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valkas [14]
3 years ago
8

Ammonia and sulfuric acid react to form ammonium sulfate. Determine the starting mass of each reactant if 20.3 g of ammonium sul

fate is produced and 5.89g of sulfuric acid remains unreacted.

Chemistry
1 answer:
strojnjashka [21]3 years ago
8 0
Hope you are able to see

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K₂SO₄(aq) + SrI₂(aq) → 2KI(aq)+ SrSO₄(s) net ionic equation
Oliga [24]

Answer:

SO4 (aq) + Sr (aq) >> SrSO4 (s)

Explanation:

K+1 SO4-2 + Sr +2 I2- >> sK +1

(you dont separate and write down the ions for Sr SO4 because its a solid)

Cancel out the same ions from both sides and write down what your left with.

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3 years ago
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Black_prince [1.1K]
Im confused here. Add the rest of the question for me to help you
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2 years ago
[Gradpoint-Question AK-3] Acid is added to a beaker containing crushed marble. What effect does stirring this mixture have on th
Evgen [1.6K]
A : The number of particle collision increases
4 0
3 years ago
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Is a forward or reverse reaction favored?
Alex Ar [27]

Answer:

(a) CH_3COOH \rightleftharpoons CH_3COO^-+ H^+: reverse reaction is favored.

(b) AgCl \rightleftharpoons Ag^+ + Cl^-: reverse reaction is favored.

(c) Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-: reverse reaction is favored.

(d) A+B\rightleftharpoons C: forward reaction is favored.

Explanation:

Hello,

(a) CH_3COOH \rightleftharpoons CH_3COO^-+ H^+:

In this case, since the Ka is lower than 1, we infer the reverse reaction is favored since the reactant (acetic acid) will tend to have a higher concentration.

(b) AgCl \rightleftharpoons Ag^+ + Cl^-:

In this case, since the Ka is lower than 1, we infer the reverse reaction is favored since the reactant (silver chloride) will tend to have a higher concentration.

(c) Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-

In this case, since the Ka is lower than 1, we infer the reverse reaction is favored since the reactant (aluminium hydroxide) will tend to have a higher concentration.

(d) A+B\rightleftharpoons C

In this case, since the Ka is greater than 1, we infer the forward reaction is favored since the product (C) will tend to have a higher concentration.

Regards.

6 0
3 years ago
Use the definition of molarity to calculate the concentration of 12.34 g of CaSO4 completely dissolved in water, with a solution
Ede4ka [16]

Answer:

[CaSO₄] = 36.26×10⁻² mol/L

Explanation:

Molarity (M) → mol/L → moles of solute in 1L of solution

Let's convert the volume from mL to L

250 mL . 1L/1000 mL = 0.250L

We need to determine the moles of solute. (mass / molar mass)

12.34 g / 136.13 g/mol = 0.0906 mol

M → 0.0906 mol / 0.250L = 36.26×10⁻² mol/L

8 0
3 years ago
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