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BabaBlast [244]
2 years ago
8

Please help me on #2 giving BRAINLIEST!

Chemistry
1 answer:
Amiraneli [1.4K]2 years ago
3 0

Answer:

basic

Explanation:

because H is hydrogen

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Please help. I will post another sheet because I need help, thanks!
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The vapor pressure of pure water at 296 K is 2778.5 Pa. The vapor forms an ideal gas. 1) In some oil, the equilibrium concentrat
Murljashka [212]

Explanation:

It is given that vapor pressure of pure water at 296 K is 2778.5 Pa.These vapors will result in the formation of an ideal gas.

Now, as water is covered with oil and contains only 1% molecules of water. Hence, the vapor pressure of this mixture will also be equal to the vapor pressure of pure water.

So, vapor pressure of mixture = 1% vapor pressure of pure water

Therefore, \text{(Vapor pressure)}_{mixture} = \frac{1}{100} \times 2778.5 Pa

                                                 = 27.785 Pa

Thus, we can conclude that the equilibrium vapor pressure of water above the oil layer is 27.785 Pa.

3 0
3 years ago
What is the final product of the electron transport chain
Anna71 [15]

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3 0
2 years ago
What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actua
notka56 [123]

Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Na_2S

Given mass = 15.5 g

Molar mass of Na_2S = 78.0452 g/mol

<u>Moles of Na_2S = 15.5 g / 78.0452 g/mol = 0.1986 moles</u>

Given: For CuSO_4

Given mass = 12.1 g

Molar mass of CuSO_4 = 159.609 g/mol

<u>Moles of CuSO_4 = 12.1 g / 159.609 g/mol = 0.0758 moles</u>

According to the given reaction:

Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS

1 mole of Na_2S react with 1 mole of CuSO_4

0.1986 mole of Na_2S react with 0.1986 mole of CuSO_4

Available moles of CuSO_4 = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, CuSO_4 is limiting reagent. (0.0758 < 0.1986)

The formation of the product is governed by the limiting reagent. So,

1 mole of CuSO_4 gives 1 mole of CuS

0.0758 mole of CuSO_4 gives 0.0758 mole of CuS

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

Theoretical yield = 7.2473 g

Given experimental yield = 3.05 g

% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

<u>Option B is correct.</u>

6 0
3 years ago
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