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ehidna [41]
2 years ago
8

When aqueous solutions of potassium cyanide and hydrobromic acid are mixed, an aqueous solution of potassium bromide and hydrocy

anic acid results. Write the net ionic equation for the reaction.
Chemistry
1 answer:
mrs_skeptik [129]2 years ago
8 0

Answer:

hey there

Explanation:

CN- (aq) + H+ (aq) → HCN(I)

The reactants are aqueous solutions:

NaCN(aq) and HBr(aq)

When you mix these compounds you make pure HCN (I)

The molecular equation is:

NaCN(aq) + HBr(aq) → NaBr(aq) + HCN(I)

When you dissociate the reactants, you have: Nat(aq) +CN¯(aq) + H*(aq) + Br−(aq) → Nat(aq) + ->

Br (aq) + HCN(I)

Sodium bromide, it is a salt, that can also be

dissociated in the solution

To make, the net ionic equation you remove the repeated ions

CN- (aq) + H+ (aq) → HCN(I)

If you like my answer please mark as brainliest

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What the sample will look like after the temperature is increased from 300 K to 450 KK at constant pressure?
sleet_krkn [62]

Answer:

The sample will look expanded and occupy more space.

Explanation:

Since, the pressure is constant here, but the temperature is changed. Therefore, according to Charles' Law Volume is directly proportional to Temperature, provided the pressure is kept constant. Mathematically:

V1/T1 = V2/T2

V1 = (T1/T2)(V2)

V1 = (300 k/450 k)(V2)

<u>V1 = (0.67)V2</u>

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Lead is listed below zinc in the activity series.
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Read 2 more answers
A mixture containing nitrogen and hydrogen weighs 3.49 g and occupies a volume of 7.45 L at 305 K and 1.03 atm. Calculate the ma
-BARSIC- [3]

Answer:

Mass percent N₂ = 89%

Mass percent H₂ = 11%

Explanation:

First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:

  • 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
  • n = 0.307 mol

So now we know that

  • MolH₂ + MolN₂ = 0.307 mol

and

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g

So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:

Express MolH₂ in terms of MolN₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ = 0.307 - MolN₂

Replace that value in the second equation:

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
  • (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
  • 0.614 - 2MolN₂ + 28molN₂ = 3.49
  • 0.614 + 26MolN₂ = 3.49
  • MolN₂ = 0.111 mol

Now we calculate MolH₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ + 0.111 = 0.307
  • MolH₂ = 0.196 mol

Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:

  • N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
  • H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂

Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%

Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%

5 0
3 years ago
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