Ask question

Ask question

All categories

borishaifa [10]

2 years ago

9

Would an alkaline earth metal make a good replacement for tin in a tin can?

1 answer:

hichkok12 [17]2 years ago

4 0

Answer:

Definitely not. This is because the alkaline earth metal [basically group 2 elements] are reactive and will react with the contents of the can, contaminating it.

You might be interested in

When two metals touch in the mouth, a small shock is created. this is known as<br> a?

Over [174]

<span>When two metals touch in the mouth, a small shock is created. this is known as a </span>galvanic action

7 0

3 years ago

ʕ•ᴥ•ʔ Which type of reaction occurs between an acid and a base and results in the formation of a salt and water? ʕ•ᴥ•ʔ

Nataliya [291]

The answer would be D. because when an acid and base mix the start to cancel each other out causing it to neutralize

3 0

3 years ago

Read 2 more answers

How many elements are in N2 and how many atoms are in N2?

Lady bird [3.3K]

Answer:

2

Explanation:

7 0

2 years ago

Read 2 more answers

(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

An exhaled air bubble underwater at 290.

Vinil7 [7]

The answer for the following problem is mentioned below.

<u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 290 kPa

Final pressure ( $P_{2}$ ) = 104 kPa

Initial volume ( $V_{1}$ ) = 18.9 ml

To find:

Final volume ( $V_{2}$ )

We know;

From the ideal gas equation;

P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of gas

n represents the no of the moles

R represents the universal gas constant

T represents the temperature of the gas

So;

P × V = constant

P ∝ $\frac{1}{V}$

From the above equation;

$\frac{P_{1} }{P_{2} } = \frac{V_{2} }{V_{1} }$

$P_{1}$ represents the initial pressure of the gas

$P_{2}$ represents the final pressure of the gas

$V_{1}$ represents the initial volume of the gas

$V_{2}$ represents the final volume of the gas

Substituting the values of the above equation;

$\frac{290}{104}$ = $\frac{V_{2} }{18.9}$

$V_{2}$ = 52.7 ml

<u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

6 0

3 years ago