; \text{mol}$ of $\text{H}_2 \; (g)$ and
$2 \; \text{mol}$ of $\text{NH}_3 \; (g)$

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$\begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.926 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.723 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{NH}_3) &= & n \cdot R \cdot T / V \\ & = & (2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 2.037 \times 10^{2} \; \text{kPa} \end{array}$

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