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3 years ago

Now liters at the same temperature and pressure: liters of H2 + liters of O2 → liters of H2O

Chemistry

1 answer:

iragen [17]3 years ago

5 0

The answer is 2H2 + O2----> 2H2O

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PLEASE HELPP!!!!

Natasha_Volkova [10]

Answer:

The combined law is:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

See the derivation below.

Explanation:

1. Boyle's law:

$PV=constant\\\\P_1V_1=P_2V_2$

2. Charles' law:

$\dfrac{V}{T}=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

3. Gay-Lussac’s law

$\dfrac{P}{T}=constant\\\\\\\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

4. Summary:

PV = K

V/T = K'

P/T = K''

Mulitply PV by V/T and P/T

$PV\times \dfrac{V}{T}\times \dfrac{P}{T}=K\times K'\times K''\\\\\\\dfrac{P^2V^2}{T^2}=constant\\\\\\\sqrt{\dfrac{P^2V^2}{T^2}}=\sqrt{constant}\\\\\\\dfrac{PV}{T}=constant$

Thus:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

Which is the combined law.

4 0

4 years ago

How many grams of C2H6 are needed to react with 54 grams of O2?

Dmitrij [34]

Answer:

Explanation:

in a combustion of ethane 2 moles of ethane react with 7 moles of O2

now no of moles in 54 gram of O2=mass/ molar mass

moles =54/32=1.7 moles

if 7 moles of O2 required 2 moles of ethane then 1.7 mole required=?

7 moles of O2=2 moles of C2H6

1.7 moles of O2=1.7*2/7=0.5 moles of C2H6

0.5 moles of C2H6 contain how much grams=?

mass= moles*molar mass=0.5*30=15

4 0

3 years ago

How many moles is 5.05 x 10^24 atoms of magnesium?

Bess [88]

Answer: 9.55 moles

Explanation: Moles are the base unit of atoms, molecules, ions, or any other substance. It is exactly 6.022 × 10²³ entities of any substance. The mole is equal to the ratio of the given mass to the molar mass.

Hope this helps :) If possible give brainliest

7 0

3 years ago

Calculate the number of grams of solid aluminium chloride that will form when a mixture containing 0.150 g of aluminum powder an

Alinara [238K]

Answer:

the answer is B:) have a good day

5 0

4 years ago

The molar mass of a certain gas is 49 g. What is the density of the gas in g/L at STP?

snow_tiger [21]

Answer:

$\boxed{\text{2.2 g/L}}$

Explanation:

We can use the Ideal Gas Law to calculate the density of the gas.

pV = nRT

n = m/M Substitute for n

pV = (m/M)RT Multiply both sides by M

pVM = mRT Divide both sides by V

pM = (m/V) RT

ρ = m/V Substitute for m/V

pM = ρRT Divide each side by RT

$\rho = \frac{pM }{RT}$

Data:

p = 1.00 bar

M = 49 g/mol

R = 0.083 14 bar·L·K⁻¹mol⁻¹

T = 0 °C = 273.15 K

Calculation:

ρ = (1.00 × 49)/(0.083 14 × 273.15) = 2.2 g/L

The density of the gas is $\boxed{\text{2.2 g/L}}$ .

8 0

4 years ago