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azamat
2 years ago
15

The temperature of 6.24 l of a gas is increased from 25.0°c to 55.0°c at constant pressure. the new volume of the gas is

Chemistry
1 answer:
bixtya [17]2 years ago
6 0
At constant pressure, temperature is directly proportional to the volume and vise versa. The formula will be

V1/T1 = V2/T2

where V1 and T1 are the initial volume and temperature and V2 and T2 are the final volume and temperature. The temperature is in Kelvin and to convert Celsius to Kelvin add 273.

so, 6.24L/298 = V2/328
      =6.87L
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Which of the following statements is true?
Rom4ik [11]
I think that its D.
6 0
2 years ago
Read 2 more answers
For double-helix formation, change in Gibbs free energy, ΔG, can be measured to be −54 kJ⋅mol−1 (−13 kcal⋅mol−1) at pH 7.0 in 1
77julia77 [94]

Answer:

Explanation:

Entropy change in the system : --

ΔG =   −54 kJ⋅mol−1 (−13 kcal⋅mol−1)  =   −54 kJ⋅mol−1 (−13 x 4.2  kJ⋅mol−1)

= - 108.6  KJ / mol

ΔH =  -251 kJ/mol (-60 kcal/mol) =  -251 kJ/mol (-60 x 4.2  kJ/mol)

= - 503  KJ / mol

ΔG = ΔH - TΔS

ΔS = ( ΔH - ΔG ) / T

=  - 503 + 108.6 / ( 273 + 25 ) KJ / mol k⁻¹

= - 1323.48 J / mol k⁻¹

Entropy change in the surrounding

+ 1323.48 J / mol k⁻¹

7 0
1 year ago
If the composition of the reaction mixture at 400 k is [brcl] = 0.00415 m, [br2] = 0.00366 m, and [cl2] = 0.000672 m, what is th
Scorpion4ik [409]
Q is unlike K value it describes the reaction that is not at equilibrium.
by considering this reaction:
aA+ bB⇄ cC
and our reaction is:
Br2 + Cl2 ⇄ 2 BrCl

According to Q low:
Q= concentration of products/concentration of reactants
but this equation in the gaseous or aqueous states only.
∴ Q = [BrCl]^2 / [Br2] [Cl2]

and we have [Br2] = 0.00366 m  [Cl2]= 0.000672 m  [BrCl] = 0.00415 m

by substitution:
                          = [0.00415]^2 / ( [0.00366] * [0.000672])
             ∴   Q   = 7
7 0
2 years ago
NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te
makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y <em>(1)</em>

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol×\frac{1mol}{80,043g}X

Y = 321J/g X <em>(2)</em>

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

<em>X = 35,2g</em>

<em />

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

6 0
2 years ago
The density of silver metal bar is 10.5 g/cm3 and it occupies a volume space of 0.5l. What is its mass in kilograms (kg).
mojhsa [17]

Density is equal to the ratio of mass to the volume.

The mathematical expression is given as:

Density = \frac{mass}{volume}

Density of silver metal bar=10.5 g/cm^{3}

Convert g/cm^{3} into g/L

1 cm^{3} = 0.001 L

Thus, density  = \frac{10.5}{10^{-3}} g/L

= 10.5\times 10^{3}g/L

Volume = 0.5 L

Put the values,

10.5\times 10^{3}g/L= \frac{mass}{0.5 L}

m= 10.5\times 10^{3}g/L\times 0.5 L

=5.25 \times 10^{3}g

Now, convert gram into kg

1 g = 0.001 kg

Therefore, mass in kg=  \frac{5.25 \times 10^{3}}{10^{-3}}

= 5.25 kg

Thus, mass of silver metal bar in kg=5.25 kg



4 0
2 years ago
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