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3 years ago

What is a way that we can slow down climate change?

Chemistry

1 answer:

Wittaler [7]3 years ago

8 0

Answer:

I would guess B, because trees make oxygen, not carbon dioxcide, and coal is actually a very harmful source of energy.

hope this helps

Explanation:

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barxatty [35]

Answer:

is 5

Explanation:

Because the baking soda has the same kilograms

4 0

3 years ago

Help me as soon as possible I’m gonna dieeee

ioda

I think it’s D I’m not sure .........

7 0

3 years ago

Read 2 more answers

When the coffee is brewed according to directions, a pound of coffee beans yields 50 cups of coffee (4 cups = 1 quart). How many

ycow [4]

About one kilogram of coffee beans is required

3 0

3 years ago

Propane C3H8 is a hydrocarbon tht is commonly used as a fuel

Mariulka [41]

Answer:C3H8 +5O2-- >3CO2+4H2O, 330L of air,Hof C3H8= 2323.7 KJ/mol,dT=75.30

Explanation:

First We need to find a balanced equation to depict the combustion of the propane:

C3H8 +5O2-- >3CO2+4H2O

$25g C3H8*\frac{1mol C3H8}{44gC3H8} *\frac{5molO2}{1molC3H8}* \frac{32gO2}{1mol O2}*\frac{1L air}{0.275g}$

330L of air

25 g C3H8 are equal to 44g and 5 mol of O2 will react with a mol of C3H8, each mol of O2 weigh 32g and for each L of air contains 0.275g of O2

c) $Hof C3H8= -(-285.8\frac{KJ}{mol}*4mol +-393.5*3mol$

the enthalpy of formation of propane will be the negative of the sum of the enthalpy of formation multiplied by the mols of the H2O and CO2

Hof C3H8= 2323.7 KJ/mol

d) We know that each 44g of C3H8(1mol) transfer 2219.2 KJ then:

$25g\frac{2219.2KJ}{44g C3H8}$ = 1260.9KJ

We use the equation of heat transfered

Q=mCpdT

1260.9KJ= 4KJ *4.186KJ/Kg*°C *dT

We clear the equation

Q/mCp=dT

dT=75.30

4 0

3 years ago

Sulfuryl chloride is in equilibrium with sulfur dioxide and chlorine gas: so2cl2(g) so2(g) + cl2(g) a system with a volume of 1.

Bess [88]

Answer:

Sulfuryl chloride decreases by -1/21 (-4.76%) (option c)

Explanation:

Denoting

sc= so2cl2(g)

s=so2(g)

c=cl2(g)

Assuming that the compression is an isothermal process , then reaction equilibrium constant in terms of pressure does not change

Kp= psc/ps*pc =

where p= partial pressures

Assuming ideal behaviour , then from Dalton's law,

Xsc₁=psc₁/P₁= psc₁/P₁ = 1 bar/(1 bar + 0.1 bar + 0.1 bar) = 5/6

Xs=ps₁/P₁ = 0.1/1.2=1/12

Xc=pc₁/P₁ = 0.1/1.2=1/12

since Xs=Xc → the reaction started as pure Sulfuryl chloride . Then representing ξ as the extent of reaction and n as the moles

nsc=nsc₀*(1-ξsc) , ns=nsc₀*ξsc , nc=nsc₀*ξsc → n=nsc +ns +nc = nsc₀*(1+ξsc)

therefore

Xs₁=ns₁/n₁=ξsc₁/(1+ξsc₁) → Xs₁*ξsc₁+Xs₁=ξsc₁ → ξsc₁=Xs₁/(1-Xs₁) = (1/12)/(11/12)= 1/11

then from the ideal gas law

ps₁*V₁=ns₁*R*T

after the reduction

ps₂V₂=ns₂*R*T

dividing both equations

(ps₂/ps₁)*(V₂/V₁)=(ns₂/ns₁)=nsc₀*ξsc₂/(nsc₀*ξsc₁) = ξsc₂/ξsc₁

ps₂ = ps₁ * (V₁/V₂) * (ξsc₂/ξsc₁)

since

psc₁*V₁=nsc₁*R*T , psc₂V₂=nsc₂*R*T → psc₂ = psc₁ * (V₁/V₂) * (1-ξsc₂)/(1-ξsc₁)

also knowing that

Kp= psc₁/ps₁² = psc₂/ps₂²

psc₂/ps₂² = psc₁/ps₁² * (V₁/V₂) * (1-ξsc₂)/(1-ξsc₁) / [(V₁/V₂) * (ξsc₂/ξsc₁) ]² =

1 = (V₂/V₁)(1-ξsc₂)*ξsc₁/ [(1-ξsc₁)*ξsc₂]

replacing ξsc₁= 1/11

1 = (V₂/V₁)(1-ξsc₂)/ξsc₂ *(1/10)

10 = (V₂/V₁)* (1/ξsc₂-1) → ξsc₂ = 1/(10*(V₁/V₂)+1)

therefore the extent of reaction varies with the volume reduction according to

ξsc₂ = 1/(10*(V₁/V₂)+1)

since V₁/V₂=2

ξsc₂ = 1/(10*2+1) = 1/21

therefore the decrease in moles of Sulfuryl chloride is

Δnsc/nsc₁ = (ξsc₂-ξsc₁)/(1-ξsc₁) = (1/21-1/11)/(10/11)= (11/21-1)/10 = -1/21 (-4.76%)

6 0

3 years ago