<span> x + 2 = 6 x + 2 = 6 x+2=6x, plus, 2, equals, 6 has a variable in it.</span>
R in the equation is a gas constant. It is .0821 liters x ATM over Moles x degrees kelvin.
I think the correct answer from the choices listed above is option A. The three components of air are all <span>classified as pure substances since they are not chemically bonded so they can be separated by certain processes and be present as a pure substance. Hope this answers the question.</span>
Freezing point, boiling point, melting point, smell, attraction or repulsion to magnets, colour change, and many more examples.
Answer is: 550,021 kWh of energy is needed to heat the water
V(water) = 51 gal = 51·3,78 = 189,3 L.
ΔT(water) = 25°C.
d(water) = 1000 g/L.
m(water) = V(water) · d(water)
m(water) = 189,3 L · 1000 g/L
m(water) = 189300 g.
Q = m(water) · ΔT(water) · C(water)
Q = 189300 g · 25°C · 4,184 J/°C·g
Q = 19800780 J = 19800,78 kJ ÷ 3600 = 550,021 kWh.
