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Ilya [14]
2 years ago
11

what is the net ionic equation with its physical states? (NH4)2CO3(aq)+Ca(ClO4)2(aq)⟶CaCO3(s)+2NH4ClO4(aq)

Chemistry
1 answer:
Temka [501]2 years ago
6 0

Answer: The net ionic equation is CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

(NH_4)_2CO_3(aq)+Ca(ClO_4)_2(aq)\rightarrow CaCO_3(s)+2NH_4ClO_4(aq)

The equation can be written in terms of ions as:

2NH_4^+(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2ClO_4^{-}(aq)\rightarrow CaCO_3(s)+2NH_4^{+}(aq)+2ClO_4^-(aq)

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

The ions which are present on both the sides of the equation are ammonium and chlorate ions and hence are not involved in net ionic equation.

Hence, the net ionic equation is :

CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)

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Which of the following statements is true? (Multiple answers possible) Group of answer choices:
Sophie [7]

Answer:The endpoint does not correspond exactly to the equivalence point

At the endpoint, a change in a physical quantity associated with the equivalence point occurs.

At the equivalence point, the mole number of equivalents of reagent added is equal to the mole number of equivalents of analyte present.

Explanation:

The end point is always indicated by some physical property that changes such as colour. At the equivalence point, the mole number of equivalents of reagent added is equal to the mole number of equivalents of analyte present. The equivalence point cannot be physically observed but can be deduced after a titration curve is plotted.

3 0
2 years ago
Consider the following thermochemical equation: C(s) + O2(g) → CO2(g) ΔH = −393 kJ CO(g) + ½O2(g) → CO2(g) ΔH = −294 kJ What is
Oksi-84 [34.3K]

Answer:

B. –99 kJ.

Explanation:

We have the following information:

1. C(s) + O₂(g) → CO₂(g);

ΔH = -393 kJ

2. 2CO(g) + O₂ → 2CO₂(g);

ΔH = -588 kJ

Using Hess's Law, Our target equation has C(s) on the left hand side, so we re-write equation 1:

1. C(s) + O₂(g) → CO₂(g);

ΔH = -393 kJ

So, we reverse equation 2 and divide by 2, we have equation 3:

3. CO₂(g) → CO(g) + ½O₂;

ΔH = +294 kJ

That is, change the sign of ΔH and divide by 2. Then we add equations 1 and 3 and their ΔH values.

This gives:

C(s) +½O₂(g) → CO(g);

ΔH = +294 - 393 kJ

= -99 kJ

The standard enthalpy of formation of carbon monoxide is -99 kJ/mol.

5 0
2 years ago
What is the conjugate acid of ClCH2CO2-?
Burka [1]
A conjugate acid<span>, within the </span>Brønsted–Lowry acid–base theory<span>, is a </span>species<span> formed by the </span>reception of a proton<span> (</span><span>H+</span><span>) by a </span>base<span>—in other words, it is a base with a </span>hydrogen<span> ion added to it. The conjugate acid would be as follows:

</span><span>Cl = HCl
CH2 = </span>H2O<span>
CO2- = </span><span>HCO3(1-)</span>
6 0
2 years ago
What is the mass % of acetonitrile in a 2.17 M solution of acetonitrile (MM = 41.05 g/mol) in water? The density of the solution
jonny [76]

why how what what what

5 0
2 years ago
Which elements do not strictly follow the octet rule when they appear in the Lewis structure of a molecule?
Virty [35]

Thee question is incomplete; the complete question is;

Which elements do not strictly follow the octet rule when they appear in the Lewis structure of a molecule?

Select one or more:

A: Chlorine

B: Carbon

C: Hydrogen

D: Sulfur

E: Fluorine

F: Oxygen

Answer:

chlorine

sulphur

Explanation:

The octet rule states that, for  atoms to be stable, they must have eight electrons on their outermost shells.

This rule is not strictly followed  by some elements such as sulphur and chlorine. The atoms of these elements can sometimes expand their octet by utilizing the d-orbitals found in the third principal energy level and beyond.

These leads to formation of compounds in which the central atom has more than eight electrons in its outermost shell.

5 0
2 years ago
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