The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of
2.94 kg of this oil from 23 °C to 191 °C?
1 answer:
Answer:
Explanation:
The formula for the heat transferred to or from an object is
q = mCΔT
Data:
m = 2.94 kg
C = 1.75 J·°C⁻¹g⁻¹
T₁ = 23 °C
T₂ = 191 °C
Calculations:
(a) Convert the mass to grams
m = 2.94 kg = 2940 g
(b) Calculate ΔT
ΔT = T₂ - T₁ = 191 – 23 = 168 °C
(c) Calculate q
q = 2940 × 1.75 × 168 = 864 000 J = 864 kJ
You must provide
.
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360 mg / 1000 => 0.36 g
molar mass => 180 /mol
number of moles:
mass of solute / molar mass
0.36 / 180 => 0.002 moles
Volume solution = 200 mL / 1000 => 0.2 L
M = n / V
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hope this helps!
Answer:
This molecule is AgN3
Explanation:
If one were to match the ratio of atoms of the elements found in this molecular formula of artificial sweetener it would be :
Carbon - 7 atoms
Hydrogen - 5 atoms
Nitrogen - 1 atom
Oxygen - 3 atoms.
Iodine has an electronegativity of 2.5, and potassium has an electronegativity of 0.8, so the difference is:
2.5 - 0.8 = 1.7
