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elena-14-01-66 [18.8K]
3 years ago
10

An object starts at position 12 on a horizontal line with a reference point of o. What is the position of the object if it moves

14
units to the left?
0-26
O-2
OOO
Chemistry
1 answer:
eimsori [14]3 years ago
6 0

Answer:

-2

Explanation:

Consider object is starting 12 units right from the reference point which is 0.

Assign the right direction positive sign.

when object is moving 14 units on left direction the position of object will be two units to the left side of reference point.

Assign the left direction negative sign position will be -2.

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Hello!

In this case, since the described combustion reaction is:

C_6H_{14}(l)+\frac{19}{2} O_2(g)\rightarrow 6CO_2(g)+7H_2O(g)

Thus, since 5.2 g of hexane (86.2 g/mol) is reacted with 33.0 g of oxygen (32.0 g/mol) we can compute the mass of hexane that was actually consumed via stoichiometry with oxygen (1:19/2 mole ratio):

m_{C_6H_{14}}^{consumed \ by\ O_2}=33.0gO_2*\frac{1molO_2}{32.0gO_2}*\frac{1molC_6H_{14}}{\frac{19}{2}gO_2 }  *\frac{86.2gC_6H_{14}}{1molC_6H_{14}} \\\\m_{C_6H_{14}}^{consumed \ by\ O_2}=9.36gC_6H_{14}

It is proved then than the hexane won't have any leftover but oxygen does, as shown below:

m_{O_2}^{consumed \ by\ C_6H_{14}}=5.2gC_6H_{14}*\frac{1molC_6H_{14}}{86.2gC_6H_{14}} *\frac{\frac{19}{2}molO_2 }{1molC_6H_{14}} *\frac{32.0gO_2}{1molO_2} \\\\m_{O_2}^{consumed \ by\ C_6H_{14}}=18g

It means the leftover of oxygen is:

m_{O_2}^{leftover}=33g-18g\\\\m_{O_2}^{leftover}=15g

Regards!

6 0
3 years ago
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