Answer:
We should add 15.15 grams of ethylen glycol.
Explanation:
Step 1: Data given
Mass of water = 125 grams
Temperature change = 1.0 °C
The boiling point elevation constant, Kbp, for water is 0.5121 °C/m
Step 2: Calculate mass needed
ΔT = i* Kb * m
⇒ with i = the van't Hoff factor = 1
⇒ with Kb = 0.5121 °c/ kg/mol
⇒ with m = molality = moles / mass
ΔT = i* Kb * m ⇒ 1°C = 0.5121 °C / kg/mol * (X/0.125kg)
X = 0.2441 moles
Step 3: Calculate mass of ethylen glycol
Mass ethylen glycol = moles * molar mass
Mass ethylen glycol = 0.2441 moles * 62.068 g/mol
Mass of ethylen glycol = 15.15 grams
We should add 15.15 grams of ethylen glycol.
Answer:
A
Explanation:
In this question, we are to calculate the enthalpy of change of the reaction. ΔH
To be able to do that, we use the Hess’ law and it involves the subtraction of the summed heat reaction of the reactants from that of the product.
Thus, mathematically, the enthalpy of change of the reaction would be;
[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4) + 4 ΔH(Cl2)]
We can see that we multiplied some heat change by some numbers. This is corresponding to the number of moles of that compound in question in the reaction.
Also, for diatomic gases such as chlorine in the reaction above, the heat of reaction is zero.
Thus, we can have the modified equation as follows;
[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4)]
Substituting the values we have according to the question, we have;
-95.98 + 4(-92.3) -(-17.9)
= -95.98 - 369.2 + 17.9
= -447.28 KJ/mol
Answer:
The atom that is oxidized is Zn.
Explanation:
Reducing agents reduce other atoms or species by being oxidized themselves. Oxidation is a loss of electrons. In the equation, the Zn atom originally has an oxidation state of zero (neutral). On the right side of the equation, Zn is part of the ionic compound zinc sulfate, which is composed of the Zn(2+) cation and the SO4(2-) anion. The Zn(2+) ion is positive, which means it must have lost electrons (which have a negative charge), and was therefore oxidized from the neutral atom Zn(s).
So let's convert this amount of mL to grams:

Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):

Then we need to convert moles to atoms using Avogadro's number:
![\frac{6.022*10^{23}atoms}{1mole} *[8.135*10^{-2}mol]=4.90*10^{22}atoms](https://tex.z-dn.net/?f=%20%5Cfrac%7B6.022%2A10%5E%7B23%7Datoms%7D%7B1mole%7D%20%2A%5B8.135%2A10%5E%7B-2%7Dmol%5D%3D4.90%2A10%5E%7B22%7Datoms%20)
So now we know that in 1.2 mL of liquid mercury, there are
present.