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olganol [36]
3 years ago
7

Name the five methods used for the formation of salts??

Chemistry
1 answer:
Viefleur [7K]3 years ago
5 0

Explanation:

In chemistry, a salt is a chemical compound consisting of an ionic assembly of cations and anions.[1] Salts are composed of related numbers of cations (positively charged ions) and anions (negatively charged ions) so that the product is electrically neutral (without a net charge). These component ions can be inorganic, such as chloride (Cl−), or organic, such as acetate (CH

3CO−

2); and can be monatomic, such as fluoride (F−) or polyatomic, such as sulfate (SO2−

4).

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PLEASE ANSWER ASAP. Which of the following represents the chemical reaction of hydrobromic acid and chlorine to produce hydrochl
Anvisha [2.4K]

Answer:

b i think

Explanation:

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3 years ago
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Based on solubility rules what ions in water might interfere with the analysis of calcium ions by precipitation of calcium carbo
Pavlova-9 [17]
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.
6 0
3 years ago
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SOLVE The density of a gas is 0.68 g/mL. What amount of space would 23.8 g of this gas occupy?
Contact [7]

Answer:

<h2>The answer is 35 mL</h2>

Explanation:

Density of a substance can be found by using the formula

Density(\rho) =  \frac{mass}{volume}

From the question we are finding the amount of space the gas will occupy that's the volume of the gas

Making volume the subject we have

volume =  \frac{mass}{Density}

From the question

mass = 23.8 g/mL

Density = 0.68 g/mL

Substitute the values into the above formula and solve

That's

volume =  \frac{23.8}{0.68}

We have the final answer as

<h3>35 mL</h3>

Hope this helps you

6 0
3 years ago
When heat is applied to 80 grams of CaCO3, it yields 39 grams of B. Determine the percentage of the yield.
EleoNora [17]

The question is incomplete, the complete question is:

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CaO Determine the percentage of the yield.

CaCO3→CaO + CO2

<u>Answer:</u> The % yield of the product is 87.05 %

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

We are given:

Given mass of CaCO_3 = 80 g

Molar mass of CaCO_3 = 100 g/mol

Putting values in equation 1, we get:

\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8mol

For the given chemical reaction:

CaCO_3\rightarrow CaO+CO_2

By stoichiometry of the reaction:

If 1 mole of CaCO_3 produces 1 mole of CaO

So, 0.8 moles of CaCO_3 will produce = \frac{1}{1}\times 0.8=0.8mol of CaO

We know, molar mass of CaO = 56 g/mol

Putting values in above equation, we get:

\text{Mass of CaO}=(0.8mol\times 56g/mol)=44.8g

The percent yield of a reaction is calculated by using an equation:

\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100 ......(2)

Given values:

Actual value of the product = 39 g

Theoretical value of the product = 44.8 g

Plugging values in equation 2:

\% \text{yield}=\frac{39 g}{44.8g}\times 100\\\\\% \text{yield}=87.05\%

Hence, the % yield of the product is 87.05 %

6 0
3 years ago
Sodium electron configuration
Alex73 [517]

Full:

1s² 2s² 2p⁶ 3s¹

Abbreviated:

[Ne] 3s¹

4 0
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