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3 years ago

What is the oxidation number of Mn in kmno4

2 answers:

7 0

KMnO₄ has an overall neutral charge, so that means that the total oxidation numbers should sum up to 0.

Oxygen will almost always have an oxidation number of -2, and K, because it is a group 1A metal, has an oxidation number of +1

There are 4 oxygens so the total oxidation number is 4*-2 = -8
And K has an oxidation number of +1

We are trying to find out the oxidation number of Mn. Remember since our compound is neutral, the sum of the oxidation numbers should be 0. So if we call the oxidation number of Mn as X, we then know that

X + (-8) + 1 = 0
X = -1 + 8
X = 7

Therefore, Mn has an oxidation number of 7.

RideAnS [48]3 years ago

5 0

Since O almost always has a charge of -2, the oxygen atoms have an overall charge of -8 (-2*4)

Since K has a charge of +1, that brings the overall charge to -7 (-8+1)

Then, in order for the overall charge to be 0, Mn would have to have a change of +7 (-7+7=0)

Hope this helps

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When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o

Feliz [49]

Answer:

The net ionic equation for the given reaction :

$3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)$

Explanation:

$3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)$ ...[1]

$MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)$ ..[2]

$Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)$ ...[3]

$NaI(aq)\rightarrow Na^+(aq)+I^-(aq)$

Replacing $MnI_2(aq)$ , NaI and $Na_3PO_4(aq)$ in [1] by usig [2] [3] and [4]

$3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)$

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

$3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)$

8 0

3 years ago

19.3 g of cadmium hydroxide reacted with 15.21 g of hydrobromic acid. How many grams of water can be made?

Alecsey [184]

Answer:

$m_{H_2O}=3.384gH_2O$

Explanation:

Hello,

In this case, the chemical reaction is:

$Cd(OH)_2+2HBr\rightarrow CdBr_2+2H_2O$

Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:

$n_{H_2O}^{by\ Cd(OH)_2}=19.3gCd(OH)_2*\frac{1molCd(OH)_2}{146.4gCd(OH)_2}*\frac{2molH_2O}{1molCd(OH)_2}=0.264molH_2O\\\\n_{H_2O}^{by\ HBr}=15.21gHBr*\frac{1molHBr}{80.9gHBr}*\frac{2molH_2O}{2molHBr}=0.188molH_2O$

In such a way, since HBr yields less water than cadmium hydroxide, we infer that HBr is the limiting one, therefore, the yielded mass of water are:

$m_{H_2O}=0.188molH_2O*\frac{18gH_2O}{1molH_2O}\\ \\m_{H_2O}=3.384gH_2O$

Regards.

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