18. this is a chemical that reflects heat fro...
18. this is a chemical that reflects heat from the sun, warming the planet:\na) greenhouse gas\nb) noble gas\nc) reactive alkali element\nd) none of the above\nbalance the following equations:\n19. ___fe + ___cl2 = ___fecl3\n20. ___fe + ___o2 = ___fe2o3\n21. ___nh3 + ___o2 = ___no + ___h2o\n22. ___na + ___cl2 = ___nacl\n23. ___n2 + ___h2 = ___nh3\n24. ___zn + ___hcl = ___zncl2 + ___h2\n25. ___n2+ ___o2 + ___h2o = ___hno3\ncalculate percent yield\n26. a reaction with a theoretical yield of 2 grams has a percent yield of 50%, what is the actual yield?
Answer
### 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:
\(1\space g\)