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kirza4 [7]
3 years ago
15

Mass of 12 helium atom​

Chemistry
1 answer:
atroni [7]3 years ago
4 0
48.024 grams is the correct mass, because helium has an atomic mass of 4.002g, so all you have do is multiply 12 by 4.002 to get your mass of 12 helium atoms.
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QUESTION 18
Levart [38]

Answer:

-43.3 °C

Explanation:

To find the temperature, you need to use the Ideal Gas Law equation. The equation looks like this:

PV = nRT

In this formula,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas Law constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

By plugging the given values into the equation and simplifying, you can find the temperature. After you get a temperature, you need to convert it into Celsius.

P = 2.88 atm                              R = 0.08206 atm*L/mol*K

V = 3.76 L                                   T = ? K

n = 0.574 moles

PV = nRT

(2.88 atm)(3.76 L) = (0.574 moles)(0.08206 atm*L/mol*K)T

10.8288 = (0.04710244)T

230. K = T

Kelvin - 273.15 = Celsius

230 K - 273.15 = -43.3 °C

8 0
2 years ago
Table salt is considered a(n) ______________ because it conducts electricity in water. A) electrolyte B) electrolysis C) nonelec
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A) electrolyte the others either describe a process, the opposite, or a whole different thing
4 0
4 years ago
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What are the boiling points and freezing points (in oC) of a solution of 50.3 g of I2 in 350 g of chloroform? The kb = 3.63 oC/m
patriot [66]

Answer:

Boiling point: 63.3°C

Freezing point: -66.2°C.

Explanation:

The boiling point of a solution increases regard to boiling point of the pure solvent. In the same way, freezing point decreases regard to pure solvent. The equations are:

<em>Boiling point increasing:</em>

ΔT = kb*m*i

<em>Freezing point depression:</em>

ΔT = kf*m*i

ΔT are the °C that change boiling or freezing point.

m is molality of the solution (moles / kg)

And i is Van't Hoff factor (1 for I₂ in chloroform)

Molality of 50.3g of I₂ in 350g of chloroform is:

50.3g * (1mol / 253.8g) = 0.198 moles in 350g = 0.350kg:

0.198 moles / 0.350kg = 0.566m

Replacing:

<em>Boiling point:</em>

ΔT = kb*m*i

ΔT = 3.63°C/m*0.566m*1

ΔT = 2.1°C

As boiling point of pure substance is 61.2°C, boiling point of the solution is:

61.2°C + 2.1°C = 63.3°C

<em>Freezing point:</em>

ΔT = kf*m*i

ΔT = 4.70°C/m*0.566m*1

ΔT = 2.7°C

As freezing point is -63.5°C, the freezing point of the solution is:

-63.5°C - 2.7°C = -66.2°C

7 0
3 years ago
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