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4 years ago

What is the molar mass for Mn3O4?

Chemistry

1 answer:

Leviafan [203]4 years ago

6 0

Answer: 228.7g/mole

Explanation:

Molar mass of Mn3O4

This shows that,

There are 3Mn

And 4 oxygen

Molar mass of Manganese is 54.9

Molar mass of oxygen is 16

Molar mass of Mn304

=3(54.9)+4(16)

=164.7+64

=228.7g/moles

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What is the mass of one mole of product? (Remember the law of conservation of mass)

denis23 [38]

Answer:

To calculate the no. of moles you must know mass and molar mass of the product

Explanation:

# of moles = mass/molar mass

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2 years ago

A laboratory requires 2.0 L of a 1.5 M solution of hydrochloric acid (HCl), but the only available HCl is a 12.0 M stock solutio

ira [324]

The amount of the solute is constant during dilution. So the mole number of HCl is 2*1.5=3 mole. The volume of HCl stock is 3/12=0.25 L. So using 0.25 L stock solution and dilute to 2.0 L.

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3 years ago

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snow_lady [41]

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3 years ago

I need help with the first question what the answer?

Margaret [11]

Answer:

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Explanation:

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3 years ago

What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)

nikdorinn [45]

Answer : The volume of $O_2$ consumed are, $0.75\times 2\times 22.4$ L.

Explanation :

The balanced chemical reaction will be:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

First we have to calculate the moles of Al.

$\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}$

Molar mass of Al = 27 g/mole

$\text{Moles of }Al=\frac{55g}{27g/mol}=2mole$

Now we have to calculate the moles of $O_2$ .

From the reaction we conclude that,

As, 4 mole of Al react with 3 moles of $O_2$

So, 2 mole of Al react with $\frac{3}{4}\times 2=0.75\times 2$ moles of $O_2$

Now we have to calculate the volume of $O_2$ consumed.

As we know that, 1 mole of substance occupies 22.4 liter volume of gas.

As, 1 mole of $O_2$ occupies 22.4 liter volume of $O_2$ gas

So, $0.75\times 2$ mole of $O_2$ occupies $0.75\times 2\times 22.4$ liter volume of $O_2$ gas

Therefore, the volume of $O_2$ consumed are, $0.75\times 2\times 22.4$ L.

3 0

3 years ago