Answer:
6CO2 + 6H20 + ENERGY = C6H12O6 + 6O2
Explanation:
Carbon dioxide + water + energy from light produces glucose and oxygen
Problem One
You will use both m * c * deltaT and H = m * heat of fusion.
Givens
m = 12.4 grams
c = 0.1291
t1 = 26oC
t2 = 1204
heat of fusion (H_f) = 63.5 J/grams.
Equation
H = m * c * deltaT + m * H_f
Solution
H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5
H = 1660.1 + 787.4
H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.
Problem Two
Formula and Givens
t1 = 14.5
t2 = 50.0
E = 5680
c = 4.186
m = ??
E = m c * deltaT
Solution
5680 = m * 4.186 * (50 - 14.5)
5680 = m * 4.186 * (35.5)
5680 = m * 148.603 * m
m = 5680 / 148.603
m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.
m = 38.2 to 3 sig digs.
Answer:
The number of moles of xenon are 1.69 mol.
Explanation:
Given data:
Number of moles of xenon = ?
Volume of gas = 37.8 L
Temperature = 273 K
Pressure = 1 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will put the values in formula.
1 atm × 37.8 L = n × 0.0821 atm.L/ mol.K ×273 K
37.8 atm.L = n × 22.413 atm.L/ mol.
n = 37.8 atm.L / 22.413 atm.L/ mol.
n = 1.69 mol
The number of moles of xenon are 1.69.
This is a simple chemical change due to what it produces and how it is added together. Hope this helps.
