Answer:
Groundwater pollution
Explanation:
Pesticides, when sprayed on crop plants, are able to flow below the surface of the ground, reaching water-bearing aquifers, thereby contaminating groundwater, making it unsuitable for both human and agricultural uses.
4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.
(a) The <em>balanced chemical equation </em>is
2B + 3Cl2 → 2BCl3
(b) Convert kilograms of BCl3 to moles of BCl3
MM: B = 10.81; Cl = 35.45; BCl3 = 117.16
Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3
(c) Use the <em>molar ratio</em> of Cl2:BCl3 to calculate the moles of Cl2.
Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2
Answer:
6.17 g/cm³
Explanation:
Data given:
one side of cube = 0.53 cm
mass of the cube is 0.92 g
density of the cube = ?
Solution:
First we will calculate for volume the cube
As we know all the sides or edges of a cube are equal so volume equation will be
So,
V = length x width x height
V = e³
as on side = 0.53 cm
then
V = (0.53 cm)³
V = 0.149 cm³
Now we will calculate density of cube
To calculate density, formula will be used
d = m/v . . . . . (1)
where
d = density
m = mass
v = volume
put values in above formula 1
d = 0.92 g / 0.149 cm³
d = 6.17 g/cm³
so. the density of cube = 6.17 g/cm³
The molarity is moles/liters.
First, convert 4,000 mL to L:
4000 mL --> 4 L
Now, you must convert the 17 g of solute to moles by dividing the number of grams by the molar mass. The molar mass of AgNO3 is <span>169.87 g/mol:
17 / 169.87 = .1
Now that you have both the number of moles and the liters, plug them into the initial equation of moles/liters:
.1/4 = .025</span>
