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3 years ago

How many grams of NH4Cl (ammonium chloride) should dissolve in 50mL of water at 20 degrees Celsius?

Chemistry

1 answer:

Anna71 [15]3 years ago

3 0

Answer:

about 19 or 20 g

Explanation:

To do this, is neccesary to watch a solubility curve of this compound. This is the only way that you can know how many grams are neccesary to dissolve this compound in 50 mL of water to a given temperature.

Now, if you watched the attached graph, you can see the solubility curve of many compounds in 100 g of water (or 100 mL of water). So, to know how many do you need in 50 mL, it's just the half.

So watching the curve, you can see that at 20 °C, we simply need between 35 g and 40 g. Let's just say we need 38 grams of NH4Cl to be dissolved in 100 mL of water.

So, in 50 mL, it's just the half. So, we only need 19 g or 20 g of NH4Cl at 20 °C, to dissolve this compound in water.

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5. A gas has a pressure of 310 kPa at 237 degrees C. What will its pressure be at 23 degrees C?)

slavikrds [6]

Answer:

The pressure of the gas at 23 C is 179.92 kPa.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

In short, when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T} =k$

Studying two states, one initial 1 and the other final 2, it is satisfied:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

P1= 310 kPa
T1= 237 C= 510 K (being 0 C= 273 K)
P2= ?
T2= 23 C= 296 K

Replacing:

$\frac{310 kPa}{510 K} =\frac{P2}{296 K}$

Solving:

$P2=296 K*\frac{310 kPa}{510 K}$

P2= 179.92 kPa

The pressure of the gas at 23 C is 179.92 kPa.

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