Answer:
ºC
Explanation:
We have to start with the variables of the problem:
Mass of water = 60 g
Mass of gold = 13.5 g
Initial temperature of water= 19 ºC
Final temperature of water= 20 ºC
<u>Initial temperature of gold= Unknow</u>
Final temperature of gold= 20 ºC
Specific heat of gold = 0.13J/gºC
Specific heat of water = 4.186 J/g°C
Now if we remember the <u>heat equation</u>:
We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:
Now we can <u>put the values into the equation</u>:
Now we can <u>solve for the initial temperature of gold</u>, so:
ºC
I hope it helps!
Answer:
x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates
Explanation:
Recall that , depression present in freezing point is calculated with the formulae = solute particles Molarity x KF
0.3473 = m * 1.86
Solving, m = 0.187 m
Moles of HClO2 = mass / molar mass = 5.85 / 68.5 = 0.0854 mol
Molality = moles / mass of water in kg = 0.0854 / 1 = 0.0854 m
Initial molality
Assuming that a % x of the solute dissociates, we have the ICE table:
HClO2 H+ + ClO2-
initial concentration: 0.0854 0 0
final concentration: 0.0854(1-x/100) 0.0854x/100 0.0854x / 100
We see that sum of molality of equilibrium mixture = freezing point molality
0.0854( 1 - x/100 + x/100 + x/100) = 0.187
2.1897 = 1 + x / 100
x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates
Answer:
Rb
Explanation:
This is because they are in the same group which means they share similar properties.
<span>The generalized reaction for chemical decomposition is: AB → A + B
NaOH is sodium hydroxide. When sodium and water is combined it makes sodium hydroxide and hydrogen
When sodium hydroxide decomposes under thermal decomposition, it breaks down into sodium oxide and water.
Thus, </span><span>C) 2NaOH Na2O + H2O</span>
