In this case, according to the given information about the oxidation numbers anf the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.
Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.
Next, we can write the following 
, since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:
Next, we multiply each anion's oxidation number by the subscript, to obtain the following:
Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.
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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:
185 gms o NaBr
Explanation:
Na Br mole weight = 22.989 +79.904 = <u>102.893 gm/mole</u>
1.083 x 10^24 molecules / 6.022 x 10^23 molecules / mole = <u>1.798 moles</u>
102.893 * 1.798 = 185 gms
We have to calculate the number of carbon atoms present in 2-carat pure diamond.
The number of carbon atoms present in 2-carat pure diamond is: 0.19 X 10²³ number of carbon atom.
Diamond is one allotrope of carbon. Atomic mass of carbon is 12.
Mass of one mole carbon atom is 12 g. One mole carbon contains Avogadro's number i.e, 6.023 X 10²³ number of atoms.
0.40 g diamond contains 0.40/12 moles= 0.033 moles of carbon atom.
So, number of carbon atoms present in 0.40 g diamond (i.e, 0.033 mole diamond) is 0.033 X 6.023 X 10²³= 1.98 X 10²²=0.19 X 10²³ .
Therefore, 0.19 X 10²³ number of carbon atoms are in a 2-carat pure diamond that has a mass of 0.40 g
Answer:
See below
Explanation:
PV = n RT R = .082057 L - Atm / (K-Mole)
.97 V = .118 (.082057)(305)
V = 3.04 liters
Will not change for argon.....will be the same for all ideal gases
