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3 years ago

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Chemistry

2 answers:

Galina-37 [17]3 years ago

8 0

Answer:

Relative humidity

Explanation:

A psychrometer measures the relative humidity in the atmosphere with the help of two thermometers:

1) A dry bulb thermometer:

It is used to measure the temperature by being exposed to the air.

2) A wet bulb thermometer:

It measures temperature by having the bulb dipped in a liquid.

Vesna [10]3 years ago

6 0

Answer:

The answer is Relative Humidity.

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How many moles of tungsten carbide will be produced if plenty of tungsten is reacted with 1.25 moles of carbon?

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<h3>1.25 moles of tungsten carbide</h3>

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Read 2 more answers

Compared to the freezing point of 1.0 M KCl(aq) at standard pressure, the freezing point of 1.0 M CaCl2(aq) at standard pressure

Galina-37 [17]

Answer : The correct option is, (1) lower

Explanation :

Formula used for lowering in freezing point :

$\Delta T_f=i\times k_f\times m$

where,

$\DeltaT_b$ = change in freezing point

$k_b$ = freezing point constant

m = molality

i = Van't Hoff factor

According to the question, we conclude that the molality of the given solutions are the same. So, the freezing point depends only on the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) The dissociation of $KCl$ will be,

$KCl\rightarrow K^++Cl^-$

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(b) The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^-$

So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3

The freezing point depends only on the Van't Hoff factor. That means higher the Van't Hoff factor, lower will be the freezing point and vice-versa.

Thus, compared to the freezing point of 1.0 M $KCl(aq)$ at standard pressure, the freezing point of 1.0 M $CaCl_2(aq)$ at standard pressure is lower.

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3 years ago