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ASHA 777 [7]
3 years ago
14

A gas at constant pressure and a temperature of 293K has a volume of 8.0 L. If the temperature of the gas is increased to 314K,

what is the volume?​
Chemistry
1 answer:
aalyn [17]3 years ago
8 0

V2 = 8.6 L

Explanation:

Given:

T1 = 293K. T2 = 314K

V1 = 8.0 L V2 = ?

Using Charles's law and solving for T2,

V1/T1 = V2/T2

V2 = (T2/T1)V1

= (314K/293K)(8.0 L)

= 8.6 L

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While 1 gram of fat provides 9 calories, 1 gram of glucose provides 4 calories. Why is that?
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6 0
3 years ago
A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m
kogti [31]

Answer:

The molar mass of the unknown gas is \mathbf{ 51.865 \  g/mol}

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

R \  \alpha  \ \dfrac{1}{\sqrt{d}}

R = \dfrac{k}{d}  where K = constant

If we compare the rate o diffusion of two gases;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}

Since the density of a gas d is proportional to its relative molecular mass M. Then;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}

Rate is the reciprocal of time ; i.e

R = \dfrac{1}{t}

Thus; replacing the value of R into the above previous equation;we have:

\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}

We can equally say:

{\dfrac{t_2}{t_1}}=  {\sqrt{\dfrac{M_2}{M_1}}

{\dfrac{6.34}{4.98}}=  {\sqrt{\dfrac{M_2}{32}}

M_2 = 32 \times ( \dfrac{6.34}{4.98})^2

M_2 = 32 \times ( 1.273092369)^2

M_2 = 32 \times 1.62076418

\mathbf{M_2 = 51.865 \  g/mol}

7 0
3 years ago
In a 28 g serving of cheese curls there are 247mg of sodium. How much sodium is in a 12.5 ounce bag
Marina86 [1]

Answer:

There should be about 3136.9mg of Na in a 12.5oz bag of cheese curls.when rounding if not rounded it would be 3126.04082143mg

Explanation:

12.5oz=354.4g

354.4/28=12.7

12.7*247=3136.9

6 0
3 years ago
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