Answer:
A. The flow diagram is in the attachment
B. Degree of freedom = 0
C. 0.486
D. Energy, power, cost of production
Explanation:
1/3 of water = 0.3333
Sugar = 1-1/3 = 2/3
B.) 45/55 = 9/11 when divided by 5
m1/m2 = 9/11
Cross multiply and make m1 subject of the formula
m1/m2 = 0.818
m1 = 0.818m2 ----1
0.15m1 + m2 = 1 x 0.667
0.15m1 + m2 = 0.667
0.15(0.818m2) = 0.667
1.1227m2 = 0.667
M2 = 0.5941
We put value of m2 in equation 1
M1 = 0.818 x 0.5941
M1 = 0.486
So 0.486 pound of strawberry is what we need in order to make 1 pound of jam.
D. We need great heat for water evaporation
The technical and economic factors would be
1. Energy which is power consumption
2. Cost of production for equipments and materials
3. Power
4. And hazards
Answer:
Oceanic crust, the outermost layer of Earth’s lithosphere that is found under the oceans and formed at spreading centres on oceanic ridges, which occur at divergent plate boundaries.
Explanation:
Answer:
Polly did an experiment with marbles in a glass bowl to show the movement of particles in solids, liquids, and gases. The experimental set-up is shown below: A glass bowl is shown with four marbles inside it. ... Add water to the bowl so that the marbles start sliding past one another
Explanation:
Answer:
Explanation:
endergonic
A chemical reaction that has a positive ΔG is correctly described as A) endergonic.
The molar mass of a, b and c at STP is calculated as below
At STP T is always= 273 Kelvin and ,P= 1.0 atm
by use of ideal gas equation that is PV =nRT
n(number of moles) = mass/molar mass therefore replace n in the ideal gas equation
that is Pv = (mass/molar mass)RT
multiply both side by molar mass and then divide by Pv to make molar mass the subject of the formula
that is molar mass = (mass x RT)/ PV
density is always = mass/volume
therefore by replacing mass/volume in the equation by density the equation
molar mass=( density xRT)/P where R = 0.082 L.atm/mol.K
the molar mass for a
= (1.25 g/l x0.082 L.atm/mol.k x273k)/1.0atm = 28g/mol
the molar mass of b
=(2.86g/l x0.082L.atm/mol.k x273 k) /1.0 atm = 64 g/mol
the molar mass of c
=0.714g/l x0.082 L.atm/mol.K x273 K) 1.0atm= 16 g/mol
therefore the
gas a is nitrogen N2 since 14 x2= 28 g/mol
gas b =SO2 since 32 +(16x2)= 64g/mol
gas c = methaneCH4 since 12+(1x4) = 16 g/mol
