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Alenkinab [10]
2 years ago
14

CaCl2 can be melted to produce calcium metal and give off chlorine gas. The equation for this is CaCl2(l) Ca(s) + Cl2(g). If 277

.45 g CaCl2 were melted, how many grams of Ca(s) would be formed? (
Chemistry
2 answers:
Vikki [24]2 years ago
8 0

Answer:

m_{Ca}=99.982gCa

Explanation:

Hello, in this case, considering that the undergoing chemical reaction is:

CaCl_2(l) \rightarrow Ca(s) + Cl_2(g)

As we see a one-to-one molar relationship between calcium chloride and calcium, by stoichiometry and the given molar masses, the mass of calcium that would be formed is:

m_{Ca}=277.45gCaCl_2*\frac{1molCaCl_2}{111gCaCl_2} *\frac{1molCa}{1molCaCl_2} *\frac{40gCa}{1molCa} \\m_{Ca}=99.982gCa

Best regards.

Bess [88]2 years ago
4 0

Answer:

82g

Explanation:

CaCl2 : Ca 111g : 40g

227.45 : xg

xg= (40×227.5)/111

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3 years ago
A 19.0 L helium tank is pressurized to 26.0 atm. When connected to this tank, a balloon will inflate because the pressure inside
vesna_86 [32]

Answer:

The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.

The balloon expands by am additional 475 L.

Explanation:

Assuming Helium behaves like an ideal gas and temperature is constant.

According to Boyle's law for ideal gases, at constant temperature,

P₁V₁ = P₂V₂

P₁ = 26 atm

V₁ = 19.0 L

P₂ = 1 atm (the balloon is said to expand till the pressure matches the pressure of the atmpsphere; and the pressure of the atmosphere is 1 atm)

V₂ = ?

P₁V₁ = P₂V₂

(26 × 19) = 1 × V₂

V₂ = 494 L (it is assumed the balloon never bursts)

The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.

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6 0
3 years ago
What is the volume in liters of 5.25 moles of He gas?​
serious [3.7K]

Answer:

\boxed {\boxed {\sf 117.6 \ L \ He}}

Explanation:

Regardless of the type of gas, 1 mole at standard temperature and pressure (STP) occupies a volume of 22.4 liters. In this case the gas is helium (He).

We can set up a ratio.

\frac { 22.4 \ L \ He}{ 1 \ mol \ He}

Multiply by the given number of moles.

5.25 \ mol \ He *\frac { 22.4 \ L \ He}{ 1 \ mol \ He}

The moles of helium will cancel.

5.25 *\frac { 22.4 \ L \ He}{ 1 }

5.25 * { 22.4 \ L \ He}

Multiply.

117.6 \ L \ He

5.25 moles of helium gas at STP is 117.6 liters of helium.

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2 years ago
A student has a mixture of solid sand and nacl(aq)in a flask to obtain solid nacl from this mixture the student should
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Next, the filtrate should be left with NaCl (aqueous state). To seperate NaCl with the liquid, the student can either do evaporation or crystallization, depending on how pure or fast he/she wants the results to be. Evaporation involves heating the beaker or whatever apparatus under the bunsen burner until all the liquid has evaporated. Then, some white powder should be left, they're NaCl solid. For crystallization, the student should just put the beaker on a room condition environment, and wait. They might have to wait a month or so for the liquid to completely evaporate itself and left with clear and pure NaCl crystals.
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