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Deffense [45]
3 years ago
6

Which option describes the most cost-effective way for a skateboard company to purchase plywood?

Chemistry
2 answers:
marta [7]3 years ago
6 0
A pallet of 500 sheets for $3,500 is the most cost-effective option.
This is because:
- if you buy 100 for $1,000, the price per sheet is $10.
- If you buy 200 for $1,800, the price per sheet is $9
- If you buy 1,000 for $8,000, the price per sheet is $8
- If you buy 500 for $3,500, the price per sheet is only $7.
Sati [7]3 years ago
6 0

Answer: C

Explanation:

The unit rate of the first problem is $10

The unit rate of the second problem is $9

The unit rate of the third problem is $7

The unit rate of the fourth problem is $8

The answer would be C because it has the lowest unit rate.

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(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
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Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

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