18. this is a chemical that reflects heat from the sun, warming the planet:\na) greenhouse gas\nb) noble…

18. Multiple - choice question

Brief Explanations:

• Greenhouse Gas: Greenhouse gases (such as (CO_2), (CH_4) etc.) trap heat (not reflect heat) from the sun, warming the planet.
• Noble Gas: Noble gases are generally inert and do not play a role in warming the planet by reflecting or trapping heat in the context described.
• Reactive Alkali Element: Reactive alkali elements (like (Na), (K) etc.) are not chemicals that reflect heat from the sun to warm the planet.

Answer:

D. None of the above

19. Balancing (Fe + Cl_2=FeCl_3)

Step - by - Step Format:

Explanation:

Step1: Balance Cl atoms

The left - hand side has (2) Cl atoms ((Cl_2)) and the right - hand side has (3) Cl atoms ((FeCl_3)). The least common multiple of (2) and (3) is (6). So, we put a coefficient of (3) in front of (Cl_2) and (2) in front of (FeCl_3): (Fe+3Cl_2 = 2FeCl_3)

Step2: Balance Fe atoms

Now, on the right - hand side, we have (2) Fe atoms. So, we put a coefficient of (2) in front of (Fe) on the left - hand side.

Answer:

(2Fe + 3Cl_2=2FeCl_3)

20. Balancing (Fe + O_2 = Fe_2O_3)

Step - by - Step Format:

Explanation:

Step1: Balance O atoms

The left - hand side has (2) O atoms ((O_2)) and the right - hand side has (3) O atoms ((Fe_2O_3)). The least common multiple of (2) and (3) is (6). So, we put a coefficient of (3) in front of (O_2) and (2) in front of (Fe_2O_3): (Fe + 3O_2=2Fe_2O_3)

Step2: Balance Fe atoms

Now, on the right - hand side, we have (4) Fe atoms. So, we put a coefficient of (4) in front of (Fe) on the left - hand side.

Answer:

(4Fe+3O_2 = 2Fe_2O_3)

21. Balancing (NH_3+O_2 = NO + H_2O)

Step - by - Step Format:

Explanation:

Step1: Balance H atoms

Let's first balance the H atoms. There are (3) H atoms in (NH_3) and (2) H atoms in (H_2O). The least common multiple of (3) and (2) is (6). So, we put a coefficient of (2) in front of (NH_3) and (3) in front of (H_2O): (2NH_3+O_2=NO + 3H_2O)

Step2: Balance N atoms

Since we have (2) N atoms on the left ((2NH_3)), we put a coefficient of (2) in front of (NO): (2NH_3+O_2 = 2NO+3H_2O)

Step3: Balance O atoms

On the right - hand side, we have (2 + 3=5) O atoms. On the left - hand side, we have (2) O atoms ((O_2)). We multiply the entire equation by (2) to get rid of the fraction. So, (4NH_3+5O_2=4NO + 6H_2O)

Answer:

(4NH_3+5O_2=4NO + 6H_2O)

22. Balancing (Na+Cl_2 = NaCl)

Step - by - Step Format:

Explanation:

Step1: Balance Cl atoms

The left - hand side has (2) Cl atoms ((Cl_2)) and the right - hand side has (1) Cl atom ((NaCl)). So, we put a coefficient of (2) in front of (NaCl): (Na+Cl_2=2NaCl)

Step2: Balance Na atoms

Now, on the right - hand side, we have (2) Na atoms. So, we put a coefficient of (2) in front of (Na) on the left - hand side.

Answer:

(2Na+Cl_2 = 2NaCl)

23. Balancing (N_2+H_2=NH_3)

Step - by - Step Format:

Explanation:

Step1: Balance N atoms

The left - hand side has (2) N atoms ((N_2)) and the right - hand side has (1) N atom ((NH_3)). So, we put a coefficient of (2) in front of (NH_3): (N_2+H_2=2NH_3)

Step2: Balance H atoms

Now, on the right - hand side, we have (6) H atoms. So, we put a coefficient of (3) in front of (H_2) on the left - hand side.

Answer:

(N_2 + 3H_2=2NH_3)

24. Balancing (Zn+HCl=ZnCl_2+H_2)

Step - by - Step Format:

Explanation:

Step1: Balance Cl atoms

The left - hand side has (1) Cl atom ((HCl)) and the right - hand side has (2) Cl atoms ((ZnCl_2)). So, we put a coefficient of (2) in front of (HCl): (Zn + 2HCl=ZnCl_2+H_2)

Step2: Check other atoms

Zn is balanced ((1) on each side), and H is also balanced ((2) on each side)

Answer:

(Zn+2HCl=ZnCl_2 + H_2)

25. Balancing (N_2+O_2+H_2O=HNO_3)

Step - by - Step Format:

Explanation:

Step1: Balance N atoms

The left - hand side has (2) N atoms ((N_2)) and the right - hand side has (1) N atom ((HNO_3)). So, we put a coefficient of (2) in front of (HNO_3): (N_2+O_2+H_2O=2HNO_3)

Step2: Balance H atoms

Now, on the right - hand side, we have (2) H atoms. So, we put a coefficient of (1) in front of (H_2O) (already balanced for H)

Step3: Balance O atoms

On the right - hand side, we have (6) O atoms ((2HNO_3)). On the left - hand side, we have (2 + 1=3) O atoms. So, we put a coefficient of (\frac{5}{2}) in front of (O_2). Multiply the entire equation by (2) to get (2N_2+5O_2 + 2H_2O=4HNO_3)

Answer:

(2N_2+5O_2+2H_2O = 4HNO_3)

26. Calculating percent yield

Step - by - Step Format:

Explanation:

Step1: Recall the percent - yield formula

The percent - yield formula is (\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100%) We are given that (\text{Percent Yield}=50%=\frac{50}{100} = 0.5) and (\text{Theoretical Yield}=2\space g)

Step2: Solve for the actual yield

From (\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}), we can rewrite it as (\text{Actual Yield}=\text{Percent Yield}\times\text{Theoretical Yield}) Substitute the values: (\text{Actual Yield}=0.5\times2\space g)

Answer:

(1\space g)