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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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A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c
gulaghasi [49]

Answer:

The enthalpy of combustion of ethanol in kJ/mol is -1419.58  kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat  capacity and the temperature change. Working with this equation, and assuming no heat is lost to  the surroundings, we write :

qcal= Ccal × ΔT= 490 J/K × 276.7 K= <u>135,583 J</u> = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

<u> Now that we have the heat of combustion, we need to calculate the molar heat.  </u>

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write  the <u>conversion factor as 135.58 kJ/ 4.40 g</u>.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58  kJ/mol.

3 0
3 years ago
1 Mothers ages 8 years more than twice her daughter's
Anit [1.1K]
<h3>♫ :::::::::::::::::::::::::::::: // Hello There ! //  :::::::::::::::::::::::::::::: ♫</h3>

➷  We can use the information to form an equation, which we can solve later

M = 2x + 8

Simply substitute the daughter's age into x:

M = 2(56) + 8

Now solve:

M = 120

<h3><u>❄️</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

8 0
3 years ago
Please help! Please try and give me a long answer. Thanks!
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I found this for your question

5 0
3 years ago
CH3COOH
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Answer:

the water which are very healthy is known as hydrated water

8 0
3 years ago
A sample originally contained 1.28 g of a radioisotope. It now contains 1.12 g of its daughter isotope.
dexar [7]
The answer is 3.

<span>The relation between number of half-lives (n) and decimal amount remaining (x) can be expressed as:

</span>(1/2) ^{n} =x

We need to calculate n, but we need x to do that. To calculate what p<span>ercentage of a radioactive species would be found as daughter material, we must calculate what amount remained:
1.28 -</span> 1.12 = 0.16

If 1.28 is 100%, how much percent is 0.16:
1.28 : 100% = 0.16 : x
x = 12.5% 
Presented as decimal amount:
x = 0.125


Now, let's implement this in the equation: 

<span>(1/2) ^{n} =0.125
</span>
Because of the exponent, we will log both sides of the equation:
n * log(1/2) = log(0.125)
n = \frac{log(0.125)}{log(1/2)}
<span>n = \frac{log(0.125)}{log(0.5)}
</span>n= \frac{-0.903}{-0.301}
n = 3

Therefore, 3 half-lives have passed <span> since the sample originally formed.</span>
4 0
3 years ago
Read 2 more answers
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