Answer:
Explanation:
we know that
ΔH=m C ΔT
where ΔH is the change in enthalpy (j)
m is the mass of the given substance which is water in this case
ΔT IS the change in temperature and c is the specific heat constant
we know that given mass=2.9 g
ΔT=T2-T1 =98.9 °C-23.9°C=75°C
specific heat constant for water is 4.18 j/g°C
therefore ΔH=2.9 g*4.18 j/g°C*75°C
ΔH=909.15 j
Answer:
Pressure, P = 67.57 atm
Explanation:
<u>Given the following data;</u>
- Volume = 0.245 L
- Number of moles = 0.467 moles
- Temperature = 159°C
- Ideal gas constant, R = 0.08206 L·atm/mol·K
<u>Conversion:</u>
We would convert the value of the temperature in Celsius to Kelvin.
T = 273 + °C
T = 273 + 159
T = 432 Kelvin
To find the pressure of the gas, we would use the ideal gas law;
PV = nRT
Where;
- P is the pressure.
- V is the volume.
- n is the number of moles of substance.
- R is the ideal gas constant.
- T is the temperature.
Making P the subject of formula, we have;

Substituting into the formula, we have;


<em>Pressure, P = 67.57 atm</em>
I believe you are referring zero as the exponent. <span>Any number (except 0) with exponent 0 is defined to mean 1.
</span>
For one thing, there is a rule:
<span> a^m/ a^m = a^m-m = a^0
</span>But (when a is not equal to <span>0),
</span>
a^m/ a^m = 1
Therefore, we must define a^0 as 1.