Which of the following is the best example of how Earth's geosphere interacts with the biosphere?

`Determining the limiting reactant is an extremely important concept in chemistry. To do this, you need to know all of the follo

mezya [45]

Answer:

C the mass of each product formed

Explanation:

To the determine the limiting reactant, it is essential that we have the balanced equation of the reaction from which we can calculate the stochiometry mole ratio of the reactant. After this, we need to calculate the molar mass of the reactants, using the mole from the balanced equation we can calculate each mass of each reactant needed. Finally we need the mass of each reactant using proportion we can calculate the amount needed for the reaction from the masses of the reactant by comparing the mass given against the mass calculated from the balanced equation. After this, the mass that is exhausted or that is finished will be the limiting reactant which is the reactant that finished and caused the reaction to stop.

7 0

3 years ago

Place whole number coefficients in the blanks to balance the chemical reaction.

mote1985 [20]

It is already balanced

4 0

3 years ago

Read 2 more answers

What does the symbol "kg" represent in the metric system?

agasfer [191]

$\huge\mathfrak\red {Answer:}$

It represents

<h2>A) 1,000 grams</h2>

(1 kilogram = 1000 grams)

$\mathfrak\purple {Hope\: this\: helps\: you...}$

3 0

3 years ago

Read 2 more answers

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

<em>B) Radius</em>

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

3 0

3 years ago

The summary or ending of your experiment

Katyanochek1 [597]

Answer:

Conclusion

Explanation:

I believe you were asking for the term that best matches with the description given. Typically the conclusion summarizes your experiment in a 1 to 2 paragraph format.

7 0

3 years ago