Answer: Option B. 76.83L
Explanation:
1 mole of a gas occupy 22.4L at stp. This implies that 1mole of Radon also occupy 22.4L at stp.
If 1 mole of Radon = 22.4L
Therefore, 3.43 moles of Radon = 3.43 x 22.4 = 76.83L
The specific heat : c = 0.306 J/g K
<h3>Further explanation</h3>
Given
Heat = 35.2 J
Mass = 16 g
Temperature difference : 7.2 K =
Required
The specific heat
Solution
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
Input the value :
c = Q / m.∆T
c = 35.2 / 16 x 7.2
c = 0.306 J/g K
Hello!
The reaction that the graph represents is
A. Exothermic because Hrxn=-167 kJTo calculate Hrxn we apply the following equation:

Looking at the graph, and at the result of the calculations, we can see that the enthalpy of the products is
lower than the enthalpy of the reagents, because the sign is negative. That means that the reaction
releases energy in the form of heat and that the reaction is
exothermic.
Have a nice day!
Answer:
D. To ensure the cooling process is not affected by surrounding temperature
Explanation:
The conical flask acts as a <u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u><u>j</u><u>a</u><u>c</u><u>k</u><u>e</u><u>t</u><u>.</u>
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87