Answer:
A group of cells ready to form roots a stem and the first leaves
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Answer:
Q sln = 75.165 J
Explanation:
a constant pressure calorimeter:
∴ m sln = m Ba(OH)2 + m HCl
∴ molar mass Ba(OH)2 = 171.34 g/mol
∴ mol Ba(OH)2 = (0.06 L)(0.3 mol/L) = 0.018 mol
⇒ mass Ba(OH)2 = (0.018 mol)(171.34 g/mol) = 3.084 g
∴ molar mass HCl = 36.46 g/mol
∴ mol HCl = (0.06 L)(0.60 mol/L) = 0.036 mol
⇒ mass HCl = (0.036 mol)(36.46 g/mol) = 1.313 g
⇒ m sln = 3.084 g + 1.313 g = 4.3966 g
specific heat (C):
∴ C sln = C H2O = 4.18 J/g°C
∴ ΔT = 26.83°C - 22.74°C = 4.09°C
heat absorbed (Q):
⇒ Q sln = (4.3966 g)(4.18 J/g°C)(4.09°C)
⇒ Q sln = 75.165 J
Answer: In Galileo’s time, what was considered the “center of everything”? The Earth! All of the planets and even the Sun went around “us”. Of course, when Galileo saw the moons of Jupiter passing in front of the planet, and disappearing to show up again, it was clear to him that these moons went around Jupiter like our moon goes ‘round the Earth.
That did not set well with the Beliefs of the day, and that is at least one answer!
Explanation: