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2 years ago

12

Floor 19 is an isotope of fluorine It has different number of?—————

1 answer:

vitfil [10]2 years ago

5 0

Try B, but i don’t know

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Why are atoms more stable when combined?

shepuryov [24]

Answer:

D

Explanation:

3 0

3 years ago

As a chemical reaction occurs, the thermometer in the container records a drop in temperature. What is true of the reaction?

nadya68 [22]

<span>When a chemical reaction occurs and the thermometer in the container records a drop in temperature t</span>he reaction is exothermic because heat was released by the reaction.

<span>B. The reaction is exothermic because heat was released by the reaction. is your answer.

Hoped I helped!</span>

4 0

3 years ago

Read 2 more answers

Really need help now please help!!

Zielflug [23.3K]

I'm sorry but the picture isn't clear enough

7 0

2 years ago

How many grams of aluminum is produced when 82.4 grams of aluminum chloride

yKpoI14uk [10]

Answer:

16.6 g of Al are produced in the reaction of 82.4 g of AlCl₃

Explanation:

Let's see the decomposition reaction:

2AlCl₃ → 2Al + 3Cl₂

2 moles of aluminum chloride decompose to 2 moles of solid Al and 3 moles of chlorine gas.

We determine the moles of salt:

82.4 g . 1mol/ 133.34g = 0.618 moles

Ratio is 2:2. 2 moles of salt, can produce 2 moles of Al

Then, 0.618 moles of salt must produce 0.618 moles of Al.

Let's convert the moles to mass → 0.618 mol . 26.98g /mol = 16.6 g

6 0

3 years ago

(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

2 years ago