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creativ13 [48]
2 years ago
5

A 2.005 sample of xCuSO4 yH2O was heated to drive off water from the hydrated salt. The anhydrous salt weighed 1.283 grams.

Chemistry
1 answer:
sweet [91]2 years ago
7 0

Answer:

yo amam fat and can fast both sides of the bed

Explanation:

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If two gases, A and B, in separate 1 liter containers exert
babunello [35]

Answer:

5Atm

Explanation:

I just guess and it’s right

5 0
3 years ago
What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)
lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

\Delta T = i \times Kf \times m =   1 \times 6.90 \°C/m  \times 1.324m = 9.14  \°C

where,

  • i: van 't Hoff factor (1 for non-electrolytes)
  • Kf: cryoscopic constant
  • m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

T = 80.26 \° C - 9.14 \° C = 71.12 \° C

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

3 0
2 years ago
Hydrocarbons separated by fractional distillation of petroleum can be cracked to make
lana66690 [7]

Answer:

c.hg cannot be cracked for fractional distillation as there is only one of each

Explanation:

8 0
3 years ago
The specific heat capacity of gold is 0.128 J/g C. How much would be needed to warm 250.0 grams of gold from 25.0 C to 100.0 C.
madam [21]

Answer:

3,200 joules

Explanation:

q = mcΔT = (250.0 g)(0.128 j/g°C)(100°C -25°C) = 3,200 joules

7 0
3 years ago
The "air bags" that are currently installed in automobiles to prevent injuries in the event of a crash are equipped with sodium
stellarik [79]

Answer:

0.0177 L of nitrogen will be produced

Explanation:

The decomposition reaction of sodium azide will be:

2NaN_{3}(s)--->2Na(s)+3N_{2}(g)

As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas

The molecular weight of sodium azide = 65 g/mol

The mass of sodium azide used = 100 g

The moles of sodium azide used = \frac{mass}{molarmass}=\frac{100}{65}=1.54mol

so 1.54 moles of sodium azide will give = \frac{3X1.54}{2}=2.31mol

the volume will be calculated using ideal gas equation

PV=nRT

Where

P = Pressure = 1.00 atm

V = ?

n = moles = 2.31 mol

R = 0.0821 L atm / mol K

T = 25 °C = 298.15 K

Volume = \frac{P}{nRT}=\frac{1}{2.31X0.0821X298.15}=0.0177L

3 0
3 years ago
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