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3 years ago

Calculate the molarity of 0.700 mol of na2s in 1.05 l of solution.

1 answer:

6 0

Molarity is the amount of solute molecule (in moles) per 1L of solvent. In this case, the solute is 0.7mol Na2S and the solvent volume is 1.05L. Since the unit in this problem is already mol and L then you don't need to do any conversion of the units. The calculation would be:

molarity = mol of solute / (1L/ volume of solvent)
molarity = 0.7 mol/ (1L/ 1.05L)= 0.67M

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A sample of ammonia ^NH3h gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool. If the total pressu

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Answer : The partial pressure of $N_2$ and $H_2$ is, 216.5 mmHg and 649.5 mmHg

Explanation :

According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.

Formula used :

$p_i=X_i\times p_T$

$X_i=\frac{n_i}{n_T}$

So,

$p_i=\frac{n_i}{n_T}\times p_T$

where,

$p_i$ = partial pressure of gas

$X_i$ = mole fraction of gas

$p_T$ = total pressure of gas

$n_i$ = moles of gas

$n_T$ = total moles of gas

The balanced decomposition of ammonia reaction will be:

$2NH_3\rightarrow N_2+3H_2$

Now we have to determine the partial pressure of $N_2$ and $H_2$

$p_{N_2}=\frac{n_{N_2}}{n_T}\times p_T$

Given:

$n_{N_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg$

$p_{N_2}=\frac{1}{4}\times (866mmHg)=216.5mmHg$

and,

$p_{H_2}=\frac{n_{H_2}}{n_T}\times p_T$

Given:

$n_{H_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg$

$p_{H_2}=\frac{3}{4}\times (866mmHg)=649.5mmHg$

Thus, the partial pressure of $N_2$ and $H_2$ is, 216.5 mmHg and 649.5 mmHg

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It’s ionic bond based on electrons gain/loss.

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3 years ago

A person with a body temperature of 37°C holds an ice cube with a temperature of 0°C in a room where the air temperature is 20.°

Arlecino [84]

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4 years ago

Someone answer the limited regent...

OleMash [197]

Answer:

20 g Ag

General Formulas and Concepts:

Chemistry - Stoichiometry

Using Dimensional Analysis

Chemistry - Atomic Structure

Reading a Periodic Table

Explanation:

Step 1: Define

[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)

[Given] 10 g Cu

Step 2: Identify Conversions

[RxN] 1 mol Cu = 1 mol Ag

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Molar Mass of Ag - 197.87 g/mol

Step 3: Stoichiometry

$10 \ g \ Cu(\frac{1 \ mol \ Cu}{63.55 \ g \ Cu})(\frac{1 \ mol \ Ag}{1 \ mol \ Cu} )(\frac{197.87 \ g \ Ag}{1 \ mol \ Ag} )$ = 16.974 g Ag

Step 4: Check

We are given 1 sig fig. Follow sig fig rules and round.

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3 years ago