Answer is 14.5 g L⁻¹.
<em>Explanation;</em>
Here, the question says reduce the units as one.
The presented units are g/L. To reduce the units to one, what we can do is take L to the upper side.
This can be done according to the rules of indices;
1 / aˣ = a⁻ˣ
Like that, we can write 1 / L as L⁻¹.
Hence, the reduced unit is g L⁻¹.
But remember to keep a space between when writing two different units.
Actually, this is an unit for density.
Answer:
b
Explanation:
The number of vibrations per second is known as the frequency
One mole is always the same number: 6.02 * 10^ 23.
So, one mole of cars = 6.02 * 10 ^23 cars; one mole of pencils = 6.02 * 10^23 pencils; one mole of atoms = 6.02 * 10^23 atoms; one mole of molecules = 6.02 * 10^23 molecules.
So, all the options are correct: one mole of calcium ions has 6.02 * 10^23 representative particles, such as one mole of calcium nuclei and one of calcium atoms.
Correct question
The density of liquid mercury is 13.6 g/mL. What is its density in units of lb/in3? (2.5 cm = 1 in., 2.205 lbs= 1 kg., 1000 g =1 kg, 1 mL = 1 cm³)
Answer:
Explanation:
Given that;-
The density = 13.6 g/mL
Also, 1 kg = 2.205 lb
1 kg = 1000 g
So, 1000 g = 2.205 lb
1 g = 0.002205 lb
Also,
1 in = 2.54 cm
1 in³ = 16.39 cm³
1 cm³ = 1 mL
So, 1 in³ = 16.39 mL
1 mL = 0.061 in³
The expression for the calculation of density is shown below as:-
Thus,
Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g
