Answer: 15.1 grams
Given reaction:
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
Mass of Na2CO3 = 20.0 g
Molar mass of Na2CO3 = 105.985 g/mol
# moles of Na2CO3 = 20/105.985 = 0.1887 moles
Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH
# moles of NaOH produced = 0.1887*2 = 0.3774 moles
Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol
Mass of NaOH produced = 0.3774*39.996 = 15.09 grams
Explanation:
The substances present before the reaction are the reactants. (As the reaction goes through, the substances that are produced are called the products of the reaction).
0.24J/g*degC * 4.37g * 2.5degC = 2.622J
The 2.5 degC is the difference between 25 and 27.5 deg C.
Answer:
m = 4450 g
Explanation:
Given data:
Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)
Initial temperature = 23.0°C
Final temperature = 57.8°C
Specific heat capacity of water = 1 cal/g.°C
Mass of water in gram = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 57.8°C - 23.0°C
ΔT = 34.8°C
4450 cal = m × 1 cal/g.°C × 34.8°C
m = 4450 cal / 1 cal/g
m = 4450 g
Answer:
light with a high enough intensity
Explanation:
