The answer is 0.59 M.
Molar mass (Mr) of MgCl₂ is the sum of the molar masses of its elements.
So, from the periodic table:
Mr(Mg) = 24.3 g/l
Mr(Cl) = 35.45 g/l
Mr(MgCl₂) = Mr(Mg) + 2Mr(Cl) = 24.3 + 2 · 35.45 = 24.3 + 70.9 = 95.2 g/l
So, 1 mol has 95.2 g/l.
Our solution contains 55.8g in 1 l of solution, which is 55.8 g/l
Now, we need to make a proportion:
1 mole has 95.2 g/l, how much moles will have 55.8 g/l:
1 M : 95.2 g/l = x : 55.8 g/l
x = 1 M · 55.8 g/l ÷ 95.2 g/l ≈ 0.59 M
Answer:
Here's what I get
Explanation:
Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We must calculate the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's gather all the information in one place.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
![Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}](https://tex.z-dn.net/?f=Q_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHI%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BH%24_%7B2%7D%24%5D%5BI%24_%7B2%7D%24%5D%7D%7D%20%3D%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.30%20%5Ctimes%200.15%7D%20%3D%20%205.56%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.30%20%5Ctimes%200.15%20%5Ctimes%205.56%20%3D%200.250%5C%5Cx%20%3D%20%5Csqrt%7B0.250%7D%20%3D%20%5Ctextbf%7B0.50%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20initial%20concentration%20of%20HI%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.50%20mol%2FL%7D%7D%24%7D)
3. Plot the initial points
The graph below shows the initial concentrations plotted on the vertical axis.
Answer:
According to Le Chatelier's principle, increasing the reaction temperature of an exothermic reaction causes a shift to the left and decreasing the reaction temperature causes a shift to the right.
Explanation:
C6H12O6(s) + 6O2(g) ⇌6CO2(g) + 6H2O(g)
We are told that the forward reaction is exothermic, meaning heat is removed from the reacting substance to the surroundings.
According to Le Chatelier's principle,
1. for an exothermic reaction, on increasing the temperature, there is a shift in equilibrium to the left and formation of the product is favoured.
2. if the temperature of the system is decreased, the equilibrium shifts to right and the formation of the reactants is favoured.
3. if the reaction temperature is kept constant, the system is at equilibrium and there is no shift to the right nor to the left.
Answer:
Two half lives.
Explanation:
It is known that the decay of isotopes and radioactive material obeys first order kinetics.
Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
That means for a sample 100% to decay to 50 % it will take one half-life, and to decay the remaining 50% to 25% it will take another half-life.
So, for a sample has 25% parent and 75% daughter it will have two half-lives.
Answer:The 1st and 2nd reactions are the example of oxidation -reduction.
Explanation:
Oxidation is basically when a species loses electrons and reduction is basically when the species gains electrons.
A reaction is known as an oxidation -reduction reaction only if oxidation and reduction simultaneously occur in the reaction. It basically means if a species is getting oxidized in the reaction then the other species present in the system must be reduced in the reaction.
Oxidation-reduction reactions are also known as redox reactions.
In the 1st reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of chlorine is 0 in reactants and in products is -1 so chlorine is reduced. Hence Na is oxidized and Cl is reduced so the reaction is a example of oxidation-reduction.
2Na(s)+Cl₂(g)→2NaCl(s)
In the second reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of Cu is +1 in reactant and 0 in products so Cu is reduced. Hence Na is oxidized and Cu is reduced so the reaction is an example of oxidation-reduction.
Na(s)+CuCl(aq)→NaCl(aq)+Cu(s)
In the third reaction the oxidation state of Na changes from +1 to +1 and that of Cu also changes from +1 to +1. So there is no change in oxidation state of the species present in reactants and products. Hence this reaction is not an example of oxidation and reduction.
