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
Learn more about temperature here: brainly.com/question/11464844
#SPJ9
Answer:
Volume of the gass will decrease by three times of the original volume
Explanation:
Volume is inversly propotional to the pressure applied on it.
Answer:
Explanation:
IT'S FOR NOW, PLEASE DELIVER IT TODAY !!! I NEED HELP IT'S CHEMICAL WORK...
Answer:
% composition O = 19.9%
% composition Cu = 80.1%
Explanation:
Given data:
Total mass of compound = 3.12 g
Mass of copper = 2.50 g
Mass of oxygen = 3.12 - 2.50 = 0.62 g
% composition = ?
Solution:
Formula:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition Cu = (2.50 g / 3.12 g)×100
% composition Cu = 0.80 ×100
% composition Cu = 80.1%
For oxygen:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition O = (0.62 g / 3.12 g)×100
% composition O = 0.199 ×100
% composition O = 19.9%
