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3 years ago

A gas has

Chemistry

2 answers:

larisa86 [58]3 years ago

5 0

The answer you're looking for would be C. no definite shape or definite volume.

Vaselesa [24]3 years ago

3 0

Gases take the shape of their container and have no definite volume because they can diffuse indefinitely. Answer C.

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A 7.028 gram sample of a sodium sulfate hydrate is heated. After driving off all the water, 3.100 grams of the anhydrous salt is

Nady [450]

Answer:

$NaSO_4.10H_2O$

Explanation:

Given that:-

Mass of the hydrated salt = 7.028 g

Mass of the anhydrous salt = 3.100 g

Mass of water eliminated = Mass of the hydrated salt - Mass of the anhydrous salt = 7.028 - 3.100 g = 3.928 g

Moles of water:

Mass of water = 3.928 g

Molar mass of $H_2O$ = 18 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles are:

$moles= \frac{3.928\ g}{18\ g/mol}$

$moles_{water}= 0.2212\ mol$

Moles of anhydrous salt:

Amount = 3.100 g

Molar mass of $NaSO_4$ = 142.04 g/mol

Thus, moles are:

$moles= \frac{3.100\ g}{142.04\ g/mol}$

$moles_{CaSO_4}= 0.02182\ mol$

The simplest ratio of the two are:

$NaSO_4:H_2O$ =0.02182 : 0.2212 = 1 : 10

Hence, the formula for hydrate is:- $NaSO_4.10H_2O$

7 0

3 years ago

The name of the compound with the formula ZnBr2 is

Brilliant_brown [7]

Zinc-bromide is its name. Since you have a metal and a non metal you don’t need to use the prefix “di” for the Br.

4 0

3 years ago

Read 2 more answers

Assuming 54.81 grams of Al are consumed in the presence of excess copper II chloride dihydrate, how many grams of AlCl3 can be p

Alchen [17]

Answer:

181.39g of AlCl3 is produced

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3CuCl2•2H2O + 2Al → 2AlCl3 + 6H2O + 3Cu

Next, we shall determine the mass of Al that reacted and the mass of AlCl3 produced from the balanced equation. This is illustrated below:

Molar mass of Al = 27g/mol

Mass of Al from the balanced equation = 2 x 27 = 54g

Molar mass of AlCl3 = 27 + (3x35.5) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Summary:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Next, we shall determine the theoretical yield of AlCl3. This can be achieved as shown below:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Therefore, 54.81g of Al will react to produce = (54.81 x 267)/54 = 271.01g of AlCl3.

Therefore, the theoretical yield of AlCl3 is 271.01g.

Finally, we shall determine the actual yield of AlCl3 produced from the reaction.

This can be obtain as follow:

Percentage yield of AlCl3 = 66.93%

Theoretical yield of AlCl3 = 271.01g

Actual yield of AlCl3 =?

Percentage yield = Actual yield/Theoretical yield x 100

66.93% = Actual yield /271.01g

Actual yield = 66.93% x 271.01

Actual yield = 66.93/100 x 271.01g

Actual yield = 181.39g.

Therefore, 181.39g of AlCl3 is produced from the reaction.

8 0

4 years ago

Elzmall.org Bookmark

soldier1979 [14.2K]

I believe the answer is PD 6

4 0

3 years ago

I will give brailiest to whoever needs it!

wolverine [178]

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3 0

3 years ago

Read 2 more answers