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3 years ago

What is the chemical name of the compound Cu(NO3)2? Use the list of polyatomic ions and the periodic table to help you answer.

Chemistry

1 answer:

vlabodo [156]3 years ago

8 0

Answer:

C. Copper (II) nitrate

Explanation:

It's an inorganic compound that forms a blue crystalline solid

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Classify each of the following as elements (E), compounds (C) or Mixtures (M). Write the letter X if it is none of these. ______

Lana71 [14]

Answer:

Diamond - E

Sugar- C

Milk - M

Iron- E

Air- M

Sulfuric Acid- C

Gasoline - M

Electricity- X

Krypton- E

Bismuth - E

Uranium - E

Kool-Aid -M

Water - C

Alcohol - C

Pail of Garbage - M

Ammonia - C

Salt - C

Energy - X

Gold - E

Wood - M

Bronze - M

Ink - M

Pizza - M

Dry Ice - C

Baking Soda - C

Explanation:

An element is the smallest part of a substance that is capable of independent existence. An element cannot be broken down into any other substance. Krypton, Bismuth etc are all elements.

A compound is a combination of two or more elements which are chemically combined together e.g NH3, CO2 etc.

A mixture is any combination of substances that are not chemically combined together. E.g Pizza, milk etc

5 0

3 years ago

An aqueous solution is 0.467 m in hcl. what is the molality of the solution if the density is 1.23 g/ml?

Reika [66]

Answer:
molarity = 0.385 moles/kg

Explanation:
Assume that the volume of the aqueous solution given is 1 liter = 1000 ml
Now, density can be calculated using the following rule:
density = mass / volume
Therefore:
mass = density * volume = 1.23 * 1000 = 1230 grams
Now, 0.467 m/L * 1L = 0.467 moles of HCl
We will get the mass of the 0.467 moles of HCl as follows:
mass = molar mass * number of moles = (1+35.5)*0.467 = 17.0455 grams
Now, we have the mass of the solution (water + HCl) calculated as 1230 grams and the mass of the HCl calculated as 17.0455 grams. We can use this information to get the mass of water as follows:
mass of water = 1230 - 17.0455 = 1212.9545 grams
Finally, we will get the molarity as follows:
molarity = number of moles of solute / kg of solution
molarity = (0.467) / (1212.9594*10^-3)
molarity = 0.385 mole/kg

Hope this helps :)

5 0

3 years ago

Read 2 more answers

Which of the following is a mixture? steel water oxygen gold

Sholpan [36]

Steel because it is a physical combination of different metals.

3 0

3 years ago

Read 2 more answers

Select the answer that completes each statement.

Law Incorporation [45]

Answer:

1 colonists

2 England

3 slaves

4 the House of Burgesses

7 0

3 years ago

Read 2 more answers

A solution contains 10.20 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL

AveGali [126]

The question is incomplete, here is the complete question:

A solution contains 10.20 g of unknown compound dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.21°C. The mass percent composition of the compound is 60.98% C , 11.94% H , and the rest is O.

What is the molecular formula of the compound?

Answer: The molecular formula for the given organic compound is $C_6H_{14}O_2$

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 50.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{50.0mL}\\\\\text{Mass of water}=(1g/mL\times 50.0mL)=50g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and the freezing point of solution

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=i\times K_f\times m$

Or,

$\text{Freezing point of pure solution}-\text{freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution (water) = 0°C

Freezing point of solution = -3.21°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal boiling point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute = 10.20 g

$M_{solute}$ = Molar mass of solute = ?

$W_{solvent}$ = Mass of solvent (water) = 50.0 g

Putting values in above equation, we get:

$(0-(-3.21))^oC=1\times 1.86^oC/m\times \frac{10.20\times 1000}{M_{solute}\times 50}\\\\M_{solute}=\frac{1\times 1.86\times 10.20\times 1000}{3.21\times 50}=118.2g$