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3 years ago

When This Occurs, Particles Spread Out

Chemistry

1 answer:

Reil [10]3 years ago

8 0

When any sort of solid or liquid becomes a gas.
HOPE THIS HELPS BRAINLIEST

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Based on the visible light spectrum, which of the following has the longest wavelength?

bezimeni [28]

Answer:

I think the answer is D a ray of violet light

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3 years ago

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The name of a binary compound ends with _____. A.-ine B.-ary C.-ed D.-ide

inn [45]

Cant really explain, for say, but its D.-ide

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3 years ago

If 0.0106 g of a gas dissolves in 0.792 L of water at 0.321 atm, what quantity of this gas (in grams) will dissolve at 5.73 atm?

Alex73 [517]

Answer:

0.189 g.

Explanation:

This problem is an application on Henry's law.

Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.

Solubility of the gas ∝ partial pressure

If we have different solubility at different pressures, we can express Henry's law as:

S₁/P₁ = S₂/P₂,

S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm

S₂ = ??? g/L and P₂ = 5.73 atm

So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.

The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.

8 0

3 years ago

The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = $\frac{0.570 g}{122.12 g/mol}=0.004667 mol$

Energy released by 0.004667 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

$Q=C\times \Delta T$

$15,066.8 J=C\times 2.053^oC$

$C=7,338.92 J/^oC$

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = $\frac{2.900 g}{180.16 g/mol}=0.016097 mol$

Energy released by the 0.016097 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$44,749.1 J=7,338.92 J/^oC\times \Delta T'$

$\Delta T'=6.097^oC$

The temperature change from the combustion of the glucose is 6.097°C.

6 0

3 years ago

Help plsssssss... 3C(s) + 2Fe2O3 (5) ► Fe

blondinia [14]

Answer:

single displacement

Explanation:

the Fe was switched out with the C. they are the only ones that switched, making it a single displacement

6 0

3 years ago

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