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azamat
2 years ago
11

“Dry ice” is the solid form of carbon dioxide. Determine the number of mass of CO2 gas in grams that are present in 61.8 L of CO

2 at STP.
_________________ g CO2. Do NOT enter unit. Report your final answer with 3 SFs.
Chemistry
1 answer:
Fofino [41]2 years ago
6 0

Answer:

Mass  = 121 g

Explanation:

Given data:

Mass in gram of CO₂ = ?

Volume = 61.8 L

Pressure = standard = 1 atm

Temperature = 273.15 K

Solution:

Formula:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

1 atm × 61.8 L = n ×0.0821 atm.L/ mol.K   × 273.15 k

61.8 L.atm = 22.42 atm.L/ mol × n

n = 61.8 L.atm /22.42 atm.L/ mol

n = 2.76 mol

Mass in gram:

Mass =  number of moles × molar mass

Mass = 2.76 mol × 44 g/mol

Mass  = 121 g

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What's the charge on Ni in the compound NiCi3? *
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Answer:

+3

Explanation:

Chlorine is anion with a -1 charge. But they are three chlorine atoms.

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So they have a -3 charge.

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-3 + 3 = 0

Furthermore, a chemical bond always has a 0 charge. Remember that.

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2 years ago
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3 years ago
In each case, decide if the change is a "chemical change" or a "physical change". a) A cup of household bleach changes the color
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8 0
3 years ago
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
3 years ago
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