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zvonat [6]
3 years ago
9

The line graph shows the heights of plants grown in fertilized and unfertilized soil. Based on this information, what will most

likely occur on day six?
f Plant A should increase about 2 cm in height.
g Neither plant should increase by 1 cm in height.
h Plant A should increase about 1 cm in height.
j Plant B should increase about 1 cm in height.

Chemistry
1 answer:
AfilCa [17]3 years ago
4 0

Answer:

  • <u><em>g) Neither plant should increase by 1 cm in height.</em></u>

Explanation:

See the graph for this question on the figure attached.

The growing of the <em>plant A</em> is represented by the line that goes above the other. At start, that line has a slope that rises about 0.75 cm ( height increase) in 1 day. From the day 2 and forward the slope of the line decreases. The line reaches its highest point about at day 4 and seems to start decreasing. Thus, you should predict that on the day six it <em>most likely </em>does not increase in height.

The growing of the <em>plant B</em> is represented by the line drawn below the other. As for the plant B, the growing decreases with the number of days. Between the days 4 and 5 the line is almost flat, which means that <em>most likely</em> this plant will not grow on the day six or grow less than 0.5 cm.

Thus, for both plants you can say that <em>on day six, most likley, neither should increase by 1 cm in height (</em>option g).

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. How many grams of Na2SO4 are required to make 2700 mL of a 2.0 M solution?
ratelena [41]

Answer:

Explanation:

From the net ionic equation

Ba2+(aq) + SO42-(aq) ==> BaSO4(s) we see that 1 mole Ba2+ reacts with 1 mole SO42- to -> 1 mol BaSO4

Find moles of Ba2+ used: 0.250 moles/L x 0.0323 L = 0.008075 moles Ba2+

Find moles SO42- present: 0.008075 moles Ba2+ x 1 mol SO42-/1 mol Ba2+ = 0.008075 mol SO42-

Find mass of Na2SO4 present: 0.008075 mol SO42- x 1 mol Na2SO4/1 mol SO42- x 142.04 Na2SO4/mole = 1.14698 g = 1.15 g Na2SO4 (to 3 significant figures)

6 0
2 years ago
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To arrive at the number of hydrogen atoms, What can be said about the relationship between the number of hydrogen atoms and the
Semenov [28]

Answer:

nothing shsheysyeggehssl

6 0
3 years ago
1. A sample of oxygen is collected over water at 22 ° C and 762 torr. What is the partial pressure of the dry oxygen? The vapor
Rom4ik [11]

Answer: The partial pressure of the dry oxygen is 742 torr

Explanation:

Dalton's Law of Partial Pressure states that the total pressure exerted by a mixture of gases is the sum of partial pressure of each individual gas present. Thus P(total)=P_1+P_2 .........

Given; Total pressure = 762 torr

partial pressure of water = 19.8 torr

partial pressure of dry oxygen = ? torr

Total pressure  = partial pressure of water + partial pressure of dry oxygen

762 torr = 19.8 torr = partial pressure of dry oxygen

partial pressure of dry oxygen = 742 torr

The partial pressure of the dry oxygen is 742 torr

8 0
3 years ago
43. A stock glucose standard has a concentration of 1,000 mg/dL. A 1/5 dilution of this standard is made. What would be the fina
Nat2105 [25]

The final concentration of the diluted standard is 0.2 mg/dL.

<h3 /><h3>What is concentration of glucose standard after 1/5 solution?</h3>

Using the dilution formula:

  • C1V1 = C2V2

where

  • C1 is initial concentration
  • V1 initial volume
  • C2 is final concentration
  • V2 is final volume.

Assuming a final volume of 100 mL, and since a 1/5 dilution is made:

C1 = 1.00 mg/dL

V1 = 20

C2 = ?

V2 = 100 mL

C2 = C1V1/V2

C2 = 20 × 1/100

C2 = 0.2 mg/dL

Therefore, the final concentration of the diluted standard is 0.2 mg/dL.

Learn more about dilution at: brainly.com/question/24881505

6 0
2 years ago
Find the pH of 8.4x10^-8 M HCl. Report the answer to the hundredths place.
Hunter-Best [27]

Answer:

pH = 7.08

Explanation:

HCl ---------> H^+ + Cl^-

It's an acid, we are using this formula

pH = -log [H]^+

H^+ = 8.4 * 10^-8

pH = - log [8.4 * 10^-8]

It can also be solved as

-log 8.4-(-8log 10)

-0.924-(-8×1)

-0.924+8

7.076

To the nearest hundredth

pH = 7.08

8 0
3 years ago
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