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3 years ago

How many grams of Nacl would you need to prepare 200 ml of a 5 M SOLUTION. ?

Chemistry

1 answer:

vazorg [7]3 years ago

8 0

Answer:

58.44 g of NaCl are needed.

Explanation:

Given data:

Mass of NaCl needed = ?

Volume of solution = 200 mL (200/1000 =0.2 L)

Molarity of solution = 5 M

Solution:

We will solve this problem through molarity formula.

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Now we will put the values.

5 M = moles of solute / 0.2 L

Moles of solute = 5 mol/L × 0.2 L

Moles of solute = 1 mol

Mass of sodium chloride:

Mass = number of moles × molar mass

Mass = 1 mol × 58.44 g/mol

Mass = 58.44 g

Thus, 58.44 g of NaCl needed.

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Atmospheric pressure at sea level is equal to a column of 760 mm Hg. Oxygen makes up 21 percent of the atmosphere by volume. The

Neporo4naja [7]

<u>Answer:</u> The partial pressure of oxygen is 160 mmHg

<u>Explanation:</u>

We are given:

Percent of oxygen in air = 21 %

Mole fraction of oxygen in air = $\frac{21}{100}=0.21$

To calculate the partial pressure of oxygen, we use the equation given by Raoult's law, which is:

$p_{O_2}=p_T\times \chi_{O_2}$

where,

$p_{O_2}$ = partial pressure of oxygen = ?

$p_T$ = total pressure of air = 760 mmHg

$\chi_{O_2}$ = mole fraction of oxygen = 0.21

Putting values in above equation, we get:

$p_{O_2}=760mmHg\times 0.21\\\\p_{O_2}=160mmHg$

Hence, the partial pressure of oxygen is 160 mmHg

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3 years ago

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nikitadnepr [17]

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2 years ago

In ionic bonding, atoms__.

Nutka1998 [239]

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2 years ago

A sample of pure NH4HS is placed in a sealed 2.0-L container and heated to 550 K at which the equilibrium constant is 3.5 x 10-3

Zolol [24]

Answer:

The mass of $NH_{3}$ in the container is 2.074 gram

Explanation:

Given:

Volume of $NH_{4} HS$ $V = 2$ lit

Equilibrium constant $k _{eq} = 3.5 \times 10^{-3}$

The reaction in which $NH_{3}$ is produced

$NH_{4} HS$ ⇄ $NH_{3} + H_{2}S$

Here equal moles of $NH_{3}$ and $H_{2}S$ is formed.

From the formula of equilibrium constant,

$k_{eq} = (NH_{3})(H_{2}S )$

$x^{2} = 3.5 \times 10^{-3}$

$x = 0.061$ M

Above value shows,

$NH_{3} = 0.061$ $\frac{moles}{L}$

So in 2 L no. moles of $NH_{3}$ = $0.061 \times 2 = 0.122$ moles.

So mass of 0.122 mole of $NH_{3}$ is = $0.122 \times 17 = 2.074$ g

Therefore, the mass of $NH_{3}$ in the container is 2.074 gram

8 0

3 years ago

Identify which sets of quantum numbers are valid for an electron. Note that each set is ordered (n,l,ml,ms).

Fynjy0 [20]

Answer:

The answer to your question is:

Explanation:

Quantum numbers have the numbers

n = from 1 to 7

l = s = 0 p = 1 d = 2 f= 3

m = ± l

s = ±1/2

Group of answer choice.

A) 2, -1. 1, -1/2 The option is incorrect "l" could not be negative

B) 2, 1, 0 , 1/2 The option is posible

C) 3, 2, 1, -1 This option is incorrect "s" can not value -1

D) 4, 3, 2, 1/2 This option is correct

E) 4, 3, -4, 1/2 This option is incorrect "m2 can not value -4

F) 1, 1, 0, 1/2 This option is incorrect, the second number is not posible

G) 1, 1, 0, -1/2 This option is incorrect the second number could not be 1.

H) 3, 0, 0, 1/2 This option is incorrect "l" can not value 0

I) 0, 2, 0, 1/2 This option is incorrect, n can not value 0

J) 3, 2, 2, 1/2 This option is correct

K) 1, 2, 0, -1/2 This option is incorrect "l" can not value 2.

3 0

3 years ago