Mass of Gold = 267.165 × 0.01552494829
⇒ 4.1477228099
The amount of heat(q) required to raise m grams of a substance-specific C from T1 to T2 is given by
q=m C (T2-T1) ........1
Given : q= 2.1200 J
the initial temperature of gold, T1 = 22.0Celcius
the final temperature of gold, T2 = 1064.4Celcius
specific heat of gold = 0.131
putting values in eq 1:
⇒ 2.1200 = m × 0.131 × (1064.4-22)
⇒ 2.1200 = m × 0.131 × 1042.4
⇒ 2.1200 / 136.5544
⇒ 0.01552494829
Since 1g= 0.01552494829 Pounds
Mass of Gold = 267.165 × 0.01552494829
⇒ 4.1477228099
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Given:
n = 0.0021 mol
M = 44 gram/mol
Required: m or the mass of the molecule
Solution:
M = m/n
Rearranging the expression,
m = M x n
m = 44 g/mol x 0.0021 mol
Evaluating the expression and cancelling units, we obtain,
m = 0.0924 grams
M is the molar mass. This value is very useful in chemistry calculations especially in calculating the amounts of reactants and products for a chemical reaction.
Answer:
1000L
Explanation:
the 1 is a sig fig and since the 0 is between the 1 and 4 its also a significant number. to round them off you look at the 0,then look back at the 4 since its less than 5 u round down. then u replace the 43 with 0's
Answer:
The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.
