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2 years ago

12 moles of sodium into grams

Chemistry

1 answer:

sladkih [1.3K]2 years ago

7 0

Answer:

12 moles sodium= 275.87724000000003 grams

Explanation:

1 moles sodium = 0.043497607849056.

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What kind of chemical reaction does the chemical equation sodium + chlorine → sodium chloride represent?

tresset_1 [31]

This would represent a synthesis reaction

7 0

2 years ago

Read 2 more answers

How would you prepare 10 ml of a 0.25% m/v hcl solution if 1% m/v hcl was available? how much 1% m/v hcl is needed? how much dis

Fed [463]

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

M1V1 = M2V2

1% x V1 = 0.25% x 10 mL

V1 =2.5 mL

Therefore, you will need to have 2.5 mL of the 1% HCl solution and 7.5 mL of distilled water. In mixing the two liquids, you should remember that the order of mixing would be acid to water. So, you use a 10 mL volumetric flask . Put small amount of distilled water and add the 2.5 mL of HCl solution. Lastly, dilute with distilled water up to the 10 mL mark.

8 0

3 years ago

A chemist determined by measurements that 0.030 moles of barium participated in a chemical reaction. Calculate the mass of bariu

schepotkina [342]

Answer: 4.1 g of barium precipitated.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}$

Given : moles of barium = 0.030

Molar mass of barium = 137 g/mol

$0.030=\frac{x}{137}$

x= 4.1 g

Thus there are 4.1 g of barium that precipitated.

3 0

2 years ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

Answer: The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

Explanation:

The given chemical equations follows:

Equation 1: $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

Equation 2: $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$