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3 years ago

15

What is h2OO??????????

1 answer:

kramer3 years ago

8 0

The answer is : H2O is water

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What is the molarity of a solution prepared by adding enough water to 2 g of acetone to make a total volume of 1.60 l?

serg [7]

Explanation:

answer is in photo above

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3 years ago

A calorimeter contains 251 g of water at 24.3 C.The temperature of the water decreases to 18.5 C when 12.5 g of potassium nitrat

vova2212 [387]

Answer:

Endothermic

Explanation:

The temperature of the water decreased.

The water lost heat.

The heat must have gone into getting the KNO₃ into solution.

So, the dissolving of KNO₃ in water is endothermic.

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4 years ago

The decomposition of SOCl2 is first-order in SOCl2. If the half-life for the reaction is 4.1 hr, how long would it take for the

qaws [65]

4.1 h = 14760 s

<span>t 1/2 = ln 2 / k </span>

<span>k = rate reaction = 4.97 x 10^-5 </span>

<span>ln 0.045 / 0.36 = - 4.97 x 10^-5 t </span>

<span>2.08 = 4.97 x 10^-5 t </span>

<span>t = 41839.9 s = 11 h 37 min 19 s</span>

3 0

3 years ago

Select the compound with the lowest lattice energy.

tatyana61 [14]

Answer:

A. CsBr(s)

Explanation:

we will get here compound with the lowest lattice energy

solution

As we know that Lattice energy is always proportional to the charge of ions and it is inversely proportional to the size of ions.

so that by the smallest charge and the largest size give us the lowest lattice energy and that charge and size is express as here as

Charge

Cs (+1), K(+1), Na (+1), Cl(-1), Br(-1), Sr(+2), Ca(+2), O(-2) .......................1

and

Size

Na+ < Ca2+ < K+ < Sr2+ < Cs+, O2- < Cl- < Br- ..........................2

so that here

correct answer is A. CsBr

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

3 years ago