The combustion of fossil fuels releases ____ into the atmosphere. (4 points)

How many grams does 1.0 liter of motor the density of motor oil is 0.80g / mL

balandron [24]

I think the answer is 1.25 grams actually i think i’m wrong

5 0

3 years ago

Can someone tell me plz :P

NemiM [27]

Answer:

There's nothing there

Explanation:

8 0

3 years ago

a sample of gas measures 5 liters at 1 atm. to change the volume to a 3.5 liters at constant temperature what pressure must be a

Scorpion4ik [409]

Answer:

0.7atm

Explanation:

3.5/5 because constant temperature

8 0

3 years ago

12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68

luda_lava [24]

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of $H_2O$ = 1.00 kg = 1000 g

Moles of $CsF$ = 3.68 mole

Molar mass of $H_2O$ = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of $H_2O$ .

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.55mole$

Now we have to calculate the mole fraction of $H_2O$

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CsF}=\frac{55.55}{55.55+3.68}=0.938$

Now we have to partial pressure of solution.

According to the Raoult's law,

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

where,

$P_{Solution}$ = vapor pressure of solution

$P^o_{H_2O}$ = vapor pressure of water = 0.692 atm

$X_{H_2O}$ = mole fraction of water = 0.938

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

$P_{Solution}=0.938\times 0.692atm$

$P_{Solution}=0.649atm$

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

5 0

3 years ago

Calculate the pH for each of the cases in the titration of 25.0 mL of 0.180 M pyridine, C 5 H 5 N ( aq ) with 0.180 M HBr ( aq )

Oxana [17]

Answer:

Explanations

Calculate the pH for each of the following cases in thetitration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 MHBr(aq):

a.Before and addition of HBr

b.After addition of 12.5ml HBr

c.After addition of 15ml HBr

d.After addition of 25ml HBr

e.After addition of 33ml HBr

SOLUTION ;;;

Kb of pyridine =1.5*10^-9

a)let the dissociation be x.so,

kb=x^2/(0.1-x)

or 1.5*10^-9=x^2/(0.1-x)

or x=1.225*10^-5

so,

[OH-]=1.225*10^-5

so,

pOH=-log([OH-])

so pH=9.088

b)now this will effectively behave as a buffer

pKb=8.82

so pOH=pKb+log(salt/acid)

=8.82+log((12.5*0.1)/(25*0.1-12.5*0.1))

=8.82

so pH=14-pOH

=5.18

c)again using the same equation as the above,

pOH=pKb+log(salt/acid)

=8.82+log((15*0.1)/(25*0.1-15*0.1))

=9

so pH=14-9

=5

d)now the base is completely neutralised.so,

concentration of the salt formed=0.1/2

=0.05 M

so,

pH=7-0.5pKb-0.5log(C)

=7-0.5*8.82-0.5*log(0.05)

=3.24

e)concentration of H+=(33*0.1-25*0.1)/(33+25)

=0.01379

so pH=-log(0.01379)

=1.86

4 0

3 years ago

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