Answer:
The amount of water converted from liquid to gas with 6,768 joules is approximately 3.035 g
Explanation:
The amount of heat required to convert a given amount of liquid to gas at its boiling point is known as the latent heat of evaporation of the liquid
The latent heat of evaporation of water, Δ
≈ 2,230 J/g
The relationship between the heat supplied, 'Q', and the amount of water in grams, 'm', evaporated is given as follows
Q = m × Δ
Therefore, the amount of water, 'm', converted from liquid to gas at the boiling point temperature (100°C), when Q = 6,768 Joules, is given as follows;
6,768 J = m × 2,230 J/g
∴ m = 6,768 J /(2,230 J/g) ≈ 3.035 g
The amount of water converted from liquid to gas with 6,768 joules = m ≈ 3.035 g.
Answer:
Final volume is 3.50L
Explanation:
It is possible to find volume of a gas using combined gas law:

<em>Where P is pressure, V is volume and T is temperature of 1: initial state and 2: final state</em>
If initial state of the gas is:
1.75L of a gas is at 700K and is under 250kPa of pressure
And final state is:
298K and 53.2kPa.
Replacing:

0.625L = 0.1785*V₂
<em>3.50L = V₂</em>
Thus, <em>final volume is 3.50L</em>
Option C, mass would be same. Only the gravitational pull will be different
Hey...
Use the molarity formula
M=moles/L and then convert to grams
0.07268*0.15=moles
<span>0.010902 mol
</span>Pb(NO3)2
1 mole=331.22g
0.010902 moles=
3.61 g