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Bumek [7]
3 years ago
12

If your end product is 1.5 moles of KMnO 4 how many moles of manganese oxide were used in the reaction? The equation for the pro

duction of potassium pemanganate is as follows 2 MnO 2 +4 KOH+O 2 2 KMnO 4 +2 KOH+H 2
Chemistry
1 answer:
vovikov84 [41]3 years ago
4 0

Answer:

1.5 moles

Explanation:

The equation of the reaction is given as:

2 MnO2 + 4 KOH + O2 --> 2KMnO 4 + 2KOH + H2

From the equation,

2 moles of MnO2 produces 2 moles of KMnO4

x moles of MnO2 would produce 1.5 moles of KMnO4

2 = 2

x = 1.5

Solving for x;

x = 1.5 * 2 / 2

x = 1.5 moles

You might be interested in
0.00000000082 -<br> scientific notation
Nadusha1986 [10]

Answer:

8.2 x 106^-11

Explanation:

To begin this problem you must remember the basic rule of scientific notation, which is, must be between 1-10. .000000000082 is much smaller than 1. However by moving the decimal 11 spots to the right, we can make it 8.2

Continue to move the decimal to the right until the value is in the 1-10 range. Make sure to count the moves to the right.

Once the decimal is in the right spot count the spots moved.

Since the number is wayyy smaller than the answer given the number will be negative 10^-11, in order to make it what is was before.

6 0
3 years ago
How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mm hg and a temperature of 21.0°c, can be made
natta225 [31]
Answer:  
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.  
1.566 g Zn x (1 mole Zn / 65.38 g Zn) = 0.02395 moles Zn  
0.02395 moles Zn x (1 mole H2 / 1 mole Zn) = 0.02395 moles H2 produced  
Now use the ideal gas law to find the volume V.  
P = 733 mmHg x (1 atm / 760 atm) = 0.964 atm  
T = 21 C + 273 = 294 K  
PV = nRT 
V = nRT/ P = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm) = 0.600 L
7 0
3 years ago
The molar solubility of silver bromide, AgBr in pure water is 0.0007350 mol/L. What is the
gayaneshka [121]

Answer:

0.000000540

Explanation:

Step 1: Make an ICE chart for the solution of AgBr

"S" represents the molar solubility of AgBr

        AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)

I                           0             0

C                          +S          +S

E                           S             S

Step 2: Write the expression for the solubility product constant (Ksp)

Ksp = [Ag⁺] [Br⁻] = S × S

Ksp = S² = (0.0007350)² = 0.000000540

7 0
2 years ago
Using stoichiometry, determine the mass of powdered drink mix needed to make 1.0 M solution of 100 mL
Rzqust [24]

Answer:

34.23 g.

M = (no. of moles of solute)/(V of the solution (L)).

6 0
3 years ago
Consider an ideal gas at 30 ∘C and 1.02 atm pressure. To get some idea how close these molecules are to each other, on the avera
Nastasia [14]

Answer: They are close to each other by 41.03 m^3

Explanation:

From Ideal gas equation, PV = nRT

Where n is negligible

R is gas constant = 8.314 J/mol.k

T = 30 + 273 = 303K

P = 1.02 * 103351.5 = 103351.5 Pascal

Then;

PV = RT

V = P/RT

V = 103351.5/(8.314*303)

V = 41.03m^3

8 0
3 years ago
Read 2 more answers
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