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3 years ago

12

Q2

1 answer:

sergey [27]3 years ago

5 0

Answer: 1.00 M

Explanation:

Moles of Solute/Liters of Solution = molarity. Then you would do 0.500 Moles/ 0.500 L and you get 1.00 M.

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When 8.00x10^22 molecules of ammonia react with 7.00x10^22 molecules of oxygen according to the chemical equation shown below, h

Pavel [41]

NH₃:

N = 8*10²²
NA = 6.02*10²³

n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol

O₂:

N=7*10²²
NA = 6.02*10²³

n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol

4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant

N₂:

n = 0.0665mol
M = 28g/mol

m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>

6 0

3 years ago

Use the equation editor or "Insert Chemistry - WIRIS editor" to write the balanced molecular chemical equation for the reaction

11Alexandr11 [23.1K]

Answer:

<u>Balanced equation:</u>

$Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)$

Explanation:

The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.

$Lead(II)Nitrate+Potassium\,carbonate \rightarrow Lead(III)\,\,carbonate+Potassium\,nitrate$

$Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)$

<u>Ionic equation:</u>

$Pb^{2+}(aq)+2NO_{3}^{-}(aq)+2K^{+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)+K^{+}(aq)+2NO_{3}^{-}$

Cancel the same ions on the both sides of the reaction.

The net ionic equation is as follows.

$Pb^{2+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)$

4 0

3 years ago

Given the following information: benzoic acid = C6H5COOH hydrocyanic acid = HCN C6H5COOH is a stronger acid than HCN (1) Write t

Maurinko [17]

Answer:

The net ionic equation is

C6H5COOH+ CN-= C6H5COO- + HCN

Explanation:

From the ionic equation

C6H5COOH + Na+ + CN- = C6H5COO- + Na+ + HCN

Only sodium is the spectator ion, so it cancels out, since C6H5COOH and HCN do not ionize completely they are left undissociated

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3 years ago

The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?

Mazyrski [523]

Last option that is none of above is right answer.

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In a colloid, solution, or suspension, particles are dispersed throughout the mixture. What is the order of these three types of

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1) solution
2) colloid
3) suspension

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