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3 years ago

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What is the volume of 0.1 mole of methane (CH4) ? (One mole of any gas occupies 22.4 L under certain conditions of temperature a

nd pressure. Assume those conditions for this question.)
2.2 L

20 L

22.4 L

44.8 L

2 answers:

Veronika [31]3 years ago

8 0

Answer:

Option A = 2.2 L

Explanation:

Given data:

volume of one mole of gas = 22.4 L

Volume of 0.1 mole of gas at same condition = ?

Solution:

It is known that one mole of gas at STP occupy 22.4 L volume. The standard temperature is 273.15 K and standard pressure is 1 atm.

For 0.1 mole of methane.

0.1/1 × 22.4 = 2.24 L

0.1 mole of methane occupy 2.24 L volume.

Firdavs [7]3 years ago

7 0

Answer:

2,2 L

Explanation:

Under normal conditions of temperature and pressure: 273 K and 1 atm. The formula is used:

PV= nRT

R= Regnault constant

1 atm x V = 0,1mol x 0,082 l atm/K mol x 273 K

V = (0,1mol x 0,082 l atm/K mol x 273 K)/ 1 atm

V= 2,2368 l

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When ethyl ether is heated with excess HI for several hours, the only organic product obtained is ethyl iodide. T/F

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Answer:

True

Explanation:

Ethers react with HI to form the corresponding alcohols and alkyl iodides.

Similarly, ethyl ether react with excess of HI to form ethanol and ethyl iodide. But in the excess of HI as mentioned in the question, ethanol too undergoes $S^{N}2$ reaction with HI to form ethyl iodide.

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Hydrogen cyanide, HCN, can be made by a two-step process. First, ammonia reacts with O2 to give nitric oxide, NO.

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Answer:

The mass of HCN is 79.65 g.

The mass of reactant which remain at the end of both reactions is 88.5 g.

Explanation:

Given that,

Mass of ammonia = 50.2 g

Mass of methane = 48.4 g

Hydrogen cyanide, HCN, can be made by a two-step process

Ammonia reacts with O₂ to give nitric oxide NO.

The reaction is,

$4NH_{3}+5O_{2}\Rightarrow 4NO+6H_{2}O$

We need to calculate the mole of NO

Using given data,

2.25 g NH_{3}=\dfrac{50.2}{17}= 2.95\ mole\ NH_{3} [/tex]

$4\ mole NH_{3}\ glose 4\ mol NO$

2.95 mol NH₃ will produced 2.95 mol NO

Then nitric oxide reacts with methane,

The reaction is,

$2NO+2CH_{4}\Rightarrow 2HCN+2H_{2}O+H_{2}$

We need to calculate the mole of methane

Using given data,

$mole\ of\ methane=\dfrac{48.4}{16}$

$mole\ of\ methane = 3.03\ moles$

2 mole NO produced 2 mole HCN

2.95 mol NO will produced $\dfrac{2.95\times3.03}{3.03}$ = 2.95 mol HCN

We need to calculate the mass of HCN

Using formula of mass

$m=N\times M$

Where, N = number of mole

M = molecular mass

Put the value into the formula

$m=2.95\times27$

$m= 79.65\ g$

The mass of HCN is 79.65 g.

We need to calculate the mass of NO

Using formula of mass

$m=N\times M$

Where, N = number of mole

M = molecular mass

Put the value into the formula

$m=2.95\times30$

$m= 88.5\ g$

Hence, The mass of HCN is 79.65 g.

The mass of reactant which remain at the end of both reactions is 88.5 g.

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Answer:

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Explanation:

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What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

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

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

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3 years ago