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Aleonysh [2.5K]
2 years ago
7

How many moles of O2 are required to react with 2.4 mol of H2 ?

Chemistry
1 answer:
Leokris [45]2 years ago
8 0

Answer:

1.2 moles

Explanation:

this is the balanced equation for the reaction of oxygen (O2) and hydrogen (H2), usually we don't write the 1 in front of O2

2H₂ + 10₂ → 2H₂O

the molar ratio of hydrogen to oxygen is 2 : 1

we are trying to react with 2.4 mol of H2 so the moles of O2 is half the number of moles of H2 = 2.4 ÷ 2 = 1.2 mol

another way to think of it:

2H₂ + 10₂

2 : 1

2.4 mol : x mol

to get from 2 to 2.4 multiply by 1.2, so do the same to the other side

1 × 1.2 = 1.2 mol

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Answer:

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3 years ago
A graduated cylinder contains 18.0ml of water. what is the new water level after 35.6g of silver metal with a density of 10.5 g/
AlekseyPX

Answer:

The answer to your question is

1.- Volume = 3.4 ml

2.- Volume = 0.61 ml

3.- Mass = 2872.8 pounds

Explanation:

Problem 1

Volume = 18 ml

mass = 35.6 g

density = 10.5 g/ml

Process

1.- Calculate the volume of silver

Formula

density = \frac{mass}{volume}

solve for volume

volume = \frac{mass}{density}

Substitution

volume = \frac{35.6}{10.5}

<u>volume = 3.4 ml</u>

2.- Problem 2

Total volume = ?

Volume = 18 + 3.4

Volume = 21.4 ml

Data

mass = 8.3 g

density = 13.6 g(ml

volume = ?

Formula

density = \frac{mass}{volume}

Solve for volume

volume = \frac{mass}{density}

Substitution

volume = \frac{8.3}{13.6}

Result

<u>volume = 0.61 ml</u>

3.- Problem 3

Data

volume = 345 gal

density = 1 g/ml

mass = ?

Formula

density = \frac{mass}{volume}

Solve for mass

mass = density x volume

Covert gal to ml

                            1 gal --------------- 3785 ml

                         345 gal -------------  x

                            x = (345 x 3785) / 1

                            x = 1305825 ml

Substitution

mass = 1 x 1305825

mass = 1305825 g

Convert g to pounds

                        1 g ------------------- 0.0022 pounds

              1305825 g ----------------   x

                        x = (1305825 x 0.0022)

                       <u> x = 2872.8 pounds</u>

5 0
3 years ago
What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso
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The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247

The pH of the buffer can be known as

pH = pK_{a} + log[\frac{[A-]}{[HA]}}]

The concentration of [A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\

Similarly, the concentration of [HA] = \frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404

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pH = 6.247 + log [ 0.304/0.404]

pH = 6.247 + log 0.304 - log 0.404=6.247-0.517+0.3936=6.1236

So, the pH of the buffer is 6.1236.

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Answer:you would pick the fluid by seeing what would work such as gas in a truck or car you need to know what would work best and the best power for the lift

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