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2 years ago

A ball thrown across a football field is an example of a(n)

Chemistry

1 answer:

Cloud [144]2 years ago

7 0

I don't know what you mean by the question, but in science it is Newton's second law.

"The second law explains how the velocity of an object changes when it is subjected to an external force."

I hope this helped!+*♡

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A tree trunk floating in the water Balanced Or Unbalanced

mars1129 [50]

Answer:

Balanced

Explanation:

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3 0

3 years ago

Write a paragraph on: how Westfield how the rust in their water formed.

Mariana [72]

Answer:

Students have investigated what's going on in the fictional town of Westfield, they learned that the mysterious reddish-brown substance in the water is actually rust, which formed because of a chemical reaction between the iron pipes and the fertilizer substance in the water.

Explanation:

8 0

2 years ago

A solution has a pOH of 8.7 so what is the pH of the solution? Is the solution acidic, basic, or neutral?

RSB [31]

PH + pOH = 14

pH = 14 - pOH

pH = 14 - 8.7

pH = 5,3

This solution is acidic.

If pH<7 - acidic
If pH=7 - neutral
If pH>7 - basic

5 0

3 years ago

Read 2 more answers

A pure substance is found to contain 53.7% fluorine and 46.3% xenon by mass. What is the empirical formula of this substance?

evablogger [386]

Answer: XF8

Explanation:

Empirical Formular shows the simplest ratio of elements in a compound.

Xe = 46.3% F = 53.7%

Divide the percentage composition of each element by the atomic mass.

Xe = 46.3/ 131.3 F= 53.7/ 19

= 0.353( approx) = 2.826 (approx)

Divide through with the smallest of the answers gotten in previous step.

Xe = 0.353 / 0.353 F = 2.826/ 0.353

= 1 = 8.0

Empirical formular = XF8

3 0

3 years ago

Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m

erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

$n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3} =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2} =0.06882molBr^-$

Now, we compute the total moles of bromide:

$n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol$

Then, the total volume in liters:

$150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L$

Therefore, the concentration of total bromide is:

$M=\frac{0.12338mol}{0.325L}\\\\M=0.380M$

Best regards!

8 0

2 years ago