Answer:
101 L
Explanation:
35.0 g KOH ÷ 56.09 g/mol KOH × (1 mol H2O/ 1 mol KOH) × 18 g/mol H2O = 11.2 g H2O
35.0 g HCl ÷ 36.45 g/mol HCl × (1 mol H2O/ 1 mol HCl) × 18 g/mol H2O = 17.3 g H2O
35.0 g KOH is the limiting reactant
Answer:
-5.51 kJ/mol
Explanation:
Step 1: Calculate the heat required to heat the water.
We use the following expression.

where,
- c: specific heat capacity
- m: mass
- ΔT: change in the temperature
The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

Step 2: Calculate the heat released by the methane
According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero
Qc + Qw = 0
Qc = -Qw = -22.0 kJ
Step 3: Calculate the molar heat of combustion of methane.
The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

Answer:
Pseudoscience
Explanation:
"Astrology has not demonstrated its effectiveness in controlled studies and has no scientific validity, thus regarded as pseudoscience."
The one on top is corrrect because the cell diagram will always contain one object or more