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AfilCa [17]
4 years ago
6

How does molecular size affect solubility

Chemistry
1 answer:
VashaNatasha [74]4 years ago
4 0

the larger the molecules of the solute are, the larger is their molecular weight and their size.

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Proportional means that
FromTheMoon [43]
Corresponding in size or amount to something else.
5 0
3 years ago
Explain why the place of aluminum in the reactivity wagon seems to be unsuitable.
Step2247 [10]

Answer:

Acids react with most metals.

When an acid reacts with a metal, the products are a salt and hydrogen.

This is the general word equation for the reaction: metal + acid → salt + hydrogen

Explanation:

5 0
2 years ago
A 0.580 g sample of a compound containing only carbon and hydrogen contains 0.480 g of carbon and 0.100 g of hydrogen. At STP, 3
Sati [7]

Answer:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C  

(58 g . 0.100) / 0.580 = 10 g of H

 48 g of C / 12 g/mol = 4 mol

 10 g of H / 1g/mol = 10 moles

7 0
3 years ago
A sample of gas at 1.10 atm has a volume of 326 mL. What is the new volume if the pressure is changed to 1.90 atm?
Brut [27]

Answer

For this we use ideal gas equation which is:

P1V1 = P2V2

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V1 = 326 ml

P2 = 1.90

V2 = ?

By rearranging the ideal gas equation:

V2 =  P1V1 ÷ P2

V2 = 1.10 × 326 ÷1.90

V2 = 358.6 ÷ 1.90

V2 = 188.7 ml

8 0
4 years ago
Suppose 200.0 mL of a 2.50 M solution of sodium hydroxide is combined with 400.0 mL of a 1.50 M solution of iron(III) nitrate. W
-BARSIC- [3]

Answer: 83%

Explanation:

The detailed solution is shown in the image attached. First we must work out the balanced reaction equation because accurate solution of the problem must be based on the stoichiometry of the reaction. From the given concentration and volume of reactants, we calculate the amount of substance reacted hence identify the limiting reactant. Lastly we use simple proportion to obtain the theoretical yield of the precipitate. This is now used to calculate the actual yield as shown in the solution attached.

6 0
3 years ago
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