The degree to which a specified material conducts electricity, calculated as the ratio of the card density in the material to the electric field that causes the flow of current. It is the reciprocal of the resistivity.
The volume of CO2 at STP =124.298 L
<h3>Further explanation</h3>
Given
Reaction
4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O
701,52 g of KMnO4
Required
volume of CO2 at STP
Solution
mol KMnO4 (MW=158,034 g/mol) :
mol = mass : MW
mol = 701.52 : 158.034
mol = 4.439
mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549
At STP 1 mol = 22.4 L, so for 5.549 moles :
=5.549 x 22.4
=124.298 L
Answer:
The final temperature of the solution is 44.8 °C
Explanation:
assuming no heat loss to the surroundings, all the heat of solution (due to the dissolving process) is absorbed by the same solution and therefore:
Q dis + Q sol = 0
Using tables , can be found that the heat of solution of CaCl2 at 25°C (≈24.7 °C) is q dis= -83.3 KJ/mol . And the molecular weight is
M = 1*40 g/mol + 2* 35.45 g/mol = 110.9 g/mol
Q dis = q dis * n = q dis * m/M = -83.3 KJ/mol * 13.1 g/110.9 gr/mol = -9.84 KJ
Qdis= -9.84 KJ
Also Qsol = ms * Cs * (T - Ti)
therefore
ms * Cs * (T - Ti) + Qdis = 0
T= Ti - Qdis * (ms * Cs )^-1 =24.7 °C - (-9.84 KJ/mol)/[(104 g + 13.1 g)* 4.18 J/g°C] *1000 J/KJ
T= 44.8 °C
The half life of carbon-14 is 5700 years. (Half life is the time taken by a radioactive isotope to decay by half of its original mass).
Let A₀ be the initial amount of carbon-14 that is found in living matter (t=0 years), to determine when there was 44.5% of A₀ left.
44.5 = 100 × (1/2)^n, where n is the number of half lives
0.5^n = 0.445
n = log 0.445/log 0.5
n = 1.168
But 1 half life is 5700 years
Therefore, the number of years will be 5700 × 1.168 = 6658.299725 years
≈ 6658.30 years
