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Olegator [25]
2 years ago
13

What is the molecular formula of manganese dioxide ?

Chemistry
2 answers:
Gre4nikov [31]2 years ago
8 0
The molecular formula for manganese dioxide is MnO₂ :)
Vadim26 [7]2 years ago
6 0

Answer:

MnO₂

Explanation:

Hope this helps you!

You might be interested in
Which one of these statements about carbon tetrachloride fire extinguishers is true?
Fofino [41]

Answer;

-Prohibited from use by OSHA

Explanation;

-Carbon tetrachloride is a clear and colorless liquid with an odor similar to that of chloroform.It is employed as a chemical reagent for a number of purposes, as a raw material in chemical manufacture and is, used very widely as a reagent to extinguish fires.

-It volatizes very easily, is a non-conductor of electricity and, as noted above, freezes at very low temperatures.

-It is effective, due to the blanketing effect of its vaporized fumes, on many Class B or volatile, flammable liquid fires and because it will mix with certain volatile, flammable liquids and form a non-burning combination.


5 0
3 years ago
GRADEC
True [87]

Answer:

it's-B.....................

8 0
2 years ago
Analysis of a compound of sulfur, oxygen and fluorine showed that it is 31.42% S and 31.35% O, with F accounting for the remaind
Leya [2.2K]

Answer:

The molecular formula is  SO2F2

Explanation:

Step 1: Data given

Suppose the mass of compound = 100 grams

The compound contains:

31.42 % S = 31.42 grams S

31.35 % O = 31.35 grams O

100 - 31.42 - 31.35 = 37.23 F

Molar mass of S = 32.065 g/mol

Molar mass F = 19.00 g/mol

Molar mass O = 16.00 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles S = 31.42 grams / 32.065 g/mol

Moles S = 0.9799 moles

Moles 0 = 31.35 grams / 16.00 g/mol

Moles 0 = 1.959 moles

Moles F = 37.23 grams / 19.00 g/mol

Moles F = 1.959 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.9799 / 0.9799 = 1

F: 1.959/ 0.9799 = 2

O : 1.959 / 0.9799 = 2

The empirical formula is SO2F2

This formula has a molecular mass of 102.06 g/mol

This means the empirical formula is also the molecular formula : SO2F2

3 0
2 years ago
For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow
lora16 [44]

Answer:

Rate constant of the reaction is 3.3\times 10^{-3} M^{-2} s^{-1}.

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

R=k[A]^a[B]^b[C]^c

Rate(R) of the reaction in trail 1 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.30 M

R=9.0\times 10^{-5} M/s

9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c...[1]

Rate(R) of the reaction in trail 2 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.90 M

R=2.7\times 10^{-4} M/s

2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c...[2]

Rate(R) of the reaction in trail 3 ,when :

[A]=0.60 M,[B]=0.30 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c...[3]

Rate(R) of the reaction in trail 4 ,when :

[A]=0.60 M,[B]=0.60 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

R=k[A]^2[B]^0[C]^1

Rate constant of the reaction = k

9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1

k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}

k=3.3\times 10^{-3} M^{-2} s^{-1}

7 0
3 years ago
What is the empirical formula for a compound if 300.00 g of it is known to contain 82.46224 g of molybdenum, 45.741 g of chlorin
Marysya12 [62]

Answer:

MoClBr₂

Explanation:

First we calculate the mass of bromine in the compound:

  • 300.00 g - (82.46224 g + 45.741 g) = 171.79676 g

Then we<u> calculate the number of moles of each element</u>, using their <em>respective molar masses</em>:

  • 82.46224 g Mo ÷ 95.95 g/mol = 0.9594 mol Mo
  • 45.741 g Cl ÷ 35.45 g/mol = 1.290 mol Cl
  • 171.79676 g Br ÷79.9 g/mol = 2.150 mol Br

Now we <u>divide those numbers of moles by the lowest number among them</u>:

  • 0.9594 mol Mo / 0.9594 = 1
  • 1.290 mol Cl / 0.9594 = 1.34 ≅ 1
  • 2.150 mol Br / 0.9594 = 2.24 ≅ 2

Meaning the empirical formula is MoClBr₂.

8 0
3 years ago
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