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lana [24]
3 years ago
11

Write the full ionic equation and net ionic equation for sodium dihydrogen phosphate + calcium carbonate, sodium oxilate + calcl

ium carbonate, and sodium hydrogen phosphate + calcium carbonate
Chemistry
1 answer:
My name is Ann [436]3 years ago
4 0

Answer:

<em>Sodium dihydrogen phosphate + calcium carbonate</em>

<u>Full ionic equation</u>

2 Na⁺(aq) + 2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)

<u>Net ionic equation</u>

2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)

<em>Sodium oxalate + calcium carbonate</em>

<u>Full ionic equation</u>

2 Na⁺(aq) + C₂O₄²⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + CaC₂O₄(s)

<u>Net ionic equation</u>

C₂O₄²⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + CaC₂O₄(s)

<em>Sodium hydrogen phosphate + calcium carbonate</em>

<u>Full ionic equation</u>

2 Na⁺(aq) + HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + 2 Na⁺(aq) + CO₃²⁻(aq)

<u>Net ionic equation</u>

HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + CO₃²⁻(aq)

Explanation:

Let's consider two kind of equations:

  • Full ionic equation: includes all ions and species that do not dissociate in water.
  • Net ionic equation: includes only ions that participate in the reaction (<em>not spectator ions</em>) and species that do not dissociate in water.
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The ka of hypochlorous acid (hclo) is 3.0 x 10-8 at . what is the % ionization of hypochlorous acid in a 0.015 m aqueous solutio
Ahat [919]

Answer is: the % ionization of hypochlorous acid is 0.14.

Balanced chemical reaction (dissociation) of an aqueous solution of hypochlorous acid:

HClO(aq) ⇄ H⁺(aq) + ClO⁻(aq).

Ka = [H⁺] · [ClO⁻] / [HClO].

[H⁺] is equilibrium concentration of hydrogen cations or protons.

[ClO⁻] is equilibrium concentration of hypochlorite anions.

[HClO] is equilibrium concentration  of hypochlorous acid.

Ka is the acid dissociation constant.

Ka(HClO) = 3.0·10⁻⁸.

c(HClO) = 0.015 M.

Ka(HClO) = α² · c(HClO).

α = √(3.0·10⁻⁸ ÷ 0.015).

α = 0.0014 · 100% = 0.14%.

5 0
3 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
Determine Δngas for each of the following reactions:(b) 2H₂(g) + O₂(g) ⇄ 2H₂O(l)
elixir [45]

2H₂(g) + O₂(g) ⇄ 2H₂O(l)

Δngas = 0 - (2 +1)

           = -3

<h3>What is Δngas?</h3>

The number of moles of gas that move from the reactant side to the product side is denoted by the symbol ∆n or delta n in this equation.

Once more, n represents the growth in the number of gaseous molecules the equilibrium equation can represent. When there are exactly the same number of gaseous molecules in the system, n = 0, Kp = Kc, and both equilibrium constants are dimensionless.

<h3>Definition of equilibrium</h3>

When a chemical reaction does not completely transform all reactants into products, equilibrium occurs. Many chemical processes eventually reach a state of balance or dynamic equilibrium where both reactants and products are present.

Learn more about Equilibrium

brainly.com/question/11336012

#SPJ4

5 0
1 year ago
How does the type of liquid used to make ice cubes affect the rate at which it melts?
3241004551 [841]
It doesn’t affect it
5 0
3 years ago
If the temperature we're to go down how would that effect the solubility of each solute
vova2212 [387]

well when there is more kinetic energy in a solution the higher its temperature is going to be so the more the molecules are able to be help together, the lower the temperature the more difficult it would be to break the molecule apart in order to bond with it.

6 0
3 years ago
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