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dmitriy555 [2]
3 years ago
6

1. 7.000 mmhg = ______ KPa? 2. 10.00 kPa = ________atm? 3. 15.00 kPa______mmHg?

Chemistry
1 answer:
Minchanka [31]3 years ago
7 0
1 kPa = 7.5 mmHg so 7.0 mmHg / 7.5 mmHg x 1 kPa = .93 kPa

101.3 kPa = 1 atm so 10 kPa / 101.3 kPa x 1 atm = .0987 atm

1 kPa = 7.5 mmHg so 15 kPa x 7.5 mmHg / 1 kPa = 112.5 mmHg
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MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

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At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

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\mathbf {i = 1.51 }

3 0
3 years ago
How many moles of N2 in 57.1 g of N2?
SpyIntel [72]

We are given –

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\qquad\pink{\bf\longrightarrow  { Molar \:mass \:of \: N_2:-}  }

\qquad\bf  \twoheadrightarrow 14\times 2

\qquad\bf \twoheadrightarrow   28

\qquad____________________

Now,Let's calculate the number of moles present in 57.1 g of \bf N_2

\qquad\purple{\bf\longrightarrow  { No \:of \:moles = \dfrac{Given \:mass}{Molar\: mass}}}

\qquad\bf   \twoheadrightarrow \dfrac{57.1}{28}

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__________________________________

7 0
2 years ago
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Dominik [7]

Answer:

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NaCl (aq) + AgNO3 (aq)

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8 0
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