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3 years ago

Write the formulas for the ionic compounds formed by the following:

Chemistry

1 answer:

PSYCHO15rus [73]3 years ago

7 0

1. KI

2. AlBr₃

3. CsNO₃

4. Al₂(CO₃)₃

Explanation:

1. potassium (K⁺) iodine (I⁻) - KI

2. aluminium (Al³⁺) bromine (Br⁻) - AlBr₃

3. caesium (Cs⁺) nitrate (NO₃⁻) - CsNO₃

4. aluminum (Al³⁺) carbonate (CO₃²⁻) - Al₂(CO₃)₃

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formulas for the ionic compounds

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In a particular redox reaction, no2– is oxidized to no3– and cu2 is reduced to cu . Complete and balance the equation for this r

lianna [129]

Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).

Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.

Balanced chemical reaction:

Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).

Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).

8 0

3 years ago

Read 2 more answers

The empirical formula of a compound is CH2O and its molecular weight is 90.09 g/mol. Determine the molecular formula.

Pachacha [2.7K]

Answer : The molecular formula of the compound will be, $C_{3}H_{6}O_3$

Explanation :

Empirical formula : It is the simplest form of the chemical formula which depicts the whole number of atoms of each element present in the compound.

Molecular formula : it is the chemical formula which depicts the actual number of atoms of each element present in the compound.

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

As we are given that the empirical formula of a compound is $CH_2O$ and the molar mass of compound is, 90.09 gram/mol.

The empirical mass of $CH_2O$ = 1(12) + 2(1) + 1(16) = 30 g/eq

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

$n=\frac{90.09}{30}=3$

Molecular formula = $(CH_2O)_n=(CH_2O)_3=C_{3}H_{6}O_3$

Thus, the molecular formula of the compound will be, $C_{3}H_{6}O_3$

7 0

4 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .