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4 years ago

How do people use bromine?

1 answer:

6 0

Bromine is used in many areas such as the following agricultural chemicals, dyestuffs, insecticides, pharmaceuticals and chemical intermediates. Some of the uses are being phased out for environmental reasons, but new uses can be found with Bromine can be used for furniture foam, plastic casings for electronics and textiles to make them less flammable

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A sample of CO2 gas at 100 degrees Celsius has a volume of 250 mL at 760 mm Hg. How many moles of CO2 are present

Fed [463]

There are 8.16 × 10-³ moles of CO2 gas at 100°C with a volume of 250 mL at 760 mm Hg.

HOW TO CALCULATE NUMBER OF MOLES:

The number of moles of a sample of gas can be calculated using the following formula:

PV = nRT

Where;

P = pressure of gas (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)

According to this question;

P = 760mmHg = 1 atm
T = 100°C = 100 + 273 = 373K
V = 250mL = 0.250L
n = ?

1 × 0.250 = n × 0.0821 × 373

0.250 = 30.62n

n = 0.250 ÷ 30.62

n = 8.16 × 10-³mol

Therefore, there are 8.16 × 10-³ moles of CO2 gas at 100°C with a volume of 250 mL at 760 mm Hg.

Learn more about number of moles at: brainly.com/question/4147359

6 0

3 years ago

How is the molar heat of sublimation related to the molar heats of vaporization and fusion? On what law are these relationships

Dovator [93]

Answer: Hsub=Hfus+Hvap

Explanation:

The molar heat of vaporization measured in kilojoules per mole, or kJ/mol is the energy needed to make vapor one mole of a liquid. .

The molar heat of sublimation measured in kilojoules per mole, or kJ/mol is the energy needed to sublime one mole of a solid,

the molar heat of fusion measured in kilojoules per mole, or kJ/mol is the energy needed to melt one mole of a solid.

Hess law helps to explain the relationship in physical chemistry stating that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps.

In this context Hess’s law helps to see the several steps involved as the heat of sublimation energy is equal to the sum of vaporization energy and fusion energy.

3 0

3 years ago

What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa? Ka = 1.8×10-5

Westkost [7]

Answer:

a) We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

b) We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

c) We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

Explanation:

a) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = pK = -log(1.8*10^-5) = 4.74

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4.74 = 4.74 + log(A-/HA)

0 = log(A-/HA)

A-/HA = 1

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1

X =0.9

We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4.74 = 4.74 + log(0.9/0.9) = 4.74

b) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 4

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4 = 4.74 + log(A-/HA)

-0.74 = log(A-/HA)

A-/HA = 0.182

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 0.182

X =0.277

We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4 = 4.74 + log(0.277/1.523)

c) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 5

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

5 = 4.74 + log(A-/HA)

0.26 = log(A-/HA)

A-/HA = 1.82

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1.82

X =1.16

We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

5 = 4.74 + log(1.16/0.64) = 5

3 0

4 years ago

Calculate the poH of a solution at 25°C which contains 1 x 10^-10 M of hydronium ions.

GenaCL600 [577]

Answer:

4.000 hope this helps :)

4 0

3 years ago

1. What are some similarities between the dogs pictured? What are some differences?

kogti [31]

1. Some differences are that they are all different kinds of dogs

2. Small dogs and big dogs

That is all I can answer

5 0

3 years ago

Read 2 more answers