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MArishka [77]
3 years ago
15

Watch the video to determine which of the following relationships correctly depict the relationship between pressure and volume

for a fixed amount of gas at constant temperature according to Boyle’s law.
A. PV∝ P
B.V∝ 1P
C. P∝ 1V
D. PV∝ V
E. P∝ V
F. V∝ P
Chemistry
1 answer:
AnnZ [28]3 years ago
5 0

Answer : The correct options are,

(B) V\propto \frac{1}{P}

(C) P\propto \frac{1}{V}

Explanation :

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

P\propto \frac{1}{V}

or,

V\propto \frac{1}{P}

The relation between the pressure and volume of two gases are:

P_1V_1=P_2V_2

where,

P_1 = initial pressure of gas

P_2 = final pressure of gas

V_1 = initial volume of gas

V_2 = final volume of gas

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The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati
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Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80M^{-1}s^{-1}

t = time taken  = 95 s

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.86 M

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

0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0
3 years ago
A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t
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Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)

So, moles of product formed are calculated as follows.

\frac{3}{2} \times 0.17 mol \\= 0.255 mol

Hence, the given data is as follows.

n_{1} = 0.17 mol,      n_{2} = 0.255 mol

V_{1} = 19.5 L,         V_{2} = ?

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}

Substitute the values into above formula as follows.

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L

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8 0
2 years ago
How many milliliters of C5H8 can be made from 366 mL C5H12 ?
zlopas [31]

The number of Ml  of C₅H₈  that can  be  made  from 366  ml  C₅H₁₂  is 314.7 ml  of C₅H₈


  <u><em>calculation</em></u>

 step  1: write  the  equation for  formation of C₅H₈

C₅H₁₂  →  C₅H₈  + 2 H₂

Step 2: find the mass of C₅H₁₂

mass = density × volume

= 0.620 g/ml × 366 ml =226.92 g

Step 3: find moles  Of  C₅H₁₂

moles  = mass÷  molar mass

from periodic table the  molar mass of  C₅H₁₂ = (12 x5) +(  1 x12) = 72 g/mol

moles = 226.92 g÷ 72 g/mol =3.152 moles

Step 4: use the  mole ratio  to determine the  moles of C₅H₈

C₅H₁₂:C₅H₈  is 1:1  from equation above

Therefore the  moles of C₅H₈  is also = 3.152  moles

Step 5: find the mass  of C₅H₈

mass = moles x molar mass

from periodic table the  molar mass of C₅H₈ = (12 x5) +( 1 x8) = 68 g/mol

= 3.152  moles x 68 g/mol = 214.34 g

Step 6: find Ml of  C₅H₈

=mass / density

= 214.34 g/0.681 g/ml = 314.7 ml



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