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3 years ago

A sample of gas occupies 17 mL at -122°C. What volume does the sample occupy at 70°C?

Chemistry

1 answer:

aev [14]3 years ago

7 0

Answer: 38.6 ml

Explanation:

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$ (At constant pressure and number of moles)

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$V_1$ = initial volume = 17 ml

$V_2$ = final volume = ?

$T_1$ = initial temperature = $-122^0C=273+(-122)=151K$

$T_2$ =final temperature= $70^0C= 70+273=343K$

$\frac{17}{151}=\frac{V_2}{343}$

$V_2=38.6ml$

Thus final volume is 38.6 ml at 70 $^0C$

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andrezito [222]

Answer:

c.The value of T at Recording Station N would be greater than T at Recording Station S.

Explanation:

We can deduce from the information provided that the value of T at recording station N would be greater than T at recording station S.

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Just like any other physical quantity, speed of a substance is directly related to time.
The larger the time taken the farther the distance it will cover assuming the velocity is constant.
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3 years ago

Arrange the following aqueous solutions in order of decreasing freezing point: 0.10 m Na3PO4, 0.35 m NaCl, 0.20 m MgCl2, 0.15 m

GarryVolchara [31]

Answer: 0.35 m $NaCl$ > 0.20 m $MgCl_2$ >0.10 m $Na3PO_4$ > 0.15 m $CH_3COOH$ > 0.15 m $C_6H_{12}O_6$

Explanation:

Depression in freezing point:

$T_f^0-T_f=i\times k_b\times m$

where,

$T_f$ = freezing point of solution

$T^o_f$ = freezing point of solvent

$k_f$ = freezing point constant

m = molality

1. For 0.10 m $Na3PO_4$

$Na_3PO_4\rightarrow 3Na^++PO_4^{3-}$

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be $4\times 0.10=0.40$

2. For 0.35 m $NaCl$

$NaCl\rightarrow Na^++Cl^-$

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be $2\times 0.35=0.70$

3. For 0.20 m $MgCl_2$

$MgCl_2\rightarrow Mg^{2+}+2Cl^-$

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be $3\times 0.20=0.60$

4. For 0.15 m $C_6H_{12}O_6$

, i= 1 as it is a non electrolyte and does not dissociate to give ions.

5. For 0.15 m $CH_3COOH$

$CH_3COOH\rightarrow CH_3COO^-+H^+$

, i= 2 as it is a electrolyte and dissociate to give 2 ions and thus have $2\times 0.15=0.30$

As concentration is highest for 0.35 m $NaCl$ , freezing point depression will be highest and thus has lowest freezing point. As concentration is lowest for 0.15 m $C_6H_{12}O_6$ , freezing point depression will be lowest and thus has highest freezing point

8 0

3 years ago

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artcher [175]

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Explanation:

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3 years ago

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nikklg [1K]

Answer:

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