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Ludmilka [50]
3 years ago
14

Riya is applying mulch to her garden. She applies it at a rate of 250,000 cn3 of mulch for every m2 of garden space. m3 At what

rate is Riya applying mulch
Chemistry
2 answers:
Lesechka [4]3 years ago
8 0

Answer:

0.25 \frac{m^{3} }{m^{2} }

Step-by-step explanation:

Since you need to convert the cm^3 into m^3 then you start off by multiplying \frac{250,000 cm^{3} }{m^{2} } to a fraction that represents m^3 over cm^3.

Like this:

\frac{250,000 cm^{3} }{m^{2} }  ×  \frac{m^{3} }{1,000,000cm^{3} }

Because cm^3 is divided by itself in the equation you cancel them both out and then your left with:

\frac{250,000m^{3} }{1,000,000m^{2} }

And when then you divide 250,000 by 1,000,000 and you get 0.25 \frac{m^{3} }{m^{2} } which is your answer.

Please mark me as Brainliest because I worked really hard on this!

Stella [2.4K]3 years ago
4 0

Mulch includes compost, grass clipping, leaves or other organic matter which increaese the fertility of soil by providing the necessary nutrients.

Volume of much used  = 250000 cm^3

To converting into m^3, divide it by 10^6

V = 250000 /10^6 = 0.25 m^3

Rate = 0.25 m^3 /m^2

Thus, riya is applying mulch at rate of 0.25 m^3 per m^2 of garden.


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The new pH is 7.69.

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The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

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Here, 100 mL  of  0.10 m TRIS buffer pH 8.3

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∴  8.3 = 8.3 + log \frac{[0.005]}{[0.005]}

<em>    </em>inverse log 0 = \frac{[B]}{[A]}

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Given; 3.0 ml of 1.0 m hcl.

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           0.003 mol of HCL.

pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69

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Learn more about pH here:

brainly.com/question/24595796

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