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3 years ago

Join me in among us if u want :) sever code-XXILYQ

Chemistry

2 answers:

myrzilka [38]3 years ago

8 0

Answer:

cool. mind giving brainliest?

maria [59]3 years ago

4 0

Answer:

im so down but where do you go?

Explanation:

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The Earth's circumference around the equator is slightly bulged as compared to the circumference around the poles. if the circum

Igoryamba

The earth in m is 510

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3 years ago

KHP is a monoprotic acid which provides one H+ ion. How would your results be affected if a diprotic acid (such as sulfuric acid

kari74 [83]

Explanation:

In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.

Equation of the reaction

2NaOH + H2SO4 --> Na2SO4 + 2H2O

Number of moles of H2SO4 = molar concentration × volume

= 0.75 × 0.0105

= 0.007875 moles.

By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875

= 0.01575 moles.

Molar concentration of NaOH = number of moles ÷ volume

= 0.01575 ÷ 0.0175

= 0.9 M of NaOH.

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3 years ago

Ocean water Choose one: A. is less dense than freshwater. B. salinity decreases as seawater evaporates. C. contains dissolved sa

Genrish500 [490]

Answer: the answer is C

Explanation:

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3 years ago

If you have 10,000 grams of a substance that decays with a half-life of 14 days, then how much will you have after 56 days?

Crank

Answer: 625 grams

Explanation:

14 goes into 56, 4 times.

10,000÷2=5,000 - first half life

5,000÷2=2,500 - second half life

2,500÷2=1,250 - 3rd half life

1,250÷2= 625 - 4th half life

Note: don't type "grams" after 625, just type "625".

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2 years ago

A certain gas is present in a 13.0 L cylinder at 1.0 atm pressure. If the pressure is increased to 2.0 atm , the volume of the g

Scorpion4ik [409]

<h3>Answer:</h3>

The gas obeys the Boyle's law

<h3>Explanation:</h3>

According to Boyle's law, the volume of a fixed mass of a gas and the pressure are inversely proportional at constant absolute temperature.
That is; $P\alpha\frac{1}{V}$
Therefore, $k=PV$ , where k is a constant
At varying volume and pressures while keeping absolute temperature constant; k = P1V1 =P2V2

In this case, we are given;

Initial Volume of 13.0 L at initial pressure of 1.0 atm

New volume of 6.5 L at new pressure of 2.0 atm

But, K = PV

Therefore,

k1 = P1V1

= 1.0 atm × 13.0 L

= 13 atm.L

k2 = P2V2

= 2.0 atm × 6.5 L

= 13 atm.L

Thus, k1=k2

Thus, the gas obeys the Boyle's law

8 0

3 years ago