When heated, KClO3 decomposes into KCl and O2. 2KClO3⟶2KCl+3O2 If this reaction produced 17.4 g KCl, how many grams of O2 were produced? maSS

Answers

Answer 1

Answer:

12.6 g O2 or

12.60 g O2 (if needed two decimal places)

Explanation:

First convert grams into moles:

17.4 g KCl × 1

74.55

= 0.233 moles of KCl

Next solve for the mole of O2:

= 0.233 × 3 moles of O2

2 moles of KCl

= 0.3495 or

= 0.350 moles of O2

Finally convert the mole to grams:

= 0.350 moles of O2 × 36g

1 mole

= 12.6 g O2 or 12.60 g O2

Answer 2

Answer:

a car traveled 20m &400k/g


Related Questions

Balance the equation ​

Answers

Sorry but I can’t help you. The image is blurry and therefore the supposition is incorrect . But 12/4 equals 3

what would happen if you put vegetable oil and ice cubes into a beaker together? what would happen if you left it on the counter overnight?

Answers

Oil is hydrophobic, which means that it repels water. The ice cube will immediately begin to melt as soon as it is placed in the oil. Because oil is hydrophobic, it will not mix with the water.

If you leave the beaker overnight, you'll notice that the water sinks to the bottom and the oil floats to the top. This is because water has a higher density than oil, causing it to sink.

Hope this helps!

Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.
A solution has a pH of 5.4. Write the formula you will use to calculate the [H+] and then show all your work leading to the determination of [H+].

Answers

Question:-

A solution has a pH of 5.4, the determination of [H+].

Given :- pH:- 5.4 pH = - log[H+]

To find :- concentration of H+

Answer:- Antilog(-5.4) or 4× 10-⁶

Explanation:- Formula:- pH = -log H+

Take negative to other side

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

[tex]Antilog (-6) = {10}^{ - 6} , \\ Antilog (0.6) = 4 [/tex]

put the value in equation

[tex] {10}^{ - 6} \times 4 = [H+] \\ 4 \times {10}^{ - 6} = [H+][/tex]

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