Answer:
904.014 j/kgk
Explanation:
Mass of metal = 45g
Temperature of metal = 85.6°
Mass of water = 150
Temperature of water = 24.6
Final temperature of system = 28.3
Heat lost by metal = Heat gained by water
m1 * c1 * dt = m2 * c2 * dt
Q = quantity of heat
Q = m*c*dt
dt = change in temperature
dt of water = 28.3 - 24.6 = 3.7
dt of metal = 85.6 - 28.3 = 57.3
Specific heat capacity of water, c = 4200
(45 * 10^-3) * c * 57.3 = (150 * 10^-3) * 4200 * 3.7
2.5785c1 = 2331
c1 = 2331 / 2.5785
= 904.01396
= 904.014 j/kgk
Answer:
4.767 grams of KCl are produced from 2.50 g of K and excess Cl2
Explanation:
The balanced equation is
2 K+ Cl2 --->2 KCI
Here the limiting agent is K. Hence, the amount of KCl will be calculated as per the mass of 2.50 gram of K
Mass of one atom/mole of potassium is 39.098 grams
Number of moles is 2.5 grams =
So, 2 moles of K produces 2 moles of KCL
0.064 moles of K will produces 0.064 moles of KCl
Mass of one molecule of KCl is 74.5513 g/mol
Mass of 0.064 moles of KCl is 4.767 grams
NaHCO₃ + HCl → NaCl + H₂O + CO₂
<u>Explanation:</u>
NaHCO₃ + HCl → NaCl + H₂O + CO₂
When Sodium bi carbonate (NaHCO₃) reacts with hydrochloric acid (HCl), it forms table salt (NaCl), water ( H₂O ) and Carbon di oxide (CO₂) gas is evolved.
Here base reacts with acid to form salt and it is the neutralization reaction.
Answer:
Concept: Application of Theory
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