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Ludmilka [50]
3 years ago
10

The black, granular material that fills a dry cell in a common flashlight (between the carbon rod and the zinc shell) is mangane

se dioxide, MnO2. The oxidation number of manganese in MnO2 is:
a. +2
b. -2
c. +4
d. -4
Chemistry
1 answer:
mestny [16]3 years ago
7 0

Answer: c. +4

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.  

Mn:25:1s^22s^22p^63s^23p^64s^23d^{5}

Mn^{4+}:21:1s^22s^22p^63s^23p^63d^1

O:8:1s^22s^22p^4

O^{2-}:10:1s^22s^22p^6

Here manganese is having an oxidation state of +4 called as  cation and oxygen forms an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged to give Mn_2O_4 which written in simplest whole number ratios to give neutral MnO_2

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Substance X is a compound containing 632mg of manganese and 368mg of oxygen. Substance X is shown
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The empirical formula : MnO₂.

<h3>Further explanation</h3>

Given

632mg of manganese(Mn) = 0.632 g

368mg of oxygen(O) = 0.368 g

M Mn = 55

M O = 16

Required

The empirical formula

Solution

You didn't include the pictures, but the steps for finding the empirical formula are generally the same

  • Find mol(mass : atomic mass)

Mn : 0.632 : 55 = 0.0115

O : 0.368 : 16 =0.023

  • Divide by the smallest mol(Mn=0.0115)

Mn : O =

\tt \dfrac{0.0115}{0.0115}\div \dfrac{0.023}{0.0115}=1\div 2

The empirical formula : MnO₂

8 0
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Calculate the mass in grams of calcium carbonate present in a 50.00 mL sample of an aqueous calcium carbonate standard, assuming
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3 years ago
The addition of 250.0 J to 30.0 g of copper initially at 22.0°C will change its temperature to what final value? (Specific heat
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Answer:

Final temperature = 43.53^{\circ} C

Explanation:

Given that,

Heat added, Q = 250 J

Mass, m = 30 g

Initial temperature, T₁ = 22°C

The Specific heat of Cu= 0.387 J/g °C

We know that, heat added due to the change in temperature is given by :

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So, the final temperature is equal to 43.53^{\circ} C.

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