Answer:
Cp = 0.237 J.g⁻¹.°C⁻¹
Explanation:
Amount of energy required by known amount of a substance to raise its temperature by one degree is called specific heat capacity.
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 640 J
m = mass = 125 g
Cp = Specific Heat Capacity = <u>??</u>
ΔT = Change in Temperature = 43.6 °C - 22 °C = 21.6 °C
Solving eq. 1 for Cp,
Cp = Q / m ΔT
Putting values,
Cp = 640 J / (125 g × 21.6 °C)
Cp = 0.237 J.g⁻¹.°C⁻¹
Answer: 300g
Explanation:
first we write the given values on top
224L. x
3 NO2 (g) + H2O (l) = 2HNO3 (l) + NO (g)
22.4L 30g
then we form a formula
224L/22.4L= x/30g
224*30/22.4
6720/22.4= 300g
A standard drink of beer is 12 ounces
The integrated rate law for a second-order reaction is given by:
![\frac{1}{[A]t} = \frac{1}{[A]0} + kt](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%20%20%5Cfrac%7B1%7D%7B%5BA%5D0%7D%20%2B%20kt%20)
where, [A]t= the concentration of A at time t,
[A]0= the concentration of A at time t=0
<span>k =</span> the rate constant for the reaction
<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min
Hence, ![\frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%2B%20%280.0265%20X%20180%29%20)
<span> = 4.858</span>
<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>
<span>
</span>
<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M
The temperature of a liquid can exceed its boiling point. An example is water. Although at ordinary pressure of 1 atm, the boiling point is 100 degrees, water can still exist in higher temperatures but this time in another state. Superheated steam is the term used for water whose temperature has higher than the boiling point
