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blondinia [14]
4 years ago
11

7.66 Write balanced equations for the following reactions: (a) potassium oxide with water, (b) diphosphorus trioxide with water,

(c) chromium(III) oxide with dilute hydrochloric acid, (d) selenium dioxide with aqueous potassium hydroxide
Chemistry
1 answer:
White raven [17]4 years ago
6 0

Explanation:

(a) potassium oxide with water

K_2O(s)+H_2O(l)\rightarrow 2KOH(aq)

According to reaction,1 mole of potassium  oxide reacts with 1 mole of water to give 1 mole of potassium hydroxide.

(b) diphosphorus trioxide with water

P_2O_3(s)+3H_2O(l)\rightarrow 2H_3PO_3(aq)

According to reaction,1 mole of diphosphorus trioxide reacts with 2 moles of water  to give 2 moles of phosphorus acid.

(c) chromium(III) oxide with dilute hydrochloric acid,

Cr_2O_3(s)+6HCl(aq)\rightarrow 2CrCl_3(aq)+3H_2O(l)

According to reaction,1 mole of chromium(III) oxide reacts with 6 moles of hydrochloric acid to give 2 moles of chromium(III) chloride and 3 moles of water.

(d) selenium dioxide with aqueous potassium hydroxide

SeO_2 (s)+2KOH (aq)\rightarrow K_2SeO_3(aq)+H_2O(l)

According to reaction,1 mole of selenium dioxide reacts with 2 moles of potassium hydroxide to give 1 mole of potassium selenite and 1 mole of water.

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If 17. 6 g of hcl are used to produce a 14. 5 l solution, what is the ph of the solution?.
kati45 [8]

This problem is providing us with the mass of hydrochloric acid and the volume of solution and asks for the pH of the resulting solution, which turns out to be 1.477.

<h3>pH calculations</h3>

In chemistry, one can calculate the pH of a solution by firstly obtaining its molarity as the division of the moles of solute by the liters of solution, so in this case for HCl we have:

M=\frac{17.6g*\frac{1mol}{36.46g} }{14.5L} \\\\M=0.0333 M

Next, due to the fact that hydrochloric acid is a strong acid, we realize its concentration is nearly the same to the released hydrogen ions to the solution upon ionization. Thereby, the resulting pH is:

pH=-log(0.0333)\\\\pH=1.477

Which conserves as much decimals as significant figures in the molarity.

Learn more about pH calculations: brainly.com/question/1195974

3 0
2 years ago
Which of the following cannotbe considered a single phase? a heterogeneous mixture a pure solid a homogeneous mixture a pure liq
MrRa [10]
A heterogeneous mixture cannot be considered a single phase
4 0
3 years ago
An ordered list of chemical substances is shown. Chemical Substances 1 CuO 2 O2 3 CO2 4 NO2 5 Fe 6 H2O Which substances in the l
a_sh-v [17]

Answer:

reactants: 2 O2

products: 3 CO2, 4 NO2, 6 H2O

Explanation:

In a combustion, a combustible material, which generally is composed of C, H, O, N, and S, is combusted, that is, react with oxygen after a spark was produced; obtaining fire, heat and subproducts, including ashes and gases.

Oxygen is always one of the reactants of a combustion.

If Nitrogen was present in the combustible, NO2 (or other nitrogen oxides) will be produced.

If Carbon was present in the combustible, CO2 will be produced (also CO can be produced).

If Hydrogen was present in the combustible, H2O will be produced.

7 0
4 years ago
Identify the precipitate (if any) that forms when KOH and Cu(NO3)2 are mixed
iris [78.8K]
When KOH and Cu(NO3)2 are mixed it yields copper(II) hydroxide and potassium nitrate. The balanced reaction is:

Cu(No3)2(aq) + 2 KOH(aq)<span> = Cu(OH)2(s) + 2 KNo3(aq)
</span>

As we can see from the equation a solid is formed therefore a precipitate is formed which is the copper (II) hydroxide which has a blue to purple appearance. This can be observed since copper (II) hydroxide has a low solubility in aqueous solution.

3 0
4 years ago
8g of aqueous Sodium Hydroxide reacts with 4g of aqueous Aluminum Chloride to produce aqueous Sodium Chloride and solid Aluminum
slava [35]

Answer:

The limiting reactant is AlCl₃ and the excess reactant is NaOH.

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>3NaOH(aq) + AlCl₃(aq) → 3NaCl(aq) + Al(OH)₃(s)↓,</em>

It is clear that 3.0 moles of NaOH(aq) react with 1.0 mole of AlCl₃(aq) to produce 3.0 moles of NaCl(aq) and 1.0 mole of Al(OH)₃(s).

  • Firstly, we need to calculate the no. of moles of (8.0 g) of NaOH and (4.0 g) of AlCl₃:

no. of moles of NaOH = mass/molar mass = (8.0 g)/(40.0 g/mol) = 0.2 mol.

no. of moles of AlCl₃ = mass/molar mass = (4.0 g)/(133.34 g/mol) = 0.03 mol.

  • From stichiometry; NaOH reacts with AlCl₃ with (3: 1) molar ratio.

∴ 0.09 mol of NaOH (the remaining 1.1 mol is in excess) reacts completely with 0.03 mol of AlCl₃.

  • So,

<em>the limiting reactant is AlCl₃ and the excess reactant is NaOH.</em>

<em></em>

6 0
3 years ago
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