1426.58 J and 340.90 calories heat in joules and in calories is required to heat at 28.4g( 1 oz) ice cube from -23C TO 1.0C.
Explanation:
Data given:
mass = 28.4 gram
initial temperature = -23 degrees
final temperature = 1degress
change in temperature ΔT = Tfinal - Tinitial
ΔT = 1 -(-23)
ΔT = 24 degrees
specific heat capacity of ice cube c = 2.093 J/g C
Formula used:
q = mc ΔT
putting the values in the equation:
q= 28.4 x 2.093 x 24
= 1426.58 J
ENERGY IN CALORIES:
340.90 calories is the energy is required in the process.
C. Ninhydrin is used to detect prints on a non-porous surface
They are living organisms in the ecosystem and make up part of the food chain.<span />
The water evaporates into the clouds
Answer:
0.65 M
Explanation:
Molarity (M) is the amount of moles of a substance per liter.
In other words, M = moles of substance / liters of solution
Since you're already given the moles of NaCl and the volume of the solution, you can plug those into the equation.
M = moles of NaCl / liters of solution
M = 1.5 mol NaCl / 2.3 L
M = 0.65 mol/L (or 0.65 M)
