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MakcuM [25]
3 years ago
14

What is the volume of 4.5 moles of H2 at STP?

Chemistry
1 answer:
lions [1.4K]3 years ago
6 0
1.0 mole ---------- 22.4 at STP
4.5 moles --------- ?

4.5 * 22.4 / 1.0

= 100.8 L of H₂
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At 27.00 °c a gas has a volume of 6.00 L . What will the volume be at 150.0 °c
mojhsa [17]
T₁ = 27°C = 27 + 273 = 300K,  V₁ = 6 L,

T₂ = 150°C = 150 + 273 = 423K,  V₂ = ?,

By Charles' Law:      V₁/T₁= V₂/T₂

                                 6/300 = V₂/423

                                  423*(6/300) = V₂

                                  8.46 = V₂

           Volume at 150°C =8.46 L.
3 0
3 years ago
At this pressure, how many molecules of air would there be in a 20 ∘C experimental chamber with a volume of 0.020 m3 ?
jonny [76]

Answer:

Explanation: The lowest pressure in a laboratory is 4.0×10^-11Pa

Using Ideal gas equation

PV = nRT

P= 4.0×10^-11Pa

V= 0.020m^3

T= 20+273= 293k

n=number of moles = m/A

Where m is the number of molecules and A is the Avogradro's number=6.02×10²³/mol

R=8.314J/(mol × K)

PV= m/A(RT)

4.0×10^-11 ×0.020 = m/6.02×10²³(8.314×293)

m = 4.0×10^-11×0.020×6.02×10^23 / (8.314×293)

m = 1.98×10^8 molecules

Therefore,the number of molecules is 1.98×10^8

8 0
3 years ago
Steel:<br> Propane:<br> Calcium chloride:<br> Water:
kondor19780726 [428]
What is the question
4 0
3 years ago
Read 2 more answers
What is the percentage of hydrogen in c2h4
jarptica [38.1K]

Ethylene- C2H4 = 85.7% Carbon and 14.3% Hydrogen


Find the atomic masses for each element and multiply it by the number of atoms in the compound, then add.


C- 12.0 * 2= 24.0


H- 1.00 * 4= 4.00


-----------------------


28.0


Take the masses for each element and divide it by the total mass. Then change the answer to get the percent.


C 24.0 / 28.0= .857 = 85.7%


H 4.00 / 28.0= .143 = 14.3%


<h3>Ethylene is 85.7% Carbon and 14.3% Hydrogen </h3>
8 0
3 years ago
Calculate the new boiling point of a solution if 10.00 g of a non-ionizing compound (C3H5(OH)3) is dissolved in 90.00 g of H2O.
erma4kov [3.2K]

Answer:

Boiling T° of solution = 100.6

Explanation:

Formula for elevation of boiling point is:

ΔT = Kb . m . i

where ΔT means Boiling T° of solution - Boiling T° of pure solvent

Our solute is a non ionizing compound.

i = 1, because it is a non ionizing compound. i, indicates the ions dissolved in solution.

m = molality (moles of solute dissolved in 1 kg of solvent)

90 g of solvent = 0.09 kg of solvent

We convert mass of solute to moles (by the molar mass):

10 g . 1 mol /92.09 g = 0.108 moles

m = 0.108 mol /0.09 kg = 1.21 m

Let's replace data: Boiling T° of solution - 100°C = 0.51 °C/m . 1.21 m . 1

Boiling T° of solution = 0.51 °C/m . 1.21 m . 1 + 100°C

Boiling T° of solution = 100.6

5 0
3 years ago
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