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anastassius [24]
3 years ago
5

Based on solubility rules what ions in water might interfere with the analysis of calcium ions by precipitation of calcium carbo

nate
Chemistry
2 answers:
Pavlova-9 [17]3 years ago
6 0
First, we must know what happens in the precipitation reaction. This type of reaction is a double replacement reactions. It is consists of two reactant compounds which interchange cations and anions to form two products. One of the products is an insoluble solid called a precipitate. For the precipitation of CaCO₃, there are two consecutive reactions involved:

1. Slaking of quicklime, CaO
    CaO + H₂O ⇒ Ca(OH)₂

2. Precipitation
    Ca(OH)₂ + CO₂ ⇒ CaCO₃ + H₂O

The ions that make up the H₂O molecule are H⁺ and OH⁻. According to solubility rules, the cation (positively charged ion) is likely to be attracted to an anion (negatively charged ion). Together, they form an ionic bond. This type of bond is when there is a complete transfer of electrons between the two. The Ca²⁺ cation lacks 2 electrons, while the anion OH⁻ has an excess 1 electron. In order to be stable, 1 Ca²⁺ ion and 2 OH⁻ ions must combine.

Therefore, the answer is OH⁻ ion.
RideAnS [48]3 years ago
4 0

The right answer is OH- (hydroxide).


Water can ionize in (H +) and (OH-). OH- can bind to calcium (the calcium of CaCl2 for example) to form Ca (OH) 2.


Ca++ + 2 OH- ==> Ca (OH) 2


Calcium hydroxide is a mineral chemical body, an ionic compound of the calcium cation and the hydroxide anion.

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During an experiment, the percent yield of calcium chloride from a reaction was 82.38%. Theoretically, the expected amount shoul
gregori [183]

Answer:

Actual yield = 86.5g

Explanation:

Percent yield = 82.38%

Theoretical yield = 105g

Actual yield = x

Equation of reaction,

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

Percentage yield = (actual yield / theoretical yield) * 100

82.38% = actual yield / theoretical yield

82.38 / 100 = x / 105

Cross multiply and make x the subject of formula

X = (105 * 82.38) / 100

X = 86.499g

X = 86.5g

Actual yield of CaCl₂ is 86.5g

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3 years ago
If 0.959J of heat is added to 0.88g of water, how much will its temperature increase
ss7ja [257]

Answer:

0.260 Celsius

Explanation:

q =c x m x (T2-T1)

c - specific heat of water 4.186 J/g.C

T2-T1 = q /(c x m) = 0.959 /(4.186 x 0.88) = 0.959/3.68 =0.260 C

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Which particle determines the nuclear charge?
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