Unbalanced it should be 2Zn+2Hcl=2ZnCl2+H2
Answer:
121 K
Explanation:
Step 1: Given data
- Initial volume (V₁): 79.5 mL
- Initial temperature (T₁): -1.4°C
- Final volume (V₂): 35.3 mL
Step 2: Convert "-1.4°C" to Kelvin
We will use the following expression.
K = °C + 273.15 = -1.4°C + 273.15 = 271.8 K
Step 3: Calculate the final temperature of the gas (T₂)
Assuming ideal behavior and constant pressure, we can calculate the final temperature of the gas using Charles' law.
V₁/T₁ = V₂/T₂
T₂ = V₂ × T₁/V₁
T₂ = 35.3 mL × 271.8 K/79.5 mL = 121 K
Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
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