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Tasya [4]
3 years ago
7

4. A bird left its post and

Chemistry
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

It is a right triangle

Explanation:

We are given the directions in which a bird flew to make a triangle.

We are required to determine whether the triangle is right angle.

What is a right triangle?

  • It is a triangle that has a right angle between the shorter sides or legs
  • The triangle also obeys the Pythagoras theorem, such that the square of the two legs is equal to the square of the hypotenuse.
  • Such that if the two legs are, a and b while c is the hypotenuse.

Then, a² + b² = c²

In this case;

Taking the legs as a= 21 yards and b = 72 yards and the hypotenuse, c = 75 yards.

Then we can determine whether it is a right triangle;

a² = 21² = 441

b²= 72² = 5184

c² = 75² = 5625

a² + b² = 5625

Therefore; a² + b² = c²

Hence, the triangle is right triangle.

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i am pretty sure it is compound

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2 years ago
The pOH of a solution of KOH is 11.30. What is the [H*] for this solution?​
Sloan [31]

Answer: The concentration of hydrogen ions for this solution is 1.99 \times 10^{-3}.

Explanation:

Given: pOH = 11.30

The relation between pH and pOH is as follows.

pH + pOH = 14

pH + 11.30 = 14

pH = 14 - 11.30

= 2.7

Also, pH is the negative logarithm of concentration of hydrogen ions.

pH = - log [H^{+}]

Substitute the values into above formula as follows.

pH = -log [H^{+}]\\2.7 = -log [H^{+}]\\conc. of H^{+} = 1.99 \times 10^{-3}

Thus, we can conclude that the concentration of hydrogen ions for this solution is 1.99 \times 10^{-3}.

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3 years ago
Which properties make a metal a good material to use for electrical wires? malleability and reactivity conductivity and ductilit
Schach [20]
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3 0
2 years ago
Read 2 more answers
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
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