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Kobotan [32]
2 years ago
15

Where does condensation occur?

Chemistry
1 answer:
Elanso [62]2 years ago
8 0

Answer:

A. Atmosphere

Explanation:

hope this helps

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If you are given 96.0 grams of O2, how many grams of H20 are made?
julsineya [31]

Answer:

10.66  grams

Explanation:

8 0
3 years ago
Which of these methods could be used to separate an insoluble solid from a soluble solid? Melting the mixture and then using a f
I am Lyosha [343]

Answer : The correct option is "Mixing the mixture with water, filtering out the insoluble solid, and then evaporating the water to isolate the soluble solid.

Explanation :

"The given mixture contains a soluble solid and an insoluble solid.

When water is added to this mixture, the soluble solid will get dissolved in water whereas the insoluble solid will remain as it is.

This will give us a heterogeneous mixture where we can see 2 distinct phases , one containing the dissolved solid and the other having undissolved solid.

In order to separate these 2 phases, we can use a simple filtration method.

When the mixture is filtered, we will get solution having soluble solid and the insoluble solid will remain on the filter paper.

This will separate the 2 solids.

To regenerate the dissolved solid, we can evaporate the water by heating it. This will give us back our original soluble solid.

Therefore, the only option that is consistent with this method is "Mixing the mixture with water, filtering out the insoluble solid, and then evaporating the water to isolate the soluble solid"

8 0
3 years ago
Butane (C4 H10(g), mc031-1.jpgHf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , mc031-2.jpgHf = –393.5 kJ/
ankoles [38]

The balanced chemical equation for the combustion of butane is:

2C_{4}H_{10}(g) +13 O_{2}(g)-->8CO_{2}(g)+10H_{2}O(g)

ΔH_{reaction}^{0} = Σn_{products}ΔH_{f}^{0}_{(products)}-Σn_{reactants}ΔH_{f}^{0}_{(reactants)}

                         = [{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]

                      = -5315 kJ/mol

Calculating the enthalpy of combustion per mole of butane:

1mol C_{4}H_{10}*( \frac{-5315kJ}{2mol C_{4}H_{10} })=-2657.5 \frac{kJ}{molC_{4}H_{10}}

Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol

Correct answer: -2657.5 kJ/mol

6 0
2 years ago
The table shows the amount of radioactive element remaining in a sample over a period of time.
Lunna [17]

Answer:

1. The half-life is 22 years.

2. 132 years

Explanation:

1. Determination of the the half-life.

The half-life of an element is the time taken for half the element to decay.

From the table given above, the original amount of the element 45 g. If we divide 45 by 2, we'll have 22.5 g as half the original amount of element.

Now, the time taken to obtain 22.5 g as shown from the table is 22 years.

Thus, the half-life the element is 22 years.

2. Determination of the time.

Original amount (N₀) = 308 g

Amount remaining (N) = 4.8125 g

Time (t) =?

Next, we shall the number of half-lives that has elapsed. This can be obtained as follow:

Original amount (N₀) = 308 g

Amount remaining (N) = 4.8125 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

4.8125 = 1/2ⁿ × 308

Cross multiply

4.8125 × 2ⁿ = 308

Divide both side by 4.8125

2ⁿ = 308 / 4.8125

2ⁿ = 64

Express 64 in index form with 2 as the base.

2ⁿ = 2⁶

n = 6

Thus, 6 half-lives has elapsed.

Finally, we shall determine the time. This can be obtained as follow:

Number of half-lives (n) = 6

Half-life (t½) = 22 years

Time (t) =?

n = t / t½

6= t / 22 years

Cross multiply

t = 6 ×22

t = 132 years.

Thus, the time taken is 132 years.

6 0
2 years ago
which material should you use so that the area is cool in terms of temperture O asphalt O red bricks O concrete O soil O
Sindrei [870]

Answer:

red bricks

Explanation:

3 0
3 years ago
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