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Elenna [48]
1 year ago
11

Could use some help, thanks. Complete each row of the table below by filling in the missing prefix or missing exponent.

Chemistry
1 answer:
Minchanka [31]1 year ago
3 0

Answer:

if eel u gngg ferr

Explanation:

asdads

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To answer this question, you need to convert the gallon unit into liter. For every gallon of milk, there would be 3.78541 liters of milk. Then, the calculation would be: 334 gallon x 3.78541liter/gallon= 1264.32 liters.

If you make it into scientific notation then it would be 1.26 x 10^3
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A graduated cylinder contains 19.6 mL of water
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Can we cover the earth with one mole of pennies? Explain.
mezya [45]

Answer:

No we can’t cover the earth with one mole of pennies

Explanation:

We can’t cover the earth with one mole of pennies due to the fact that making this possible will require the flattening and extra expansion before it covers the whole surface.

The extra expansion and flattening with result in the loss of intermolecular forces and won’t remain as a solid and form other states of matter which won’t be able to cover the surface.

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3 0
2 years ago
139%
GaryK [48]

<u>Answer:</u>

The percent composition of this compound is 94%

<u>Explanation:</u>

The reaction can be formed as

2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}

\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}

\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}

\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}

Based on no. of iron reacted,  

\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)

n = m/M

\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}

% composition ofFeCl_3  

=  (9.75 / 10.39)^{*} 100

= 94%

6 0
3 years ago
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