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adoni [48]
3 years ago
15

If you worked at a job for 9.00/hour, what type of correlation is the number of hours worked and your take-home pay?

Chemistry
1 answer:
WITCHER [35]3 years ago
3 0

Answer:

There is a linear correlation between the number of hours worked and take-home pay. And more exactly, a direct proportionality. C = 9\cdot t

Explanation:

From statement we deduce that amount of hours worked (t) (Independent variable, measured in hours) is directly proportional to take-home pay (C) (dependent, measured in monetary units), since proportionality ratio is constant and, therefore, there exists a linear correlation between both variables. That is:

C \propto t

C = k\cdot t (Eq. 1)

Where k is the proportionality ratio, measured in monetary units per hour. If we know that k = 9\,\frac{m.u.}{h}, then the resulting correlation is:

C = 9\cdot t

There is a linear correlation between the number of hours worked and take-home pay.

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A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.
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Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\  &=185.59\text{ g/mol} \end{aligned}

Dimensional Analysis:

\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }

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