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belka [17]
3 years ago
14

Place the

Chemistry
1 answer:
9966 [12]3 years ago
3 0
Potassium, aluminum, silicon, magnesium neon
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What is the percent composition of Br in CuBr3?
tekilochka [14]

Answer:

about 79% (79.04369332 to be exact)

Explanation:

Percent composition=(Molar mass of element x amount of it)/Molar mass of compound x 100

Br= 3 x 79.9/303.25 x100=79.04369332

6 0
3 years ago
What might go wrong while heating a substance<br> with a candle?
GaryK [48]

Answer:

the wax melts

Explanation:

because i don't know

7 0
2 years ago
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What is the mass of air in room
tigry1 [53]

Answer:

the mass of the air in the room is 4.96512 kg ( in 0°C)

3 0
3 years ago
A rigid 3.80 L sealed vessel contains 0.650 mol Ne, 0.321 mol Kr, and 0.190 mol Xe. Find the density of the mixture in g/L.
Eddi Din [679]

Answer:

17.09g/L

Explanation:

Density = total mass of elements/ volume

We need to find the mass of each mixture constituents using their molar mass:

mole = mass/molar mass

For Neon (Ne) which contains 0.650mol;

0.650 = mass/20.18

mass = 0.650 × 20.18

mass = 13.12g

For Krypton (Kr) which contains 0.321mol;

0.321 = mass/83.79

mass = 0.321 × 83.79

mass = 26.89g

For Xenon (Xe) which contains 0.190mol;

0.190 = mass/131.3

mass = 0.190 × 131.3

mass = 24.95g

Total mass = 13.12g + 26.89g + 24.95g = 64.96g

Density = total mass / volume

Density = 64.96g / 3.80L

Density of the mixture = 17.09g/L

7 0
3 years ago
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
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