The answer is 200 g.
If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M
x = 79.90 g
So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be 79.90 g.
Now, we need to prepare ten beakers with 250 mL of solutions:
10 * 250 mL = 2500 mL = 2.5 L
79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
x = 199.75 g ≈ 200 g
Answer:
c) 22
Explanation:
Let's consider the following balanced equation.
N₂(g) + 3 H₂(g) ----> 2 NH₃(l)
According to the balanced equation, 34.0 g of NH₃ are produced by 1 mol of N₂. For 170 g of NH₃:

According to the balanced equation, 34.0 g of NH₃ are produced by 3 moles of H₂. For 170 g of NH₃:

The total gaseous moles before the reaction were 5.00 mol + 15.0 mol = 20.0 mol.
We can calculate the pressure (P) using the ideal gas equation.
P.V = n.R.T
where
V is the volume (50.0 L)
n is the number of moles (20.0 mol)
R is the ideal gas constant (0.08206atm.L/mol.K)
T is the absolute temperature (400.0 + 273.15 = 673.2K)

The answer to this question is "carrying capacity." The term "growth
rate," refers to how fast a population grows, and the term "population
density," refers to the number of organisms located within a specific
area. Carrying capacity is correct because is directly addresses the
maximum number of organisms that an ecosystem can handle, as opposed to
how fast they are growing or how many there currently are.
The correct answer is B) Basic. Hope this helps.