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3 years ago

What are the steps to balance a chemical equation?

Chemistry

1 answer:

aleksandrvk [35]3 years ago

5 0

Answer:

Step 1: The Unbalanced Chemical Equation. The unbalanced chemical equation is given to you. ...

Step 2: Make a List. ...

Step 3: Identifying the Atoms in Each Element. ...

Step 4: Multiplying the Number of Atoms. ...

Step 5: Placing Coefficients in Front of Molecules. ...

Step 6: Check Equation. ...

Step 7: Balanced Chemical Equation.

Explanation:

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2. What is an example of a molecule with different types of atoms?

agasfer [191]

Answer:

a compound

Explanation:

A molecule with multiple atoms is compound like water it's made up of 1 oxygen and two hydrogen.

4 0

3 years ago

Which of these compounds would you expect

Veseljchak [2.6K]

ClBr, two nonmetals

Hope this helps you

6 0

2 years ago

How does changing the concentration of the reactants change the parts of a rate law? It changes the rate, R. It changes the rate

Sidana [21]

Answer

A. It changes the rate, R

Explanation

When we change the concentration of the reactants in a chemical reaction, it affects the rate of reaction that happens in the process. Typically, the rate of reaction will decrease with time if the concentration of the reactants decreases because the reactants will be converted to products. Similarly, the rate of reaction will increase when the concentration of reactants are increased.

4 0

3 years ago

What is the balanced NET ionic equation for the reaction when aqueous Cs₃PO₄ and aqueous AgNO₃ are mixed in solution to form sol

rusak2 [61]

Answer:

$PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)$

Explanation:

Hello!

In this case, since the net ionic equation of a chemical reaction shows up the ionic species that result from the simplification of the spectator ions, which are those at both reactants and products sides, we take into account that aqueous species ionize into ions whereas liquid, solid and gas species remain unionized. In such a way, for the reaction of cesium phosphate and silver nitrate we can write the complete molecular equation:

$Cs_3PO_4(aq)+3AgNO_3(aq)\rightarrow Ag_3PO_4(s)+3CsNO_3(aq)$

Whereas the three aqueous salts are ionized in order to write the following complete ionic equation:

$3Cs^+(aq)+PO_4^{3-}(aq)+3Ag^+(aq)+3NO_3^-(aq)\rightarrow Ag_3PO_4(s)+3Cs^+(aq)+3NO_3^-(aq)$

In such a way, since the cesium and nitrate ions are the spectator ions because of the aforementioned, the net ionic equation turns out:

$PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)$

Best regards!

7 0

2 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

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 $CO_2=13.00g$

Mass of $H_2O=2.662g$

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 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

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 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

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 :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago