Answer:
376.56
Explanation:
using the formula q=mcΔT, where m=15g, C=4.18J/g°C and ΔT=6.0°C
The class can break or if you put it in cold water it can cool down down fast.
Answer:
Q = 30284.88 j
Explanation:
Given data:
Mass of ethanol = 257 g
Cp = 2.4 j/g.°C
Chnage in temperature = ΔT = 49.1°C
Heat required = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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
Now we will put the values in formula.
Q = 257 g× 2.4 j/g.°C × 49.1 °C
Q = 30284.88 j
Answer:
4121 years
Explanation:
From;
0.693/t1/2 = 2.303/t log No/N
t1/2= half life of the carbon-14
No= count rate of the living tissue
N= count rate of the sample
t = age of the sample
0.693/5730 =2.303/t log (13.5/8.2)
1.21 * 10^-4 = 2.303/t * 0.2165
1.21 * 10^-4 = 0.4986/t
t = 0.4986/1.21 * 10^-4
t = 4121 years
Answer:
B- The polarity of the molecules and hydrogen bonding between molecules.
Explanation:
Hope this helps:)
