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Shalnov [3]
3 years ago
7

Describe what is meant by physical properties Using a salt water solution , describe how one could change two different physical

properties and not alter the chemical nature ofthe salt water
Chemistry
1 answer:
kkurt [141]3 years ago
6 0

Answer: Changing the amount of solute and volume of solvent we can change physical properties of the given salt water solution.

Explanation:

Physical properties are defined as the properties which does not cause any change in chemical composition of a substance. For example, density, mass, volume, etc are all physical properties.

When we take a salt water solution then on adding more amount of salt into the water there will occur change in density of the solution as density is mass present per liter of solution.

Also, if we add more amount of solvent then there will occur change in volume of the solution.

Hence, in both the cases physical properties of the solution are changing and no change in its chemical composition is taking place.

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What is the molarity of a solution of 58.7 grams of MgCl2 in 359 ml of solution?
jekas [21]

Answer:

1.72 M

Explanation:

Molarity is the molar concentration of a solution. It can be calculated using the formula a follows:

Molarity = number of moles (n? ÷ volume (V)

According to the information provided in this question, the solution has 58.7 grams of MgCl2 in 359 ml of solution.

Using mole = mass/molar mass

Molar mass of MgCl2 = 24 + 35.5(2)

= 24 + 71

= 95g/mol

mole = 58.7g ÷ 95g/mol

mole = 0.618mol

Volume of solution = 359ml = 359/1000 = 0.359L

Molarity = 0.618mol ÷ 0.359L

Molarity = 1.72 M

6 0
3 years ago
How much electric energy does a 40 W light bulb use if it is left on for 12 hours?
OlgaM077 [116]
480 hrs of electricity
3 0
3 years ago
g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete
Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899        0.01714        0           0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714    0.01714 (end)

0.00185  | 0 → 0.01714    0.01714 (end)  

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0
3 years ago
Read 2 more answers
10 In a bond between an atom of carbon and an atom of fluorine, the fluorine atom has a
Oksanka [162]
The best answer is (2) <span>stronger attraction for electrons, for the fluorine atom has a higher electronegativity than the carbon one, if not highest of all nonmetals.
Hope this helps~</span>
4 0
3 years ago
Compare the following sodium atoms in terms of their neutrons and mass numbers 24Na, 23Na​
olga_2 [115]

Na-24 and Na-23 have the same number of atoms=11, different neutrons(13 and 12),different mass number(24 and 23)

<h3>Further explanation </h3>

The elements in nature have several types of isotopes

Isotopes have the same number of protons and having a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

  • ₁₁²³Na

number of atoms = 11

mass number = 23

number of neutrons=23-11=12

  • ₁₁²⁴Na

number of atoms = 11

mass number = 24

number of neutrons=24-11=13

7 0
3 years ago
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