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3 years ago

In an ionic compound a

Chemistry

1 answer:

kirza4 [7]3 years ago

7 0

Answer:

In an ionic compound a metal is bonded to a nonmetal.

The element in the compound that has a positive charge is called the cation

while the element with the negative charge is called the anion.

Explanation:

In an ionic compound a metal is bonded to a nonmetal.

The element in the compound that has a positive charge is called the cation

while the element with the negative charge is called the anion.

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3 years ago

What are the hypotheses on which Dalton's atomic theory is based?

worty [1.4K]

Answer:

Dalton's theory are based on the two laws that are: Law of conservation of mass and law of constant composition. This theory basically described their properties of atoms.

This theory state that all the atoms are made up of matter which are invisible and in the elements all the atoms are identical in mass and properties.

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3 years ago

What is the proper formula for vanadium (iv) percholrate?

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Explanation:

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3 years ago

An unknown compound in the lab is either LiOH, CaF2, or LiF. If the sample has a mass of 260 grams and contains 3.33 moles, what

sergij07 [2.7K]

Answer:

CaF2

Explanation:

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3 years ago

Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670

Aliun [14]

Answer : The empirical formula of the compound is $C_3H_8O_2$

Explanation :

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.043g$

Mass of $H_2O=0.5670g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 1.043 g of carbon dioxide, $\frac{12}{44}\times 1.043=0.284g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.5670 g of water, $\frac{2}{18}\times 0.5670=0.063g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.600) - (0.284 + 0.063) = 0.253 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.284g}{12g/mole}=0.0237moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.063g}{1g/mole}=0.063moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.253g}{16g/mole}=0.0158moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0158 moles.

For Carbon = $\frac{0.0237}{0.0158}=1.5$

For Hydrogen = $\frac{0.063}{0.0158}=3.98\approx 4$

For Oxygen = $\frac{0.0158}{0.0158}=1$

The ratio of C : H : O = 1.5 : 4 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 3 : 8 : 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 8 : 2

Hence, the empirical formula for the given compound is $C_3H_8O_2$

6 0

3 years ago