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3 years ago

What is the temperature of 0.500 moles of a gas that is stored in a 2.50 L container under 13.0 atm of pressure?

Chemistry

1 answer:

faust18 [17]3 years ago

3 0

<h2>Answer:</h2>

Moles of a gas = 0.500

Volume = 2.50 L

Pressure = 13. atm

Temperature = ?

Solution:

Formula:

PV = n RT

Putting the values in formula:

T = PV/nR = 13 * 2.5 / 0.5 * 0.082057

= 32.5/0.041 = 792.68 K

T = 792.68 K

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What is the total number of oxygen atoms on the right-hand side of this chemical equation?

vivado [14]

The chemical equation is

Cu(s) +4HNO3(aq) ⇒ Cu(NO3)2(aq) + 2NO2(g) + 2H2O(g)

Answer:

Explanation:

In the right hand side of the equation, there are three compound which contains O2, which are;

Cu(NO3)2 , number of oxygen atoms =3*2 =6

2NO2, number of oxygen atoms = 2*2=4

2H2O, number of oxygen atoms =2*1=2

Total number of oxygen atoms on the right side of equation = 6+4+2 =12

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3 years ago

What is the Big Bang

Mnenie [13.5K]

<span>the rapid expansion of matter from a state of extremely high density and temperature that according to current cosmological theories marked the origin of the universe.</span>

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3 years ago

Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h

bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

We know that,

The relation between the $C_p\text{ and }C_v$ for an ideal gas are :

$C_p-C_v=R$

As we are given :

$C_p=28.253J/K.mole$

$28.253J/K.mole-C_v=8.314J/K.mole$

$C_v=19.939J/K.mole$

Now we have to calculate the entropy change of the gas.

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

$\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J$

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. $(w=-pdV)$

(C) Heat during the process will be,

$q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J$

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

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3 years ago

A 0.0000792 M sample of Compound X in a solvent has an absorbance of 0.341 at 528 nm in a 1.000-cm cuvet. The solvent alone has

natima [27]

Answer:

a) a = 3485 M⁻¹cm⁻¹

b) C = 0,000127 M

Explanation:

Lambert-Beer law says that there is a linear relationship between absorbance and concentration of a chemical substance. The formula is:

A = a×b×C

Where A is absorbance, a is molar absorptivity, b is path length and C is concentration.

a) In the problem Concentration is 0.0000792 M, b is 1,000cm and Absorbance is absorbance of sample-absorbance of blank: 0,341-0,065 = 0,276

Replacing:

0,276 = a×1,000cm×0,0000792M

b) As the experiment consist in the same compound in the same solvent, the molar absorptivity will be the same, a = 3485 M⁻¹cm⁻¹, path length will be 1,000cm and absorbance: 0,508-0,065 = 0,443

Replacing:

0,443 = 3485 M⁻¹cm⁻¹×1,000cm×C

I hope it helps!

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3 years ago

Why scientists use the word dominant

Anni [7]

Answer:

Dominant traits are traits that more likely to be transferred from a parent to an offspring so scientist use this term dominant.

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3 years ago

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