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2 years ago

Which temperature is equal to 120. K? (1) -153°C (3) +293°C (2) -120.°C (4) +393°C

1 answer:

3 0

First of all, the formula for finding Kelvin is Celsius + 273
Therefore, if we subtract 273, we get the temperature in degrees
120 - 273 = - 153

Therefore, the answer is (1), or -153 degrees Celsius

Hope this helped!! :D

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A sample of a gas (1.50 mol) is contained in a 15.0 l cylinder. the temperature is increased from 100 °c to 150 °c. what is the

pav-90 [236]

Given:

Moles of gas, n = 1.50 moles

Volume of cylinder, V = 15.0 L

Initial temperature, T1 = 100 C = (100 + 273)K = 373 K

Final temperature, T2 = 150 C = (150+273)K = 423 K

To determine:

The pressure ratio

Explanation:

Based on ideal gas law:

PV = nRT

P= pressure; V = volume; n = moles; R = gas constant and T = temperature

under constant n and V we have:

P/T = constant

(or) P1/P2 = T1/T2 ---------------Gay Lussac's law

where P1 and P2 are the initial and final pressures respectively

substituting for T1 and T2 we get:

P1/P2 = 373/423 = 0.882

Thus, the ratio of P2/P1 = 1.13

Ans: The pressure ratio is 1.13

7 0

1 year ago

Which of the following energy sources obtains its energy from gravitational potential energy?

ipn [44]

It is a geothermal power plant

6 0

1 year ago

A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?

dem82 [27]

Answer: $1.67\times 10^{-3}$

Explanation:

$ClCH_2COOH\rightarrow ClCH_2COO^-+H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given: c = 0.15 M

pH = 1.86

$K_a$ = ?

Putting in the values we get:

Also $pH=-log[H^+]$

$1.86=-log[H^+]$

$[H^+]=0.01$

$[H^+]=c\times \alpha$

$0.01=0.15\times \alpha$

$\alpha=0.06$

As $[H^+]=[ClCH_2COO^-]=0.01$

$K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}$

$K_a=1.67\times 10^{-3]$

Thus the vale of $K_a$ for the acid is $1.67\times 10^{-3}$

4 0

2 years ago

For nitrogen gas the values of Cv and Cp at 25°C are 20.8 J K−1 mol−1 and 29.1 J K−1 mol−1, respectively. When a sample of nitro

kifflom [539]

Answer:

The fraction of energy used to increase the internal energy of the gas is 0.715

Explanation:

Step 1: Data given

Cv for nitrogen gas = 20.8 J/K*mol

Cp for nitrogen gas = 29.1 J/K*mol

Step 2:

At a constant volume, all the heat will increase the internal energy of the gas.

At constant pressure, the gas expands and does work., if the volume changes.

Cp= Cv + R

⇒The value needed to change the internal energy is shown by Cv

⇒The work is given by Cp

To find what fraction of the energy is used to increase the internal energy of the gas, we have to calculate the value of Cv/Cp

Cv/Cp = 20.8 J/K*mol / 29.1 J/K*mol

Cv/Cp = 0.715

The fraction of energy used to increase the internal energy of the gas is 0.715

3 0

1 year ago

Read 2 more answers

Determine the normality of the following solutions note the species of interest is H 95 g of PO4 3- in 100mL solution

Aliun [14]

Answer : The normality of the solution is, 30.006 N

Explanation :

Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.

Mathematical expression of normality is:

$\text{Normality}=\frac{\text{Gram equivalent of solute}}{\text{Volume of solution in liter}}$

or,

$\text{Normality}=\frac{\text{Weight of solute}}{\text{Equivalent weight of solute}\times \text{Volume of solution in liter}}$

First we have to calculate the equivalent weight of solute.

Molar mass of solute $PO_4^{3-}$ = 94.97 g/mole

$\text{Equivalent weight of solute}=\frac{\text{Molar mass of solute}}{\text{charge of the ion}}=\frac{94.97}{3}=31.66g.eq$

Now we have to calculate the normality of solution.

$\text{Normality}=\frac{95g}{31.66g.eq\times 0.1L}=30.006eq/L$

Therefore, the normality of the solution is, 30.006 N

5 0

1 year ago