Given is the specific heat of water equal to 4.18 Joule per gram per *C.
This means to raise the temperature of 1 g of water by 1 degree Celsius we need 4.18 joule of energy.
Now, look at the question. We are asked that how much amount of energy would be required to raise the temperature of 25 g of water by (54-50) = 4 degree celsius.
To do so we have formula
Q = m C (temperature difference)
Have a look at pic for answer
Answer:
no, the correct answer is NaCI
Explanation:
you're welcome
Heat energy is absorbed by the substance
Answer: The density of 0.50 grams of gaseous carbon stored under 1.50 atm of pressure at a temperature of -20.0 °C is 0.867 g/L.
Explanation:
- d = m/V, where d is the density, m is the mass and V is the volume.
- We have the mass m = 0.50 g, so we must get the volume V.
- To get the volume of a gas, we apply the general gas law PV = nRT
P is the pressure in atm (P = 1.5 atm)
V is the volume in L (V = ??? L)
n is the number of moles in mole, n = m/Atomic mass, n = 0.50/12.0 = 0.416 mole.
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature in K (T(K) = T(°C) + 273 = -20.0 + 273 = 253 K).
- Then, V = nRT/P = (0.416 mol)(0.082 L.atm/mol.K)(253 K) / (1.5 atm) = 0.576 L.
- Now, we can obtain the density; d = m/V = (0.50 g) / (0.576 L) = 0.867 g/L.