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Ksivusya [100]
2 years ago
15

A gas at STP has a volume of 1.0 liter. If the pressure is doubled and the temperature remains constant, the new volume of the g

as will be...?
Chemistry
1 answer:
Aliun [14]2 years ago
5 0
PV = nRT

A gas at STP... This means that the temperature is 0°C and pressure is 1 atm.
R is the gas constant which is 0.08206 L*atm/(K*mol)

Rearranging for volume
V = nRT/P

The temperature and number of moles are held constant. This means that this uses Boyle's Law. (The ideal gas law could be manipulated to give us this result when T and n are held constant.)
PV = k
where k is a constant.

This means that
P₁V₁ = k = P₂V₂
P₁V₁ = P₂V₂

(1 atm) * (1 L) = (2 atm) * V₂
V₂ = 0.5 L

The new volume of the gas is 0.5 L.
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How many of the planets have an orbital period of less than one Earth year?
GrogVix [38]

Since orbital period depends on how far you are from the sun, planets closer to the sun have a orbital period less than one earth year.

These planets are Mercury and Venus

8 0
2 years ago
When sound molecules "stack up" in front of high-speed flights, what could
FrozenT [24]

pretty sure its B thank me later

4 0
3 years ago
How many grams of h3po4 are in 255 ml of a 4.50 m solution of h3po4?
murzikaleks [220]

H3PO4 has molecular weight of approximately 98 grams per mole. 4.50 M is equal to 4.50 mole per 1000 mL solution of H3PO4. 255 mL times 4.50 mol /1000 mL times 98 g/mol is equal to 112.455 grams. Note that I automatically equate 1 Liter to 1000 mL since the given volume is in mL for easier computation.

7 0
2 years ago
If a sample containing 18.1 g of NH3 is reacted with 90.4 g of
USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃  = 18.1g

Mass of Cu₂O  = 90.4g

Unknown:

Limiting reactant  = ?

Mass of N₂ formed  = ?

Solution:

The reaction equation is given as:

       Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

  Number of moles = \frac{mass}{molar mass}  

Molar mass of Cu₂O = 2(63.6) + 16  = 143.2g/mol

Molar mass of NH₃  = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = \frac{18.1}{143.2}   = 0.13moles

Number of moles of NH₃   = \frac{90.4}{17}   = 5.32moles

  From this reaction;

       1 mole of  Cu₂O combines with 2 mole of NH₃

So   0.13moles of  Cu₂O will combine with 0.13 x 2 mole of NH₃

                                              = 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

   Mass = number of moles x molar mass

    1 mole of Cu₂O  will produce 1 mole of N₂

    0.13 mole of Cu₂O  will produce 0.13 mole of N₂

    Mass  = 0.13 x (2 x 14) = 3.64g

5 0
3 years ago
Consider a situation in which 211 g
Stella [2.4K]

Answer:

3.00 mol

Explanation:

Given data:

Mass of P₄ = 211 g

Mass of oxygen = 240 g

Moles of P₂O₅ = ?

Solution:

Chemical equation:

P₄ + 5O₂       →     2P₂O₅

Number of moles of P₄:

Number of moles = mass/ molar mass

Number of moles = 211 g / 123.88 g/mol

Number of moles = 1.7 mol

Number of moles of O₂ :

Number of moles = mass/ molar mass

Number of moles = 240 g / 32g/mol

Number of moles = 7.5 mol

Now we will compare the moles of product with reactant.

                       O₂         :         P₂O₅

                        5          :           2

                        7.5       :        2/5×7.5 = 3.00

                       P₄          :         P₂O₅

                        1           :           2

                       1.7         :       2×1.7 = 3.4 mol

Oxygen is limiting reactant so the number of moles of P₂O₅ are 3.00 mol.

Mass of P₂O₅:

Mass = number of moles × molar mass

Mass = 3 mol ×283.9 g/mol

Mass = 852 g

3 0
2 years ago
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