Answer:
2 mol H
Explanation:
For every 2 mol of NaOH, we're reacting 2 mol of H2O. In order to figure out how many mol of H are needed, it needs to be set up stochiometrically. Starting off with the given value, 1 mol of NaOH, we can then make a mol to mol ratio. For 2 mol of NaOH, we have 2 mol of H2O. For every 2 mol of H2O, we have 4 mol of H (this is because we are multiplying the coefficient by the subscript: 2 × 2). Now, we can solve for our answer.
1 mol NaOH × (2 mol H₂O / 2 mol NaOH) × (4 mol H / 2 mol H₂O)
= 2 mol H
Thus, we get 2 mol of H are needed to completely react 1 mol of NaOH.
Answer:
17
Explanation:
Step 1: Calculate the needed concentrations
[A]i = 1.00 mol/5.00 L = 0.200 M
[B]i = 1.80 mol/5.00 L = 0.360 M
[B]e = 1.00 mol/5.00 L = 0.200 M
Step 2: Make an ICE chart
A(aq) + 2 B(aq) ⇄ C(aq)
I 0.200 0.360 0
C -x -2x +x
E 0.200-x 0.360-2x x
Then,
[B]e = 0.360-2x = 0.200
x = 0.0800
The concentrations at equilibrium are:
[A]e = 0.200-0.0800 = 0.120 M
[B]e = 0.200 M
[C]e = 0.0800 M
Step 3: Calculate the concentration equilibrium constant (K)
K = [C] / [A] × [B]²
K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17
Answer:
The 150 g Al will reach a higher temperature.
Explanation:
- The amount of heat added to a substance (Q) can be calculated from the relation:
<em>Q = m.c.ΔT.</em>
where, Q is the amount of heat added,
m is the mass of the substance,
c is the specific heat of the substance,
ΔT is the temperature difference (final T - initial T).
Since, Q and c is constant, ΔT will depend only on the mass of the substance (m).
∵ ΔT is inversely proportional to the mass of the substance.
<em>∴ The piece with the lowest mass (150.0 g) will reach a higher temperature than that of a higher mass (250.0 g).</em>
<em>So, the right choice is: The 150 g Al will reach a higher temperature.</em>
Answer:
Thermometer A, because it measures accurately to the tenths digit.
Answer: 1. 
2. 3 moles of 
: 2 moles of 
3. 0.33 moles of : 0.92 moles of
4. is the limiting reagent and is the excess reagent.
5. Theoretical yield of 
is 29.3 g
Explanation:
To calculate the moles :
The balanced chemical equation is:
According to stoichiometry :
3 moles of require = 2 moles of
Thus 0.33 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 3 moles of give = 2 moles of
Thus 0.33 moles of give = of
Theoretical yield of 
Thus 29.3 g of aluminium chloride is formed.
