Explanation:
- F Cl Br
- ionization energy decreases we move down the group
- so Br<Cl<F
2. Na K Li
- ionization energy decreases we move down the group
- so K<Na<Li
3. C N O F
- ionization energy increases as we move across the peroid
- so C<N<O<F
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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.

I believe it is B. A series circuit has one path for electrons, but a parallel circuit has more then one path.
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Answer:
0.9 mole of Fe(OH)3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
Fe(NO3)3 + 3NaOH —> Fe(OH)3 + 3NaNO3
Now, we can determine the moles of iron (III) hydroxide formed from the reaction as follow:
From the balanced equation above,
3 moles of NaOH reacted to produce 1 mole of Fe(OH)3.
Therefore, 2.7 moles of NaOH will react to produce = 2.7/3 = 0.9 mole of Fe(OH)3.
Therefore, 0.9 mole of Fe(OH)3 is produced from the reaction.