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Sonbull [250]
3 years ago
7

Waves crash on a beach one after another. Why doesn't water pile up on the beach

Chemistry
2 answers:
Klio2033 [76]3 years ago
6 0

sand

Explanation:

it is absorbing it

UNO [17]3 years ago
3 0

Answer:

Sand absorbers the water

Explanation:

On other days, however, when winds blow surface water toward land, the water can't just pile up on the beach, so water near the bottom has to compensate, moving offshore. ... One measures the speed and direction of flowing water by reflecting sound off particles carried by the currents.

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Comprehension Questions:
DaniilM [7]

Answer:

It does not change and object's mass

5 0
3 years ago
A balloon has a volume of 1.75L at a temperature of 25 °C . what will be the volume of the balloon if you take it out into the w
NemiM [27]
Boyle’s Law P1V1 = P2V2 P1 = 0.80 atm V1 = 1.8 L P2 = 1.0 atm V2 = ?? (.8 atm)(1.8 L) = (1.0 atm)(V2) 1.44 atm x L = 1 atm V2

3 0
3 years ago
Consider the following 1.0 mol L–1
Ainat [17]
(a) sodium phosphate > sulfuric acid > phosphoric acid > sucrose (sugar) 
<span>(b) sucrose (sugar) > sulfuric acid > phosphoric acid > sodium phosphate </span>
<span><u>(c) sulfuric acid > sodium phosphate > phosphoric acid > sucrose (sugar) </u></span>
<span>(d) sodium phosphate > phosphoric acid > sulfuric acid > sucrose (sugar)</span>
4 0
3 years ago
Three measurements of 34.5m, 38.4m, and 35.3m are taken. If the accepted value of the measurement is 36.7m, what is the percent
8_murik_8 [283]

Answer:

A

Explanation:

% error in 1

36.7-34.5/36.7*100=5.99%

% error in 2

38.4-36.7/36.7*100=4.63℅

% error in 3

36.7-35.3/36.7*100=3.81%

7 0
4 years ago
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of its conjugate acid and the ratio of the con
Kobotan [32]

Answer:

The correct answer is 190.5 mL of 1.00 M KH₂PO₄

Explanation:

A phosphate buffer is composed by phosphate acid (KH₂PO₄) and its conjugated base (K₂HPO₄). To obtain the relation between the concentrations of base and acid to add, we use Henderson-Hasselbach equation:

pH= pKa + log \frac{base}{acid}

We have: pH= 6.97 and pKa= 7.21. So, we replace the values in the equation:

6.97= 7.21 + log \frac{base}{acid}

6.97-7.21= log \frac{base}{acid}

-0.24= log \frac{base}{acid}

10^{-0.24}= \frac{base}{acid}

0.575 = \frac{base}{acid}

\frac{0.575}{1}= \frac{base}{acid}

It means that you have to mix a volume 0.575 times of conjugated base and 1 volume of acid. If we assume a total buffer concentration of 1 M, we have:

base + acid = 1

base= 1 - acid

We replace in the previous equation:

0.575= \frac{1-acid}{acid}

0.575 acid= 1 - acid

0.575 acid + 1 acid= 1

1.575 acid = 1

acid= 1/1,575

acid= 0.635

base= 1 - acid = 1 - 0.635 = 0.365

For a total volume of 300 ml, the volumes of both acid and base will be:

300 ml x 0.635 M = 190.5 ml of acid (KH₂PO₄)

300 ml x 0.365 M= 109.5 ml of base (K₂HPO₄)

We can corroborate our calculations as follows:

190.5 ml + 109.5 ml = 300 ml

109.5 ml / 190.5 ml = 0.575

4 0
3 years ago
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