Answer:
The total heat required is 691,026.36 J
Explanation:
Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L
Where Q: amount of heat, m: mass and L: latent heat
On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).
In this case, the total heat required is calculated as:
- Q for liquid water. This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° C
Q= c*m*ΔT

Q=103,763.2 J
- Q for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then
)
Q= m*L

Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)
- Q for temperature change from 100.0
∘
C to 154
∘
C, this is, the sensible heat of steam from 100 °C to 154°C.
Q= c*m*ΔT

Q=25,176.96 J
So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J
<u><em>The total heat required is 691,026.36 J</em></u>
B.
Delta G is negative and Delta S is positive.
Answer:
The density of this liquid is 0.320 kg/L
Explanation:
Given:
Volume of the Liquid = 0.820 L
Mass of the liquid = 2.56 kg.
To Find:
The density of the liquid in kg/L
Solution:
Density is the mass occupied by the substance in unit volume. This density is essential determining whether the substance floats or sinks. Greek letter(rho) is used to denote density
The equation of for density is


where m is the mass
v is the volume
On substituting the given values


Phosphorus can be prepared from calcium phosphate by the following reaction:

Phosphorite is a mineral that contains
plus other non-phosphorus-containing compounds. What is the maximum amount of
that can be produced from 2.3 kg of phosphorite if the phorphorite sample is 75%
by mass? Assume an excess of the other reactants.
Answer: Thus the maximum amount of
that can be produced is 0.345 kg
Explanation:
Given mass of phosphorite
= 2.3 kg
As given percentage of phosphorite
is = 



According to stoichiometry:
2 moles of phosphorite gives = 1 mole of 
Thus 5.56 moles of phosphorite give=
of 
Mass of 
Thus the maximum amount of
that can be produced is 0.345 kg