Answer:
_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)
Explanation:
Step 1:
The unbalanced equation:
AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
Step 2:
Balancing the equation.
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
The above equation can be balanced as follow:
There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)
There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)
Now the equation is balanced
Answer:
0.259 kJ/mol ≅ 0.26 kJ/mol.
Explanation:
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 100.0 g).
c is the specific heat of water (c of ice = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 21.56°C - 25.0°C = -3.44°C).
<em>∵ Q = m.c.ΔT</em>
∴ Q = (100.0 g)(4.186 J/g.°C)(-3.44°C) = -1440 J = -1.44 kJ.
<em>∵ ΔH = Q/n</em>
n = mass/molar mass = (100.0 g)/(18.0 g/mol) = 5.556 mol.
∴ ΔH = (-1.44 kJ)/(5.556 mol) = 0.259 kJ/mol ≅ 0.26 kJ/mol.
Answer:
a. 7.8*10¹⁴ He⁺⁺ nuclei/s
b. 4000s
c. 7.7*10⁸s
Explanation:
I = 0.250mA = 2.5 * 10⁻³A
Q = 1.0C
1 e- contains 1.60 * 10⁻¹⁹C
But He⁺⁺ Carrie's 2 charge = 2 * 1.60*10⁻¹⁹C = 3.20*10⁻¹⁹C
(A).
No. Of charge per second = current passing through / charge
1 He⁺⁺ = 2.50 * 10⁻⁴ / 3.2*10⁻¹⁹C
1 He⁺⁺ = 7.8 * 10¹⁴ He⁺⁺ nuclei
(B).
I = Q / t
From this equation, we can determine the time it takes to transfer 1.0C
I = 1.0 / 2.5*10⁻⁴ = 4000s
(C).
Time it takes for 1 mol of He⁺⁺ to strike the target =?
Using Avogadro's ratio,
1.0 mole of He = (6.02 * 10²³ ions/mol ) * (1 / 7.81*10¹⁴ He ions)
Note : ions cancel out leaving the value of the answer in mols.
1.0 mol of He = 7.7 * 10⁸s
