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melamori03 [73]
3 years ago
11

if .00327 g of gas dissolves in .376 L of water at 876 torr what quantity of this gas will dissolve 759 torr ?

Chemistry
1 answer:
MAXImum [283]3 years ago
4 0

Answer:- Solubility of the gas at 759 torr is 0.00753 g/L.

Solution:- From given data, 0.00327 g of gas is soluble in 0.376 L of water at 876 torr.

Solubility of gas at 876 torr pressure is = 0.00327g/0.376L = 0.00869 g/L

Solubility of gases is directly proportional to the pressure. It means, grater is the pressure, more is the solubility of gases.

So, 0.00869/876 = X/759

Where, X is the solubility of the gas at 759 torr.

X = 0.00869(759)/876

X = 0.00753 g/L

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Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,:

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where,

P_1 = initial pressure of gas = 101.3 kPa

P_2 = final pressure of gas = 94.6 kPa

V_1 = initial volume of gas = 20.0 ml

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T_1 = initial temperature of gas = 297K

T_2 = final temperature of gas = 283K

Now put all the given values in the above equation, we get the final volume of gas.

V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

V_2=20.4ml

Thus the correct numerical setup for calculating the new volume is \frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

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Explanation:

<em>Hi</em><em> </em><em>there</em><em>!</em><em>!</em>

<em>you</em><em> </em><em>asked</em><em> </em><em>to</em><em> </em><em>multiply</em><em> </em><em>these</em><em> </em><em>all</em><em> </em><em>right</em><em>,</em>

<em>you</em><em> </em><em>can</em><em> </em><em>simply</em><em> </em><em>multiply</em><em> </em><em>it</em><em> </em><em>,</em>

<em>=</em><em>3</em><em>cm</em><em> </em><em>×</em><em> </em><em>4</em><em> </em><em>cm</em><em> </em><em>×</em><em> </em><em>1</em><em>cm</em>

<em>=</em><em> </em><em>1</em><em>2</em><em>cm</em><em>^</em><em>2</em><em>×</em><em>1</em><em>cm</em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>4</em><em>×</em><em>3</em><em>=</em><em>1</em><em>2</em><em>)</em>

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<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em>

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3 years ago
(a) a 0.2 m potassium hydroxide solution is titrated with a 0.1 m nitric acid solution. (i) balanced equation: (ii)what would be
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\text{KOH} (aq) + \text{HNO}_3 (aq) \to \text{KNO}_3 (aq) + \text{H}_2\text{O} (l)

The solution shall contain only \text{KNO}_3 (aq) (and water) at the equivalence point. Both potassium hydroxide and nitric acid exist as strong electrolytes. As a result,  \text{KNO}_3 (aq), the salt derived from a reaction between the two species would undergo hydrolysis of a negligible extent. This neutralization reaction therefore be neutral at the equilibrium point.

The question states that the solution is "titrated with a ... nitric acid solution" indicating that \text{HNO}_3 is added to the initially-basic solution. PH value of the solution would keep decreasing as the volume of the acid added increases. The final solution would be acidic as it contains not only water and \text{KNO}_3 (aq), but some \text{HNO}_3 as well. Bromothymol blue would therefore demonstrates a yellow color, the color it present in an acidic solution, at the end of the titration.

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cricket20 [7]

Answer:

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Explanation:

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