Ask question

Ask question

All categories

3 years ago

11

What is the molar mass of PCL3

1 answer:

antiseptic1488 [7]3 years ago

8 0

The molar mass for PCL3 is 137.33 g/mol

You might be interested in

HELP I HAVE 14 MINUTES!!

ki77a [65]

Answer:

2 one

Explanation:

is right coz I think

4 0

3 years ago

Read 2 more answers

What is the pH of a substance with a hydrogen concentration of 1.0 x 10-13?

Nikolay [14]

Answer:

Favorite Answer

1.0 x10^-14 = (1.0 x 10^-13) (x)

x = 1.0 x 10^-1 = 0.1 M (this is the [OH-])

pOH = -log 0.1 = 1.0

Explanation:

I hope this helps :) sorry if not :(

4 0

3 years ago

Sheela wants to make a computer model of a land ecosystem. Part of her model simulates the chemical processes of photosynthesis

shusha [124]

<span>The correct answer is A, the ligt-dependent reactions. These reactions are responsible for the production of glucose molecules, by the utilization of carbon dioxide, and water along with the sunlight. Glucose is then broken down during resiration process, for the production of ATP in mitochondria.</span><span />

5 0

3 years ago

Read 2 more answers

Density is a derived unit of mass / volume.

lapo4ka [179]

Answer:

1kg/L

Explanation:

1.) convert grams to kilograms

1000g÷1000=1kg

2.)use formula to find density

$density = \frac{mass}{volume}$

= 1kg/1.0L

=1kg/L

6 0

2 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