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Sophie [7]
2 years ago
5

Weeds growing in a garden will compete with vegetable plants for sunlight, water, and nutrients.

Chemistry
1 answer:
Yuki888 [10]2 years ago
5 0

Answer:

different species

Explanation:

weeds and vegetables are both plants, but they are completely different types of plants.

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mamaluj [8]
Most scientific questions are developed from Observations.
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3 years ago
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5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
3 years ago
Copy and balance these chemical equstions​
lyudmila [28]

where are equations dear????

4 0
3 years ago
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Scientists frequently use all of the following to support the theory that organisms change over time except.
Pie

Answer:

4

Explanation:

8 0
3 years ago
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A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
Harman [31]
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
y-y.......................3y...........y

3y = 1.2
y = 0,4M

Na₃PO₄ -----> 3Na(+) + PO₄(3-)
0,4-0,4..............1,2..........0,4
0.........................1,2..........0,4

C = n/V
n = C×V
n = 0,4×0,65L
n = 0,26 mol Na₃PO₄

mNa₃PO₄: (23×3)+31+(16×4) = 164 g/mol

164g ----- 1 mol
Xg -------- 0,26 mol
X = 164×0,26
X = 42,64g Na₂SO₄
5 0
3 years ago
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