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kiruha [24]
1 year ago
6

How many moles of H2O are in 64.0 g of H2O

Chemistry
1 answer:
Alina [70]1 year ago
8 0

Answer:

Explanation:

stoichiometry of C₂H₂ to H₂O is 2:2.

Number of moles of C₂H₂ = molar mass of C₂H₂  

Since the molar mass of C₂H₂  is 26 g/mol.

Number of C₂H₂  moles reacted = 64.0 g / 26 g/mol = 2.46 mol.

according to a molar ratio of 2:2.

the number of H₂O moles formed = a number of C₂H₂  moles reacted.

Therefore the number of H₂O moles produced = 2.46 mol

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A race car is driven by a professional driver at 99 . What is this speed in and ? 1 mile = 1.61 kilometers 1 hour = 60 minutes E
slavikrds [6]

Answer: The speed is equivalent to <u>159.39 kilometers per hour </u>or <u>2.65 kilometers per minute.</u>

Explanation:

Given, The speed of a race car = 99 miles/ hour

To convert the speed into  kilometers per hour and kilometers per minute

Since   1 mile = 1.61 kilometers

So, Speed of car = (99 ) x (1.61 )

= 159.39 kilometers per hour.

Also, 1 hour = 60 minutes

Then, Speed of car = (159.39) ÷60

= 2.6565≈2.65 kilometer per minute.

Hence, the speed is equivalent to <u>159.39 kilometers per hour </u>or <u>2.65 kilometers per minute.</u>

4 0
3 years ago
What is the mechanism by which Ozone is formed in the presence of sunshine?
katovenus [111]
Stratospheric ozone is formed naturally through the interaction of UV radiation with molecular oxygen.

not sure if this is what you want but hope it helps!!!
8 0
3 years ago
how will you seperate dye from black ink? draw a neatly labelled same?? fast I will mark brainly 100%​
Nonamiya [84]

Answer:

put it through a distillery

7 0
3 years ago
g A laboratory analysis of an unknown compound found the following composition: C 75.68% ; H 8.80% ; O 15.52%. What is the empir
sammy [17]

Answer:

THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O

Explanation:

The empirical formula for the unknown compound can be obtained by following the processes below:

1 . Write out the percentage composition of the individual elements in the compound

C = 75.68 %

H = 8.80 %

O = 15.52 %

2. Divide the percentage composition by the atomic masses of the elements

C = 75 .68 / 12 = 6.3066

H = 8.80 / 1 = 8.8000

O = 15.52 / 16 = 0.9700

3. Divide the individual results by the lowest values

C = 6.3066 / 0.9700 = 6.5016

H = 8.8000 / 0.9700 = 9.0722

O = 0.9700 / 0.9700 = 1

4. Round up the values to the whole number

C = 7

H = 9

O = 1

5 Write out the empirical formula for the compound

C7H90

In conclusion, the empirical formula for the unknown compound is therefore C7H9O

7 0
3 years ago
Assume that the test tube shown started out having 10.00 g of mercury(II) oxide. After heating the test tube briefly, you find 1
anyanavicka [17]

This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.

Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:

m_{HgO}^{consumed}=10.00g-1.35 g=8.65 g

Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:

m_{O_2}=8.65g-8.00g=0.65g

Learn more:

  • brainly.com/question/14502981
  • brainly.com/question/14236219
3 0
2 years ago
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