Answer:
Percentage error = 1.88 %
Solution:
Data Given:
Mass of Sample = 20.46 g
Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL
Formula Used:
Density = Mass / Volume
Putting values,
Density = 20.46 g / 3.0 mL
Density = 6.82 g.mL⁻¹
Percentage Error:
Experimental Value = 6.82 g.mL⁻¹
Accepted Value = 6.95 g.mL⁻¹
= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %
Percentage Error = 100 % - 98.12 %
Percentage error = 1.88 %
Answer:
0.1066 hours
Explanation:
A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.
t1/2 = ln2/k
t1/2 = ln2/6.5 h⁻¹
t1/2 = 0.1066 h
The half-life of the pesticide is 0.1066 hours.
Answer:
Rubidium
Rubidium is the first element placed in period 5.
Answer:
40.4 kJ
Explanation:
Step 1: Given data
- Heat of sublimation of CO₂ (ΔH°sub): 32.3 kJ/mol
Step 2: Calculate the moles corresponding to 55.0 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
n = 55.0 g × 1 mol/44.01 g = 1.25 mol
Step 3: Calculate the heat (Q) required to sublimate 1.25 moles of CO₂
We will use the following expression.
Q = n × ΔH°sub
Q = 1.25 mol × 32.3 kJ/mol = 40.4 kJ
Answer:
The following properties are either physical or chemical. Which one is different from the rest based on those two categories? We chose all of the above