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FromTheMoon [43]
3 years ago
10

Red blood cell count typically goes down as result of training

Chemistry
1 answer:
Y_Kistochka [10]3 years ago
4 0

I believe it is not that the red blood cell count goes down, but that the white blood cell count increases.

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ILL GIVE BRAINLISTS!!!!!
sergejj [24]

The scientists responsible for testing the hypothesis are already famous

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3 years ago
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The graph below shows how the pH of the soil in
kondor19780726 [428]

Answer:

Point B

Explanation:

From 0 - 7 on the pH scale, substances are said to be acidic. At 7, the substance is said to be neutral. From 7 - 14, substances are basic.

Lime is a basic substance. Therefore, when it is added to the soil, one should expect that the soil becomes more basic. On the graph, this should be represented by the line increasing in pH. As such, the point at which lime was added was most likely Point B.

4 0
2 years ago
When a 10 mL graduated cylinder is filled to the 10 mL mark, the mass of the water was measured to be 9.925 g. If the density of
Vlada [557]
Measured volume = 10 ml 
mass = 9.925 g 
density = 0.9975 g/ml
density = \frac{mass}{volume} 
so actual volume = \frac{mass}{density} = \frac{9.925}{0.9975} = 9.95 mL
Percentage Error = \frac{measured volume - Actual volume}{actual volume} x 100 
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5 0
4 years ago
Write the net ionic equation for the acid-base reaction<br> PH3(aq) + HCl(aq) - PH4Cl(aq)
elixir [45]

Explanation:

i dont understand dont you already have the equation?

8 0
3 years ago
Is anyone good at chemistry if so Can someone help me please NO LINKS
Nookie1986 [14]

Answer:

6.25 g

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 68.8 years

Time (t) = 344 years

Original amount (N₀) = 200 g

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 68.8 years

Time (t) = 344 years

Number of half-lives (n) =

n = t / t½

n = 344 / 68.8

n = 5

Thus, 5 half-lives has elapsed.

Finally, we shall determine the amount of the Uranium-232 that remains. This can be obtained as follow:

Original amount (N₀) = 200 g

Number of half-lives (n) = 5

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2⁵ × 200

N = 1/32 × 200

N = 200 / 32

N = 6.25 g

Thus, the amount of Uranium-232 that remains is 6.25 g

4 0
3 years ago
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