Answer:
The final temperature is:- 7428571463.57 °C
Explanation:
The expression for the calculation of heat is shown below as:-
Where,
is the heat absorbed/released
m is the mass
C is the specific heat capacity
is the temperature change
Thus, given that:-
Mass of water = 1.75 mg = 0.00175 g ( 1 g = 0.001 mg)
Specific heat of water = 4.18 J/g°C
Initial temperature = 35 °C
Final temperature = x °C
![\Delta T=(x-35)\ ^0C/tex] Q = [tex]1.3\times 10^4](https://tex.z-dn.net/?f=%5CDelta%20T%3D%28x-35%29%5C%20%5E0C%2Ftex%5D%0A%3C%2Fp%3E%3Cp%3EQ%20%3D%20%5Btex%5D1.3%5Ctimes%2010%5E4)
kcal
Also, 1 kcal = 4.18 kJ = 
J
So, Q = 
J = 54340000 J
So,
Thus, the final temperature is:- 7428571463.57 °C
Answer:
g/cm³ for solids,
g/ml for liquids
g/L for gases.
Explanation:
Though SI unit of density is kg/m³, for convenience we use g/cm³ for solids, g/ml for liquids and g/L for gases. Mathematically, density is defined as mass divided by volume:
ρ=m/V
Answer:
The ideal gas law is expressed mathematically by the ideal gas equation as follows;
P·V = n·R·T
Where;
P = The gas pressure
V = The volume of the gas
n = The number of moles of the gas present
R = The universal gas constant
T = The temperature of the gas
A situation where the ideal gas law is exhibited is in the atmosphere just before rainfall
The atmospheric temperature of the area expecting rainfall drops, (when there is appreciable blockage of the Sun's rays by cloud covering) followed by increased wind towards the area, which indicates that the area was in a state of a low pressure, 'P', and or volume, 'V', or a combination of both low pressure and volume P·V
When the entry flow of air into the area is observed to have reduced, the temperature of the air in the area is simultaneously sensed to have risen slightly, therefore, the combination of P·V is seen to be proportional to the temperature, 'T', and the number of moles of air particles, 'n' in the area
Explanation:
Answer:
Number of moles = 2.8 mol
Explanation:
Given data:
Number of moles of water = ?
Volume of water = 50 mL
Density of water = 1.00 g/cm³
Solution:
1 cm³ = 1 mL
Density = mass/ volume
1.00 g/mL = mass/ 50 mL
Mass = 1.00 g/mL× 50 mL
Mass = 50 g
Number of moles of water:
Number of moles = mass/molar mass
Number of moles = 50 g / 18 g/mol
Number of moles = 2.8 mol
