Answer:
= 8350 joules
Solution and explanation:
- The heat of fusion refers to the quantity of heat released when a given amount of water freezes.
- For example, 1 g of water releases 334 J when it freezes at 0°C.
Therefore; For 25.0 g of water.
Heat released = Mass of water × heat of fusion
= 25 g × 334 J/g
= 8350 Joules
Hence, the amount of heat released when 25.0 g of water freezes at 0°C is 8350 J.
Answer:
c. Kay's rule
Explanation:
Kay's rule -
The rule is used to determine the pseudo reduced critical parameters of mixture , with the help of using the critical properties of the components of a given mixture .
The equation for Kay's rule is as follows ,
PV = Z RT
Where Z = The compressibility factor of the mixture .
Hence from the given options , the correct answer is Kay's rule .
Good grief, this stuff got caught in a black hole somewhere. It is terribly dense.
1 mL = 1 cc under normal conditions.
d = mass / volume
m = 20 kg
v = 5 mL
d = 20kg / 5 mL
d = 4 kg / mL
d = 4 kg / cc
A <<<<answer
Answer:
the volume will decrease
Explanation:
an increase in temperature will lead to an increase in effective collision which also result in an increase in volume, when the temperature decreases, the volume will decrease
Answer: C₃H₂
Explanation:
1) Since you are told naphthelene is a hydrocarbon, you know that its chemical formula only has atoms of C and H.
2) Since combustion adds O but not C nor H, you know that all the C and H present in the products come from the naphthelene sample.
3) Calculate the amount of C in 13.20 grams of CO₂:
Use atomic masses:
C: 12.01 g/mol
O: 16.00 g/mol
⇒ molar mass of CO₂ = 12.01 g/mol + 2×16.01 g/mol = 44.01 g/mol
Proportion of C in 13.20 g of CO₂
12.01 g C / 44.01 g CO₂ = x / 13.20 g CO₂ ⇒
x = 13.20 g CO₂ × 12.01 g C / 44.01 g CO₂ = 3.602 g C
3) Calculate the mass of H in 1.80 g of H₂O
Atomic masses:
H: 1.01 g/mol
O: 16.0 g/mol
Molar mass H₂O = 18.02 g/mol
Proportion of H in 1.80 g H₂O
2.02 g H / 18.02 g H₂O = x / 1.80 g H₂O
x = 1.80 gH₂O × 2.02 g H / 18.02 gH₂O = 0.202 g H.
4) Calculate the number of moles of C and H in the masses calculated above
Number of moles of C = 3.602 g C / 12.01 g/mol = 0.3 mol
Number of moles of H = 0.202 g / 1.01 g/mol = 0.2 mol
5) Divide both amounts by the smallest one to calculate the proportion in moles:
C: 0.3 / 0.2 = 1.5 mol
H = 0.2 / 0.2 = 1 mol
Multiply by 2 to obtain integers:
C: 3
H: 2
6) Then the proportion leads to the empirical formula C₃H₂
