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sergey [27]
3 years ago
7

What must be true for a substance to be able to dissolve in water?

Chemistry
1 answer:
zvonat [6]3 years ago
3 0

In order for a solute to dissolve in a solvent, the attractive forces between solute particles and the solvent particles must be stronger than the attractive forces between solute-solute and solvent-solvent particles. This is important so that the solute will remain in solution.

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Sodium sulphate reacts with calcium nitrate to produce sodium nitrate and calcium sulphate. A) identify the reactants and produc
Rudik [331]

Answer:

yes ...................

6 0
3 years ago
How many moles of H2O form when 4.5 moles O2 reacts?
mart [117]

Explanation:

Start with a balanced equation.

2H2 + O2 → 2H2O

Assuming that H2 is in excess, multiply the given moles H2O by the mole ratio between O2 and H2O in the balanced equation so that moles H2O cancel.

5 mol H2O × (1 mol O2/2 mol H2O) = 2.5 mol O2

Answer: 2.5 mol O2 are needed to make 5 mol H2O, assuming H2 is in excess.

4 0
3 years ago
Calculate the density of sulfuric acid if 35.4 mL of the acid weighs 65.14 g
Nadya [2.5K]
M/V=D
65.14/35.4≈1.84
The density of the sulfuric acid would be about 1.84g/mL
5 0
3 years ago
A salt forms in the reaction of barium with chlorine. What is the most likely formula unit of this salt?
Scrat [10]
The answer to this question would be: BaCl 2
Barium is an alkali metal with 56 atomic number. Barium located in the group 2 of the periodic table because it has 2 valence electrons. Chlorine is a nonmetal that has 1 valence electron. When react, it would need 2 chlorine for each barium as the valence electron of barium is twice the chlorine. 
6 0
3 years ago
Read 2 more answers
What is the volume of 3.00 M sulfuric aid that contain 9.809 g of H2SO4 (98.09g/mol)
Slav-nsk [51]

Given :

Molarity of sulfuric acid solution is 3.0 M.

Amount of sulfuric acid present in solution is 9.809 g.

To Find :

The volume of solution.

Solution :

We know, molarity is given by :

Molarity = \dfrac{number \ of \ moles \times 1000}{Volume\ ( ml )}\\\\M = \dfrac{w \times 1000}{M.M \times V}\\\\3 = \dfrac{9.809\times 1000}{98.09 \times V}\\\\V = \dfrac{1000}{10\times 3}\  ml\\\\V = 33.33 \ ml

Therefore, volume required is 33.33 ml .

5 0
3 years ago
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