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2 years ago

An example of polytamic anoin would be a. Ca+2 b. NH4+ c. F d. PO4-3

Chemistry

1 answer:

ella [17]2 years ago

8 0

$PO_4^{-3}$ and $NH^{4+}$ are polyatomic anions composed of two or more atoms.

<h3>What is a polyatomic anion?</h3>

A polyatomic ion is an ion composed of two or more atoms. Examples: The hydroxide cation ( $OH^-$ ) and the phosphate cation ( $PO_4^{3-}$ ) are both polyatomic ions.

$PO_4^{-3}$ and $NH^{4+}$ are the polyatomic anions because both contain more than 1 atom.

Hence, $PO_4{-3}$ and $NH^{4+}$ are polyatomic anions composed of two or more atoms.

Learn more about the polyatomic anion here:

brainly.com/question/13544727

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Given the balanced equation: 2KClO₃ ---> 2KCl + 3O₂ How many moles of O₂ are produced when 4.0 moles of KCl are produced?

Verizon [17]

Answer:

The answer to your question is 6.0 moles of O₂

Explanation:

Data

2KClO₃ ⇒ 2KCl + 3O₂

moles of O₂ = ?

moles of KCl = 4

Process

To find the number of moles of O₂, use proportions and cross multiplication.

Use the coefficients of the balanced equation.

2 moles of KCl ----------------- 3 moles of O₂

4 moles of KCl ----------------- x

x = (4 x 3) / 2

-Simplification

x = 12/2

-Result

x = 6 moles of O₂

-Conclusion

When 4,0 moles of KCl are produced, 6.0 moles of O₂ will be produced.

6 0

3 years ago

Pre-lab questions 1. a concentration gradient affects the direction that solutes diffuse. describe how molecules move with respe

VLD [36.1K]

Just like how heat moves from a region of higher temperature to a region of lower temperature, molecules also tend to move from a region of higher concentration to a region of lower concentration. This is called natural diffusion and is naturally happening to reach stability.

7 0

3 years ago

Gas laws describe and predict the behavior of gases without attempting to explain why they happen

Nutka1998 [239]

The gas laws describe and predict the behavior of gases with an explanation and experimental data

So the given statement is False.

2) The volume of gas can be calculated based on Avagadro's law

It states that the volume of a gas is directly proportional or varies with the moles of the gas. Higher the moles more the volume, condition is the pressure and temperature are constants in the two conditions

Thus as here the pressure and temperature of nitrogen gas is kept constant

V α moles

$\frac{V1}{n1}=\frac{V2}{n2}$

Where

V1 = 6 l

n1 = 0.50 mol

V2 = ?

n2 = 0.75 mol

On putting values

V2 = 6 X 0.75 / 0.5 = 9 L

so resulting volume of the gas will be 9L

5 0

3 years ago

Read 2 more answers

The measurement of grams of solute per one million grams of solution is known as

Zinaida [17]

I think it's Molality because the moles of solute over the kilograms of solvent is equal to the molality...

5 0

3 years ago

A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to

melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0

2 years ago