0.1 mol/L
. The concentration of the HCl is 0.1 mol/L
a) Write the <em>balanced chemical equation
</em>
HCl + NaOH → NaCl + H2O
b) Calculate the <em>moles of NaOH
</em>
Moles of NaOH = 0.050 L NaOH x (0.1 mol NaOH/1 L NaOH)
= 0.0050 mol NaOH
c) Calculate the <em>moles of HCl
</em>
Moles of HCl = 0.0050 mol NaOH x (1 mol HCl/1 mol NaOH)
= 0.0050 mol HCl
d) Calculate the <em>molar concentration</em> of the HCl
<em>c</em> = moles/litres = 0.0050 mol/0.050 L = 0.1 mol/L
The question is incomplete, the complete question is;
Consider the following reaction at 298K.
2 H+(aq) + 2 Cr2+(aq) =H2(g) + 2 Cr3+(aq)
Which of the following statements are correct?
Choose all that apply.
K > 1
ΔGo < 0
Eocell < 0
n = 2 mol electrons
The reaction is reactant-favored.
Answer:
K > 1
ΔGo < 0
n = 2 mol electrons
Explanation:
If we look at the reaction; 2 H+(aq) + 2 Cr2+(aq) -----> H2(g) + 2 Cr3+(aq), we will notice that the reaction is a spontaneous reaction because E°cell= 0-(-0.41) = 0.41 V
If the reaction is a spontaneous electrochemical process, we expect it to be product favoured with K>1. Also, a spontaneous reaction must have ∆G<0.
Lastly, we can see that two electrons were transferred according to the balanced equation for the electrochemical reaction.
Sodium <u>loses</u> an electron, so it forms a <u><em>positive</em></u> ion
Answer: 
.
Explanation:
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
(1)
![S_{monoclinic}+O_2(g)\rightarrow SO_2(g)/tex] [tex]\Delta H^0_2=-296.36kJ](https://tex.z-dn.net/?f=S_%7Bmonoclinic%7D%2BO_2%28g%29%5Crightarrow%20SO_2%28g%29%2Ftex%5D%20%5Btex%5D%5CDelta%20H%5E0_2%3D-296.36kJ)
(2)
The final reaction is:
(3)
By subtracting (1) and (2)
Hence the enthalpy change for the transformation S(rhombic) → S(monoclinic) is 0.3kJ
Answer:
A Family Portrait
That element is NOT in the family. When we told you about families, we said that they were groups of elements that react in similar ways. Hydrogen is a very special element of the periodic table and doesn't belong to any family. While hydrogen sits in Group I, it is NOT an alkali metal.
