A 29.0-g sample of water at 290. K is mixed with 51.0 g water at 300. K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings.
2 answers:
Let me conduct my research
answer: 296 K
steps:
mass times temperature =mass times temperature
-51 * 300 = 29 * -290
300 * 51 = 15300
290 * 29 = 8410
-51x + 15300 = 29x - 8410
51 + 29 = 80
15300 + 8410 = 23710
23710 = 80x
80x = 23710
x = 296.375
https://socratic.org/questions/a-30-0-g-sample-of-water-at-280-k-is-mixed-with-50-0-g-of-water-at-330-k-how-wou
You might be interested in
Protons and neutrons are found inside the nucleus
Answer:
The volume will be 89.6875 ml
Explanation:
So to count this we will use a single proportion.
0.0640 mol - 1000 ml
5.74×10−3 mol - x ml
x ml=5.74×10−3 mol*1000 ml/0.0640 mol=89.6875 ml
Answer: 1.46moles
Explanation:
Applying PV= nRT
P= 1atm, V= 32.6L, R= 0.082, T = 273K
Substitute and simplify
1×32.6/(0.082×273)=n
n= 1.46moles
a. ELEMENT : A substance<span> that </span>cannot<span> be decomposed (</span>broken down<span>) </span>into simpler substances by ordinary chemical<span> means.</span>
