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2 years ago

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What is the oxidation state of Mn atom

1 answer:

liraira [26]2 years ago

4 0

The oxidation state for an atom on Mn or manganese is +2. It also depends on the element it can be paired with. When it is paired with the compound KMnO4, the oxidation state of Mn is 5+. Oxidation state is useful in redox reactions.

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Which one is false PLEASE HELP

Mkey [24]

Answer:

D

Explanation:

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1 year ago

How do I calculate volume of a gas?

ale4655 [162]

The molar<span> volume of an ideal gas is therefore 22.4 dm</span>3<span> at </span>stp<span>. And, of course, you could redo this calculation to find the volume of 1 mole of an ideal gas at room temperature and pressure - or any other temperature and pressure.</span>

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2 years ago

Gifblaar is a small South African shrub and one of the most poisonous plants known because it contains fluoroacetic acid (FCH2CO

PSYCHO15rus [73]

[H_{3}O^{+}] = 0.00770 M

The equilibrium equation representing the dissociation of $FCH_{2}COOH$

$FCH_{2}COOH(aq) + H_{2}O (l) FCH_{2}COO^{-}(aq)+ H_{3}O^{+}(aq)$

Given [H_{3}O^{+}] = 0.00770 M

Let the initial concentration of acid be x and change y

So y = $[H_{3}O^{+}]$ = $[FCH_{2}COO^{-}]$ = 0.00770 M

$pK_{a} = 2.59K_{a} = 10^{-2.59} = 0.00257 M$

$K_{a} = \frac{(0.00770 M)(0.00770 M)}{x - 0.00770}$

$0.00257 = \frac{0.00005929}{x - 0.00770}$

0.00257 x - 0.00001979 = 0.00005929

x = 0.031 M

Therefore, initial concentration of the weak acid is <u>0.031 M</u>

4 0

2 years ago

What is the molar mass of magnesium hydroxide

leonid [27]

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7 0

2 years ago

Read 2 more answers

A cylinder, with a piston pressing down with a constant pressure, is filled with 2.10 moles of a gas (n1), and its volume is 46.

olga2289 [7]

Answer:

The volume is $V_2 =15.33 \ L$

Explanation:

From the question we are told that

The number of moles of the gas is $n_1 = 2.10 \ moles$

The volume of the gas is $V = 46.0 \ L$

The of moles of the gas that leaked is $n_2 = 0.700 \ mol$

Generally from the ideal gas law

$PV = nRT$

Generally at constant pressure and temperature

$\frac{V}{n} = constant$

So $\frac{V_1 }{n_1} = \frac{V_2 }{n_2}$

=> $V_2 = \frac{ V_1 * n_2 }{ n_1}$

=> $V_2 = \frac{ 46.0 * 0.700 }{ 2.10 }$

=> $V_2 =15.33 \ L$

5 0

2 years ago