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baherus [9]
3 years ago
10

In the life cycle, growth happens after:birth death decay reproduction

Chemistry
1 answer:
Vitek1552 [10]3 years ago
3 0

Answer:

Birth

Explanation:

Once your out of the wound  you develop into a kid to a toddler, then a teen and so on

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How can u describes a true solution?
zubka84 [21]
<span>True Solution is a homogeneous mixture of two or more substances in which substance dissolved (solute) in solvent has the particle size of less than 10-9 m or 1 nm. Simple solution of sugar in water is an example of true solution.</span>
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What must happen to make for an ion to from
Mazyrski [523]

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when atoms lose or gain electrons ions form

Explanation:

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Calculate the fractional saturation for hemoglobin when the partial pressure of oxygen is 40 mm Hg. Assume hemoglobin is 50%% sa
kumpel [21]

Answer:

The fractional saturation for hemoglobin is 0.86

Explanation:

The fractional saturation for hemoglobin can be calculated using the formula

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Where Y_{O_{2} } \\ is the fractional oxygen saturation

{P_{O_{2} } is the partial pressure of oxygen

P_{50} is the partial pressure when 50% hemoglobin is saturated with oxygen

and h is the Hill coefficient

From the question,

{P_{O_{2} } = 40 mm Hg

P_{50} = 22 mm Hg

h = 3

Putting these values into the equation, we get

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Y_{O_{2} } = \frac{40^{3} }{22^{3} + 40^{3}  }

Y_{O_{2} } = \frac{64000 }{10648 + 64000  }

Y_{O_{2} } = \frac{64000 }{74648 }

Y_{O_{2} } = 0.86

Hence, the fractional saturation for hemoglobin is 0.86.

4 0
3 years ago
You need to prepare an acetate buffer of pH 5.36 from a 0.900 M acetic acid solution and a 2.62 M KOH solution. If you have 680
Alexus [3.1K]

Answer:

Explanation:

pH = pKa + log [ CH₃COO⁻ ] / [CH₃COOH ]

5.36 = 4.86 + log [ CH₃COO⁻ ] / [CH₃COOH ]

log [ CH₃COO⁻ ] / [CH₃COOH ]  = .5

[ CH₃COO⁻ ] / [CH₃COOH ]  = 3.16

moles of CH₃COOH = .680 x .9 = .612 M

Let  x mole of KOH is required

x /( .612 - x ) = 3.16

x = 1.933 - 3.16 x

x = .46488

.46488 moles of KOH will be required

volume required be V

v x 2.62 =  .46488

v = .1774 L

= 177.4 mL

177.4 mL  of  2.62 M KOH will be required .

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What are biotic factors in a ecosystem
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animals, grass, and decomposers

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