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lord [1]
2 years ago
12

A given mass of air has a volume of 6.00 L at 101 kPa. At constant temperature, the pressure is decreased to 25.0 kPa. Calculate

the final volume for the gas, as described by Boyle’s law.
Chemistry
1 answer:
GenaCL600 [577]2 years ago
4 0

Answer:

24.24 L

Explanation:

Boyle’s law, also called Mariotte’s law, a relation concerning the compression and expansion of a gas at constant temperature.

This empirical relation, formulated by the physicist Robert Boyle in 1662, states that the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature; i.e., in equation form, pv = k, a constant.

Real gases obey Boyle’s law at sufficiently low pressures, although the product pv generally decreases slightly at higher pressures, where the gas begins to depart from ideal behaviour.

As, PV = k

P₁ V₁ = P₂ V₂

Given P₁ = 101 KPa

V₁ = 6 L

P₂ = 25 kPa

So, V₂ = P₁ V₁ /P₂ = 101 *6/25 = 24.24 L

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Answer:

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That is the way to find the greatest magnitude, but because I don't know your numbers so I can not answer your question, but this is the way to solve for it.

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6 0
2 years ago
How many mL of a stock 50% (w/v) KNO3 solution are needed to prepare 250 mL of a 20% (w/v) KNO3 solution?
Andre45 [30]

Answer:

100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.

Explanation:

In the given question it is mentioned that

     S1=50%

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      S2= 20%

We all know that

                     V1S1=V2S2

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When a redox reaction that takes place in an acidic solution involves an oxygen imbalance, oxygen should be balanced by adding _
olya-2409 [2.1K]

Balancing redox reactions:

Oxygen should be balanced by adding H_{2}O  as needed, while hydrogen should be balanced by adding H^{+}.

What is a redox reaction?

Redox reactions, also known as oxidation-reduction reactions, involve the simultaneous oxidation and reduction of two different reactants.

The Half-Equation Method is one technique used to balance redox processes. The equation is divided into two half-equations using this technique: one for oxidation and one for reduction.

By changing the coefficients and adding H_{2}O, H^{+}, and e^{-} in that order, each reaction is brought into equilibrium:

  1. By putting the right number of water (H_{2}O) molecules on the other side of the equation, the oxygen atoms are brought into balance.
  2. By adding H^{+} ions to the opposing side of the equation, one can balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom).
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  4. The e^{-} on either side must be made equal; if not, they must be multiplied by the lowest common multiple (LCM) in order to make them equal.
  5. One balanced equation is created by adding the two half-equations and canceling out the electrons. Additionally, common terms should be eliminated.
  6. Now that the equation has been verified, it can be balanced.

Learn more about redox reaction here,

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