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anzhelika [568]
3 years ago
10

To what volume must 175 L of air at 1.0 atm be compressed to yield a pressure of 5.2 atm? (You can assume constant temperature.

Chemistry
2 answers:
dmitriy555 [2]3 years ago
8 0

Answer:

The volume upto which a gas can be  compressed at 5.2 atm pressure is 33.65 L.

Explanation:

Under constant temperature:

Initial volume of the gas V_1=175 L

Initial pressure of the gas = P_1=1.0 atm

Final volume of the gas V_2=?

Final pressure of the gas = P_2=5.2 atm

Applying Boyle's law:

P_1V_1=P_2V_2 (constant temperature)

V_2=\frac{1.0 atm\times 175 L}{5.2 atm}=33.65 L

The volume upto which a gas can be compressed at 5.2 atm pressure is 33.65 L.

Elena L [17]3 years ago
6 0
Using the ideal gas law, for a constant temperature, pV = constant (also known as Boyle's Law)
(1 atm)(175 L) = (5.2 atm)(V) = constant
V = 33.7 L
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1N2 + 3H2 -->
Hunter-Best [27]

Answer:

28.23 g NH₃

Explanation:

The balanced chemical equation is:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

Thus, 1 mol of N₂ reacts with 2 moles of H₂ to produce 2 moles of NH₃. We convert the moles to mass (in grams) by using the molecular weight (MW) of each compound:

MW(N₂) = 2 x 14 g/mol = 28 g/mol

mass N₂= 1 mol x 28 g/mol = 28 g

MW(H₂) = 2 x 1 g/mol = 2 g/mol

mass H₂ = 3 mol x 2 g/mol = 6 g

MW(NH₃) = 14 g/mol + (3 x 1 g/mol) = 17 g/mol

mass NH₃= 2 moles x 17 g/mol = 34 g

Now, we have to figure out which is the limiting reactant. For this, we know that the stoichiometric ratio is 28 g N₂/6 g H₂. If we have 36.85 g of H₂, we need the following mass of N₂:

36.85 g H₂ x 28 g N₂/6 g H₂ = 171.97 g N₂

We have 23.15 g N₂ and we need 171.97 g. So, we have lesser N₂ than we need. Thus, the limiting reactant is N₂.

Now, we calculate the product (NH₃) by using the stoichiometric ratio 34 g NH₃/28 g N₂, with the mass of N₂ we have:

23.25 g N₂ x 34 g NH₃/28 g N₂ = 28.23 g NH₃

Therefore, the maximum amount of NH₃ that can be produced is 28.23 grams.

5 0
2 years ago
The blank <br> and the blank <br> of atoms are the same on both sides of a chemical equation.
tester [92]

Answer:

Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. Therefore, there must be the same number of atoms of each element on each side of a chemical equation.

Explanation:

7 0
2 years ago
Draw the products of the complete hydrolysis of an acetal. Draw all products of the reaction.
jolli1 [7]

Answer: the product is ketone or aldehyde

Explanation:

The first step is the conversion of acetal to hemiacetal in the presence of H3O+/ ROH, and then the final conversion of hemiacetal to ketone/aldehyde using

H3O+/ ROH...

Attached is the structural conversion

8 0
3 years ago
How many grams of barium sulfate, BaSO 4 , be precipitated when 2.25 moles of sodium sulfate , Na 2 SO 3 , reacts with an excess
gulaghasi [49]

Answer:

525.1 g of BaSO₄ are produced.

Explanation:

The reaction of precipitation is:

Na₂SO₄ (aq) + BaCl₂ (aq) →  BaSO₄ (s) ↓  +  2NaCl (aq)

Ratio is 1:1. So 1 mol of sodium sulfate can make precipitate 1 mol of barium sulfate.

The excersise determines that the excess is the  BaCl₂.

After the reaction goes complete and, at 100 % yield reaction, 2.25 moles of BaSO₄ are produced.

We convert the moles to mass: 2.25 mol . 233.38 g/mol = 525.1 g

The precipitation's equilibrium is:

SO₄⁻² (aq)  +  Ba²⁺ (aq)  ⇄  BaSO₄ (s) ↓       Kps

7 0
2 years ago
What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

3 0
3 years ago
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