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2 years ago

9

When gases are treated as real, via use of the van der Waals equation, the actual volume occupied by gas molecules ________ the

pressure exerted and the attractive forces between gas molecules ________ the pressure exerted, as compared to an ideal gas.
decreases, increases
does not affect, decreases
increases, decreases
does not affect, increases
increases, increases

1 answer:

Tresset [83]2 years ago

6 0

Answer:

Increases, decreases

Explanation:

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A stock solution of HNO3 is prepared and found to contain 13.5 M of HNO3. If 25.0 mL of the stock solution is diluted to a final

salantis [7]

Answer : The correct option is, (C) 0.675 M

Explanation :

Using neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = concentration of $HNO_3$ = 13.5 M

$M_2$ = concentration of diluted solution = ?

$V_1$ = volume of $HNO_3$ = 25.0 ml = 0.0250 L

conversion used : (1 L = 1000 mL)

$V_2$ = volume of diluted solution = 0.500 L

Now put all the given values in the above law, we get the concentration of the diluted solution.

$13.5M\times 0.0250L=M_2\times 0.500L$

$M_2=0.675M$

Therefore, the concentration of the diluted solution is 0.675 M

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2 years ago

A bomb calorimeter has a heat capacity of 783 J/oC and contains 254 g of water whose specific heat capacity is 4.184 J/goC. How

IrinaK [193]

Answer : The amount of heat evolved by a reaction is, 4.81 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $783J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 254 g

$\Delta T$ = change in temperature = $T_2-T_1=(23.73-26.01)=-2.28^oC$

Now put all the given values in the above formula, we get:

$q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]$

$q=-4208.28J=-4.81kJ$

Therefore, the amount of heat evolved by a reaction is, 4.81 kJ

7 0

3 years ago

Most atoms have no net charge because they have

Ne4ueva [31]

Atoms have no electric charge because the protons and electrons "cancel out" each others charges. Neutrons have no charge. What is the atomic number of an element? The atomic number is the number of protons in the atom's nucleus.

Hope this helps have a great day :)

6 0

3 years ago

Read 2 more answers

At 73.0 ∘c , what is the maximum value of the reaction quotient, q, needed to produce a non-negative e value for the reaction so

damaskus [11]

Here we will use the general formula of Nernst equation:

Ecell = E°Cell - [(RT/nF)] *㏑Q

when E cell is cell potential at non - standard state conditions

E°Cell is standard state cell potential = - 0.87 V

and R is a constant = 8.314 J/mol K

and T is the temperature in Kelvin = 73 + 273 = 346 K

and F is Faraday's constant = 96485 C/mole

and n is the number of moles of electron transferred in the reaction=2

and Q is the reaction quotient for the reaction
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔ Br2(aq) + SO2(g) +2H2O(l)

so by substitution :

0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q → solve for Q

∴ Q = 4.5 x 10^-26

6 0

2 years ago

If the volame ofa gas coetainer at 32°C changes froem 1.55 L to 753 ml, what will the final temperature be? Assume the pressure

motikmotik

Answer : The final temperature of gas will be, 149 K

Explanation :

Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$

or,

$\frac{V_1}{V_2}=\frac{T_1}{T_2}$

where,

$V_1$ = initial volume of gas = 1.55 L

$V_2$ = final volume of gas = 753 ml = 0.753 L

$T_1$ = initial temperature of gas = $32^oC=273+32=305K$

$T_2$ = final temperature of gas = ?

Now put all the given values in the above formula, we get the final temperature of the gas.

$\frac{1.55L}{0.753L}=\frac{305K}{T_2}$

$T_2=149K$

Therefore, the final temperature of gas will be, 149 K

7 0

3 years ago