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Vika [28.1K]
3 years ago
6

Initial Volume (VI)Final Volume (Vf) =Object's volume =​

Chemistry
1 answer:
Blizzard [7]3 years ago
3 0
Final volume minus the initial equals objects volume
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How can I balance this equation? ____ KClO3 ---> ____ KCl + ____ O2
tensa zangetsu [6.8K]
__ KClO₃ → __ KCl + __ O₂

Left Side:
1 K
1 Cl
3 O

Right Side:
1 K
1 Cl
2 O

Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.

This gives us 2KClO₃ → __ KCl + 3O₂.

However, this equation is still not balanced.

Left Side:
2 K
2 Cl
6 O

Right Side:
1 K
1 Cl
6 O

In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.

2KClO₃ → 2KCl + 3O₂
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How many moles in 3.69 x 10^30 molecules of carbon dioxide?
Sedaia [141]

0.612 \times 10^{7} \text { moles of } \mathrm{CO}_{2} are present in 3.69 \times 10^{30} \text { molecules of } \mathrm{CO}_{2}

<u>Explanation:</u>

It is known that each mole of an element is composed of avagadro's number of molecules. So if we need to determine, we need to divide the number of molecules with the avagadro's number.

So,

    1 mol of element =6.02 \times 10^{23} molecules of element

As here 3.69 \times 10^{30} molecules of carbon di oxide is given. So the moles in it will be

   No. of moles of carbon dioxide = \frac{3.69 \times 10^{30}}{6.02 \times 10^{23}}

    No. of moles = 0.612 \times 10^{7} moles of carbon dioxide.

Thus,

0.612 \times 10^{7} of carbon dioxide are present in 3.69 \times 10^{30} \text { molecules of } \mathrm{CO}_{2}.

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