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3 years ago

6

How much do 7.83 × 10^22 molecules of water weigh? Answer in units of g.

1 answer:

____ [38]3 years ago

4 0

<span>The answer I think is, 7.83E22.</span>

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Which of the following is not a common way for a lake to form?

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How often do girls normally experience menstruation a.once every 30 days b. once every 29 c.once every 28 d.once every 27

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3 years ago

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0

3 years ago

What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)

lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

$\Delta T = i \times Kf \times m = 1 \times 6.90 \°C/m \times 1.324m = 9.14 \°C$

where,

i: van 't Hoff factor (1 for non-electrolytes)
Kf: cryoscopic constant
m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

$T = 80.26 \° C - 9.14 \° C = 71.12 \° C$

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

3 0

3 years ago