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3 years ago

At STP, what is the volume of 1.0 mole of carbon dioxide?

Chemistry

2 answers:

anzhelika [568]3 years ago

8 0

Using ideal gas equation,

PV=nRT

Where P=pressure(atm), V=volume(L), n=number of moles of the gas, R=gas constant, 0.0821 atm L/moles K, T=temperature(K)

At STP:

P=1 atm

T=273K

n=1 as given

So using ideal gas equation,

V=nRT/P

So,

V=1*0.0821*273/1

=22.4 L

Ganezh [65]3 years ago

5 0

Answer : The volume of carbon dioxide at STP is 22.4 L

Explanation : Given,

Moles of carbon dioxide = 1.0 mole

STP : It stands for standard temperature and pressure.

The pressure is 1 atm and temperature is 273 K.

As we know that at STP, 1 mole of substance always contains 22.4 L volume of substance.

That means,

1.0 mole of carbon dioxide contains 22.4 L volume of carbon dioxide

Hence, the volume of carbon dioxide at STP is 22.4 L

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Magnesium burns in air with a dazzling brilliance to produce magnesium oxide: 2 Mg(s) + O2(g) → 2 MgO(s) How many moles of O2 ar

TEA [102]

Answer:

1.05 mols of O2 gas

Explanation:

For this type of problem, it's important to understand what the balanced Chemical equation tells us:

<h3><u>Balanced Chemical equations</u></h3>

Let's look at the balanced chemical equation:

$2Mg(s)+O_2(g)-- > 2MgO(s)$

This equation two sides of the reaction arrow. On the left are the reactants (things you start with, and that react during the chemical reaction), and on the right side are products (things that are produced during the chemical reaction that you end with).

The numbers in front of each compound tell how many of molecules are involved for a full reaction without anything left over. A "mol" is a large quantity ( $6.022*10^{23}$ ), in this case, of molecules , since it's unlikely you're only taking a single molecule of each substance (it would be so tiny, you wouldn't even know you were doing the reaction).

So,for every 2 moles of Magnesium used, we'll also need 1 mole of Oxygen, and it will produce 2 moles of Magnesium Oxide.

In a way, during the reaction it's almost like 2 moles of Magnesium is equal to 1 mole of Oxygen and is equal to 2 moles of Magnesium Oxide:

$2 mol Mg(s)=1mol O_2(g)=2molMgO(s)$

From here, we can build some unit ratios, to convert between the known quantity of moles we have, and find the unknown quantity of moles that are requested.

<h3><u>Finding the right unit ratio</u></h3>

We know that we are looking at 2.10 mol of Magnesium, so we want a unit ratio with moles of Mg on the bottom. <u>We want to find moles of O2</u>, <u>so we want a unit ratio of moles of O2 on top</u>.

The unit ratio we want is the middle part of the equation, divided by the left part of the equation.

$2 mol Mg(s)=1mol O_2(g)\\\frac{2 mol Mg(s)}{2 mol Mg(s)}=\frac{1mol O_2(g)}{2 mol Mg(s)}\\1=\frac{1mol O_2(g)}{2 mol Mg(s)}$

Since this quantity is 1, it is a unit ratio and can be multiplied to other things to change their units (for this problem).

<h3><u>Finding the answer</u></h3>

Starting with what we know, and multiplying by our unit ratio:

$2.10 mol Mg(s)*\frac{1mol O_2(g)}{2 mol Mg(s)}=1.05mol O_2(g)$

Notice that the units from the first quantity cancel with the units on the bottom of the fraction, leaving only the unit on top of the fraction ... the exact units we wanted!

So, 1.05mols of O2 would be consumed during the reaction if exactly 2.10moles of Magnesium are burned.

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Explanation:

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What is the molar mass of magnesium carbonate, MgCO3? 52.314 g/mol 84.313 g/mol 96.771 g/mol 102.588 g/mol

Orlov [11]

Answer:

Molar mass of MgCO3 = 84.3 g/mol

Explanation:

What is the molar mass of magnesium carbonate.

Molar mass of a substance is defined as the mass of one mole of the substance expressed in grams hence its unit is gram per mole (g/mol)

MgCO3 is a compound and it's molar mass would be the sum of the molar masses of the elements that make up the compound. MgCO3 contains 1 atom of magnesium, one atom of carbon and 3 atoms of oxygen

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