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2 years ago

9

What is haloalkanes?

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2 answers:

expeople1 [14]2 years ago

7 0

Answer:

It is a group of compound derived from alkanes containing one or more halogens .

Bad White [126]2 years ago

5 0

Answer:

compounds formed by halogens and alkanes

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A gas evolved during the fermentation of alcohol had a volume of 19.4 L at 17 °C and 746 mmHg. How many moles of gas were collec

AlexFokin [52]

Answer:- 0.800 moles of the gas were collected.

Solution:- Volume, temperature and pressure is given for the gas and asks to calculate the moles of the gas.

It is an ideal gas law based problem. Ideal gas law equation is used to solve this. The equation is:

PV=nRT

Since it asks to calculate the moles that is n, so let's rearrange this for n:

$n=\frac{PV}{RT}$

V = 19.4 L

T = 17 + 273 = 290 K

P = 746 mmHg

we need to convert the pressure from mmHg to atm and for this we divide by 760 since, 1 atm = 760 mmHg

$P=746mmHg(\frac{1atm}{760mmHg})$

P = 0.982 atm

R = $0.0821\frac{atm.L}{mol.K}$

Let's plug in the values in the equation to get the moles.

$n=\frac{0.982atm*19.4L}{0.0821\frac{atm.L}{mol.K}*290K}$

n = 0.800 moles

So, 0.800 moles of the gas were collected.

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3 years ago

Which mass of urea, CO(NH2)2, contains the same mass of nitrogen as 101.1g of potassium nitrate?

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In order to calculate the mass of nitrogen, we must first calculate the mass percentage of nitrogen in potassium nitrate. This is:
% nitrogen = mass of nitrogen / mass of potassium nitrate
% nitrogen = 14 / 101.1 x 100

The mass of nitrogen = % nitrogen x sample mass
= (14 / 101.1) x 101.1
= 14 grams

The molar weight of nitrogen is 14. Each mole of urea contains two moles of nitrogen. Therefore, for there to be 14 grams of nitrogen, there must be 0.5 moles of urea.
Mass of urea = moles urea x molecular weight urea
Mass of urea = 0.5 x 66.06
Mass of urea = 33.03 grams

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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

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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.

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