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3 years ago

Jay wants to know how the amount of acidity in rainwater influences the weathering rate of limestone.

2 answers:

5 0

Answer:- To check the reliability of our experiments we repeat the experiment many times and see if we get the same results or almost same results.

It's also good to compare the results of two or more students for the same experiment and see if they get the similar results.

So, to improve the reliability and accuracy of the experimental data, Jay needs to repeat the experiment and also ask his one or more friends to do the experiment and record the data.

creativ13 [48]3 years ago

3 0

He can repeat his experiment multiple times, and ask a friend to repeat it as well to see if they get the same answer.

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Why must humans find substitutes for many minerals found on Earth? (A) form at an extremely slow rate (B) controlled by other co

DochEvi [55]

The correct answer is C, too deep in the Earth to collect. Hope this helps!

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3 years ago

WITCH IS MADE BY WEATHERING ASH SEDEMENTS MAGMA OR MINERALS

algol [13]

Magma is made from sediments

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3 years ago

What is the formula for tricarbon heptafluoride? C2F3 CF7 C3F7 C773

bagirrra123 [75]

Answer:

C3F7

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hence F7

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3 years ago

True or False: You can make a concentrated solution more dilute by adding solvent.

Vlad [161]

That statement is true because to make a diluted solution, you have to add more solvent than the solute in a concentrated solution. This is extra information, but to make a diluted solution more concentrated, you have to add more solute since a concentrated solution has an excess amount of solute as stated before.

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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

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

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

B) Radius

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

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

$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