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Elan Coil [88]
3 years ago
14

What did timbuktu contributed to mali's important as a kingdom

Chemistry
2 answers:
frosja888 [35]3 years ago
8 0

Answer:

Timbuktu was important as an area of hardworking trade.

Explanation:

Hello,

Timbuktu was a city in the kingdom of Mali that during the 12th century, produced high-demand goods like gold, salt, and ivory as prime commodities of trade, what made it very important as a kingdom.

Mali was founded by Sundiata Keita around 1235 C.E when he gathered the tribes of the Malinke people. As the time went by, Mali empire became demanding and conquered the Ghana Empire, in such a way, the empire flourished much with the contribution of Timbuktu by strengthening the trade.

Best regards.

Alex73 [517]3 years ago
5 0
Timbuktu was considered a MASSIVE trade center in Mali. That is how all of the foreign goods came into the kingdom. It also had education centers which was very important in Mali.
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\boxed {\boxed {\sf 0.541 \  mol \ Na_2CO_3}}

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

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Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

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We will convert using dimensional analysis. Set up a ratio using the molar mass.

\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

We are converting 57.3 grams to moles, so we multiply by this value.

57.3 \ g \ Na_2CO_3} *\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

Flip the ratio so the units of grams of sodium carbonate cancel.

57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386  \ g \ Na_2CO_3}

57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }

\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3

0.5406295944 \ mol \ Na_2CO_3

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

0.541 \  mol \ Na_2CO_3

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

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