Answer:
The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams
Explanation:
The question is with regards to density calculations
The density of the given sodium carbonate solution, ρ = 0.4 g/dm³
The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³
Therefore, we have;
The mass, "m", of the sodium carbonate in = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g
The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.
Answer:
Explanation:
We're asked to calculate the number of atoms of
Ca
in
153
g Ca
.
What we must first do is convert the given mass of calcium to moles of calcium, using its molar mass (referring to a periodic table, this is
40.08
g
mol
):
153
g Ca
(
1
mol Ca
40.08
g Ca
)
=
3.82
mol Ca
Using Avogadro's number,
6.022
×
10
23
particles
mol
, we can calculate the number of atoms present:
3.82
mol Ca
(
6.022
×
10
23
atoms Ca
1
mol Ca
)
=
2.30
×
10
24
atoms Ca
Answer:
An object at position A. has all potential energy
An object at position B. has about half potential and half kinetic energy
An object at position C. has all kinetic energy
Explanation:
I already did it and got all of them correct. I hope this helped!! :)
