Answer: B. A secondary pollutant
An acid rain is an example of a secondary pollutant because it is a chemical reaction of two gases namely, the sulfur dioxide and the nitrogen dioxide. The acid rain is not also directly emitted by the atmosphere.
Answer:
The mass of ascorbic acid in 2.87×10−4 mol are 50,5 mg
Explanation:
Molar mass = 176 g/mol
Moles . molar mass = grams.
2,87 x10*-4 m . 176 g/m = 50,5 x10*-3 grams
A human can take one sweet lime and half to cover the daily requirement of vitamin C. =)
Answer:
Solution
dT = K * b
K (H2O) = 1.86 0C
b(C2H6O2) = dT/ K = 33.2 / 1.86 = 17.85(mol/kg)
b(C2H6O2)= n (C2H6O2) / m (H2O)
m (H2O) = V * = 12.2 * 1 = 12.2 (kg)
n (C2H6O2) = b (C2H6O2) * m (H2O) = 17.85 * 12.2 = 217.77 (mol)
n (C2H6O2) = m / M;
m (C2H6O2) = n * M = 217.77 * (12*2 + 1*6 + 16 *2) = 13501.74 (g) = 13.5 (kg)
V (C2H6O2) = m / = 13.5 / 1.11 = 12.16 (L)
Answer V (C2H6O2) = 12.16 L
These 2 isotopes are all of bromine that is naturally found (their percentages add up to 100%). In order to calculate the mean atomic mass, we can use the following calculation:
. This yields an average atomic mass of 79.903 amu. This procedure is general; whenever we have many types F of a specific genre G, the mean of the genre G can be calculated by:
where
is the proportion of type i in the genre G. In this example, Br79 and Br81 are the 2 types and the genre is Bromine.
