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2 years ago

The greater the mass of an object

1 answer:

6 0

The greater the mass of an object, the greater<span> its gravitational force.
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Plz chose as brainliest answer plz :) <#

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2 years ago

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2 years ago

the volume of gas in a container was originally 3.24L, while at standard pressure 1.00 atm. what will the volume be if the press

Simora [160]

Hello!

To solve this problem, we will use the Boyle's Law, which describes how pressure changes when volume changes and vice-versa. The equation for this law is the following one, and we'll clear for V2:

$P1*V1=P2*V2 \\ \\ V2= \frac{P1*V1}{P2}= \frac{3,24 L * 1 atm}{1,20 atm}= 2,7 L$

So, the final volume after increasing the pressure would be 2,7 L. That means that volume decreases when the pressure increases

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2 years ago

How is iron extracted? Describe in brief

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2 years ago

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The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5

Zigmanuir [339]

Answer:

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <--------> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <----> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. -----------------------(**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

4 0

2 years ago