Answer:
Explanation:
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
The substances with higher value of specific heat capacity require more heat to raise the temperature by one degree as compared the substances having low value of specific heat capacity.For example,
The specific heat capacity of oil is 1.57 j/g. K and for water is 4.18 j/g.K. So, water take a time to increase its temperature by one degree by absorbing more heat while oil will heat up faster by absorbing less amount of heat.
Consider that both oil and water have same mass of 5g and change in temperature is 15 K. Thus amount of heat thy absorbed to raise the temperature is,
For oil:
Q = m.c. ΔT
Q = 5 g× 1.67 j/g K × 15 K
Q = 125.25 j
For water:
Q = m.c. ΔT
Q = 5 g× 4.18 j/g K × 15 K
Q = 313.5 j
we can observe that water require more heat which is 313.5 j to increase its temperature.
Answer: It is B.) 1s^22s^12p^1
Explanation: mind giving me brainliest? :)
Answer:
B and D as well
Explanation:
All the electronic configuration of the elements of B end on s sub shell.
S energy level sub shell can hold 2 electrons and since Be, Mg, Ca Sr end in s.
We can say, they have same valence electrons, i.e 2 electrons.
Answer:
the volume of a give gas simple is directly propotional assolute temperature at constant pressure .the volume of a gavi. amount of gass is inversely propotional ot Its pressure when temperature is help constant
Answer:
1. Yes
2.The solubility of X is 34.55g/L
Explanation:
Solubility of solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm
or 1 Litre of water.
From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved
= 0.019kg*1000 = 19g.
if 19g is required to saturate 550mL at 30°C,
then
will saturate 1L
= 34.545g will saturate 1Litre
The solubility thus is 34.55g/L
