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CaHeK987 [17]
3 years ago
13

Calculate the mass of nitrogen dissolved at room temperature in an 95.0 LL home aquarium. Assume a total pressure of 1.0 atmatm

and a mole fraction for nitrogen of 0.78.
Chemistry
1 answer:
harina [27]3 years ago
8 0

Answer:

86.3 g  of N₂ are in the room

Explanation:

First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.

We apply the mole fraction:

Mole fraction N₂ = N₂ pressure / Total pressure

0.78 . 1 atm = 0.78 atm → N₂ pressure

Room temperature → 20°C → 20°C + 273 = 293K

Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K

(0.78 atm . 95L) /0.082 . 293K = n

3.08 moles = n

Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g  

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3 years ago
What is the limiting reactant when 8.4 moles of lithium react with 4.6 moles of oxygen gas?
serious [3.7K]

Answer: Lithium

Explanation: The balanced chemical equation is:

4Li+O_2\rightarrow 2Li_2O

It can be seen, 4 moles of lithium combines with 1 mole of oxygen gas to produce 2 moles of lithium oxide.

Thus 8.4 moles of lithium combines with=\frac{1}{4}\times {8.4}=2.1molesof oxygen gas to produce 4.2 moles of lithium oxide.

As, Lithium limits the formation of product, it is the limiting reagent and Oxygen gas is present in excess, it is called the excess reagent. (4.6-2.1)=2.5 moles of oxygen gas are present in excess.

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3 years ago
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3 years ago
When beryllium ion combines with carbonate ion, the numbers of each ion are??
Katyanochek1 [597]

Answer:

One of each

Explanation:

Be is in Group 2, so it loses its two valence electrons in a reaction to form Be²⁺ ions.

Carbonate ion has the formula CO₃²⁻.

We can use the criss-cross method to work out the formula of beryllium carbonate.  

The steps are

Write the symbols of the anion and cation.

Criss-cross the numbers of the charges to become the subscripts of the other ion.

Write the formula with the new subscripts.

Divide the subscripts by their highest common factor.

Omit all subscripts that are 1.

When you use this method with Be²⁺ and CO₃²⁻, you might  be tempted to write the formula for the beryllium carbonate as Be₂(CO₃)₂

However, you can divide the subscripts by their largest common factor (2).

This gives you the formula Be₁(CO₃)₁.

We omit subscripts that are 1, so the correct formula is  

BeCO₃

There is one Be²⁺ ion and one CO₃²⁻ ion in a formula unit of beryllium carbonate.

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