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When 1 mol of a fuel burns at constant pressure, it exchanges-3452 kJ of heat and does-11 kJ of workon the surroundings. What ar

Degger [83]

Answer:-3463 kJ and -3452kJ

Explanation:

ΔU is the change in internal energy of a system and its formula is;

ΔU = q + w

Where q represents heat transferred into or out of the system. Its value is positive when heat is transfer into the system and negative when heat is produced by the system.

W represents the work done on or by the system. Its value is positive when work is done on the system and negative when it is done by the system.

For the system in this question, we see that it produces heat which means heat is transferred out of the system, therefore the value of q is negative, it can also be seen that work is done by the system which means that w is also negative.

Therefore,

ΔU = -q-w

ΔU = -3452 kJ – 11kJ

= - 3463kJ

ΔH is the change in the enthalpy of a system and its formuls is;

ΔH = ΔU + Δ(PV)

By product rule Δ(PV) becomes ΔPV + PΔV

At constant pressure ΔP = 0. Therefore,

ΔH = -q-w + PΔV

w is equals to PΔV, So:

ΔH = -q

ΔH = -3452kJ

6 0

3 years ago

(34.6785 x 5.39) + 435.12

Shalnov [3]

To solve this, we should follow order of operations. To start, we should multiply the values inside of the parentheses.

(34.6785*5.39)+435.12
186.917115+435.12

Now, we should add the 2 values we are left with together.

186.917115
+435.120000
 622.037115

Using the math above, we can see that this expression is equal to 622.037115.

5 0

3 years ago

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A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? A

anygoal [31]

this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.

c1v1 = c2v2

where c1 is concentration and v1 is volume of the concentrated solution

and c2 is concentration and v2 is volume of the diluted solution to be prepared

50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL

substituting these values in the formula

1.50 M x 50.0 mL = C x 250 mL

C = 0.300 M

concentration of the final solution is A) 0.300 M

4 0

3 years ago

Read 2 more answers

If an ion had a charge of (-3), how many valence electrons could be expected

zimovet [89]

Answer:

Key Points. Ionic bonds are formed through the exchange of valence electrons between atoms, typically a metal and a nonmetal.

Explanation:

7 0

3 years ago

A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k

Sergeeva-Olga [200]

Answer: The partial pressure of oxygen gas is 2.76 bar

Explanation:

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0831\text{ L bar }mol^{-1}K^{-1}$

n = Total number of moles = ?

Putting values in above equation, we get:

$5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol$

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0

3 years ago