All categories

3 years ago

How many moles are in 22 grams of argon?

Chemistry

1 answer:

RoseWind [281]3 years ago

6 0

Moles of Ar = mass/atomic weight of Ar
= 22/39.948
= 0.55 moles

Hope this helps!

You might be interested in

A combustion reaction will always be associated with a change in entropy that will be:________.

Alex Ar [27]

Answer:

Option (A) : Positive

Explanation:

A combustion reaction will always be associated with a change in entropy that is Positive due to the gaseous products released. Hence, there is a large positive entropy change.

8 0

3 years ago

Ground water, the largest source of fresh water, is stored in bodies of rock and or sediment

zepelin [54]

D. Aquifers
Hope this helps chief.

7 0

3 years ago

What is the mass number of an atom containing 35 protons and 45 neutrons?

vodka [1.7K]

Answer:

Explanation:

1. Weight = 35 P and 45 N = 80

2. Atom: 35P = 35 electrons

3. Bromine

4. It's a non metal. It's in column 17. It will gain 1 electron (normally).

5. Ion: 35 protons, 36 electrons

6. Charge: - 1 usually, but there are exceptions

7. Anions go to the Anode. The anode attracts minus charged elements.

8. In an uncharged state, element 36 would be next. That would be Krypton which is always uncharged. It is a noble gas.

6 0

2 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$

$molH_{2} O = 0,783 mol$

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴ $MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}$

$MassOfH_{2}O = 14,09g$

5 0

3 years ago

Use the balanced equation from number four above to determine the limiting reactant if we had a mixture of 5.0 moles of Fe and 4

Tom [10]

Answer:

2Fe + 3O₂ → 2Fe₂O₃

Limiting reactant: O₂

2Fe + O₂ → 2FeO

Limiting reactant: Fe

Explanation:

2Fe + 3O₂ → 2Fe₂O₃

2Fe + O₂ → 2FeO

These are the possible reactions:

In first case, 2 moles of Fe need 3 mol of oyxgen to react

If I have 5 moles of Fe, I will need (5 .3)/2 = 7.5 moles of O₂

Then, the oxygen is my limiting ( I only have 4 moles)

3 moles of O₂ need 2 moles of Fe

If I have 4 moles of O₂, I will need (4 .2)/3 = 2.66 moles (I have 5)

Fe, is the reactant in excess.

For second case, 2 moles of Fe need 1 mol of O₂, to react.

If I have 5 moles of Fe, I will need ( 5 .1) / 2 =2.5 moles of O₂

I have 4 moles of oxygen, so now it is my excess.

1 mol of O₂ need 2 moles of Fe, to react

If I have 4 moles of O₂, I will need the double of amount, 8 moles of Fe.

I have 5 moles, then the Fe is my limtiing reactant.

8 0

2 years ago