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3 years ago

J.j thomson a british physicist was the first to identify the

Chemistry

1 answer:

Shalnov [3]3 years ago

5 0

<span>J.j thomson a british physicist was the first to identify the electron in 1987</span>

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(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

3 0

3 years ago

If 1.16 L of water is initially at 24.2 ∘C, what will its temperature be after absorption of 9.4×10−2 kWh of heat?

vitfil [10]

Answer:

The temperature will be 93.92 °C

Explanation:

To explain this we will use following equation: also Q = ∆U + W known as the NON-FLOW ENERGY EQUATION (N.F.E.E.)

With Q = heat added to the system

with ∆U = change in internal energy

⇒∆U = ( m )( Cv )( T2 - T1 )

With W = work done by the system

⇒For this situation W = 0 because there isn't work done

So we get: ∆U = ( m )( Cv )( T2 - T1 ) = Q

To find the temperature, we have to isolate T in the equation:

(T2-T1) = Q / (m)(Cv)

⇒ Since we know that m = density * volume we can calculate the mass of water.

mass = 1000g/L * 1.16 L = 1160g

Cv = heat capacity ⇒ water has a heat capacity of 4.184 J/g °C

We know the absorption of heat is 9.4x 10^-2 kWh but to know how many joule this is we should convert ( 1 joule = 3.6 x 10^6 kWh)

⇒Q = ( 0.094 kWh ) ( 3.6 x 10^6 J / kWh ) = 0.3384 x 10^6 J

For the temperature we get then: T2 -T1 = Q / (m)(Cv)

T2 - T1 = 0.3384 x 10^6 J / (( 1160g)*(4.184 J/g °C)) = 69.72 ° C

T2 = ( T2 - T1 ) + T1 ⇒ 69.72 + 24.2 = 93.92 °C

6 0

3 years ago

Which chemical equation correctly represents the reaction that takes place when nitrogen gas and hydrogen gas are formed as ammo

Leviafan [203]

Answer:

d. 2NH3(g)-N2(g)+3H2(g)

4 0

2 years ago

What is the number of oxygen atoms in 32grams of the gas

Lady bird [3.3K]

One mole of oxygen gas, which has the formula O2, has a mass of 32 g and contains 6.02 X 1023 molecules of oxygen but 12.04 X 1023 (2 X 6.02 X 1023) atoms, because each molecule of oxygen contains two oxygen atoms.

7 0

2 years ago

Where would you have the least mass?

beks73 [17]

Answer:

the moon but mass stays the same

Explanation:

7 0

3 years ago