Write a balanced equation for (a) the reaction between fluorine gas and water to give oxygen difluoride and hydrogen fluoride gases. (b) the reaction between oxygen and ammonia gases to give nitrogen dioxide gas and water. (c) the burning of gold(III) sulfide in hydrogen to give gold metal and dihydrogen sulfide gas. (d) the decomposition of sodium hydrogen carbonate to sodium carbonate, water, and carbon dioxide gas. (e) the reaction between sulfur dioxide gas and liquid hydrogen fluoride to give sulfur tetrafluoride gas and water.

In this reaction, the reactants are fluorine gas (F2) and water (H2O). The products are oxygen difluoride (OF2) and hydrogen fluoride (HF).

F2 + H2O -> OF2 + HF

To balance the equation, we need to have equal numbers of each type of atom on both sides of the equation. Let's start with fluorine. Since there are two fluorine atoms in the reactants (F2) and three in the products (OF2 + 2HF), we need to multiply F2 by 3 and HF by 2. 3F2 + H2O -> OF2 + 2HF Now, we check the hydrogen atoms. There are two hydrogens in H2O and two in 2HF, so the number of hydrogen atoms is balanced. Furthermore, there is only one oxygen atom in both H2O and OF2, so the number of oxygen atoms is also balanced. Thus, the balanced equation is: 3F2 + H2O -> OF2 + 2HF (b) Reaction between oxygen and ammonia gases to give nitrogen dioxide gas and water.

In this reaction, the reactants are oxygen gas (O2) and ammonia (NH3). The products are nitrogen dioxide (NO2) and water (H2O).

O2 + NH3 -> NO2 + H2O

We'll start by balancing the nitrogen atoms. There is one nitrogen atom in NH3 and one in NO2, but we have two nitrogen atoms (N2) in the reactant, so we need to multiply NO2 by 2. O2 + NH3 -> 2NO2 + H2O Now, let's balance the hydrogen atoms. We have three hydrogen atoms in NH3 and two in H2O, so we'll need to multiply NH3 by 2 and H2O by 3. O2 + 2NH3 -> 2NO2 + 3H2O Finally, let's balance the oxygen atoms. There are two oxygen atoms in O2, four in 2NO2, and three in 3H2O. We need to multiply O2 by 5/2 to balance the equation. (5/2)O2 + 2NH3 -> 2NO2 + 3H2O However, fractional coefficients are not typically used in chemical equations. Instead, we multiply the entire equation by 2 to eliminate the fractional coefficient. 4O2 + 4NH3 -> 4NO2 + 6H2O (c) Burning of gold(III) sulfide in hydrogen to give gold metal and dihydrogen sulfide gas.

In this reaction, the reactants are gold(III) sulfide (Au2S3) and hydrogen gas (H2). The products are gold metal (Au) and dihydrogen sulfide gas (H2S).

Au2S3 + H2 -> Au + H2S

Starting with the gold atoms, we have two in the reactants and one in the products, so we need to multiply Au by 2. Au2S3 + H2 -> 2Au + H2S Next, we'll balance the sulfur atoms. We have three sulfur atoms in Au2S3 and one in H2S, so we must multiply H2S by 3. Au2S3 + H2 -> 2Au + 3H2S Finally, we'll balance the hydrogen atoms. There are two hydrogen atoms in H2 and six in 3H2S. We need to multiply H2 by 3 to balance the equation. Au2S3 + 3H2 -> 2Au + 3H2S (d) Decomposition of sodium hydrogen carbonate to sodium carbonate, water, and carbon dioxide gas.

In this reaction, the reactant is sodium hydrogen carbonate (NaHCO3). The products are sodium carbonate (Na2CO3), water (H2O), and carbon dioxide gas (CO2).

NaHCO3 -> Na2CO3 + H2O + CO2

Start by balancing the sodium atoms. There is one sodium atom in NaHCO3 and two in Na2CO3, so we need to multiply NaHCO3 by 2. 2NaHCO3 -> Na2CO3 + H2O + CO2 Now, let's balance the hydrogen atoms. There are two hydrogen atoms in 2NaHCO3 and two in H2O, so the number of hydrogen atoms is balanced. Furthermore, there are two carbon atoms in 2NaHCO3, one in Na2CO3, and one in CO2, so the number of carbon atoms is also balanced. Lastly, there are six oxygen atoms in 2NaHCO3, three in Na2CO3, one in H2O and two in CO2, so the number of oxygen atoms is balanced. Thus, the balanced equation is: 2NaHCO3 -> Na2CO3 + H2O + CO2 (e) Reaction between sulfur dioxide gas and liquid hydrogen fluoride to give sulfur tetrafluoride gas and water.

In this reaction, the reactants are sulfur dioxide gas (SO2) and liquid hydrogen fluoride (HF). The products are sulfur tetrafluoride gas (SF4) and water (H2O).

SO2 + HF -> SF4 + H2O

First, let's balance the sulfur atoms. We have one sulfur atom in SO2 and one in SF4, so the number of sulfur atoms is balanced. Next, we'll balance the oxygen atoms. There are two oxygen atoms in SO2 and one in H2O. Therefore, we need to multiply H2O by 2. SO2 + HF -> SF4 + 2H2O Lastly, we'll balance the fluorine and hydrogen atoms. We have one fluorine atom in HF and four in SF4. Therefore we need to multiply HF by 4. SO2 + 4HF -> SF4 + 2H2O Now, the equation is balanced.