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irinina [24]
3 years ago
13

In the exercise to determine the concentration of a solution of CuSO4, for the

Biology
1 answer:
Soloha48 [4]3 years ago
5 0

Answer:

This question is incomplete

Explanation:

Experimentally, when trying to determine the concentration of an unknown sample of CuSO₄ with known absorbance, a standard curve is used. The standard curve is a scattered plot/graph in which known concentrations and there respective/corresponding absorbance are plotted on a "X and Y axis" graph (scattered plot). The absorbance is represented on the y-axis while the concentration is represented on the x-axis.

Once this known values are plotted on the standard curve, an unknown value (either absorbance or concentration) can be extrapolated from the curve. What is missing from the question is data required for the standard curve.

This question shows the absorbance of the CuSO₄ has been determined (most likely using spectrophotometer), the concentration could have been extrapolated from a standard curve.

An example of a standard curve is provided in the attachment for guidance.

Download docx
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Energy from nutrients is lost at each trophic level

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What are the current populations of each organism?
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Answer:

The correct answer is nausea and breathing trouble.

Explanation:

In the given example, the client is receiving 1000ml saline with 40mEq of KCl (Potassium Chloride) and we are looking for symptoms of hyperkalemia. Hyperkalemia is a disease observed when there is excess amounts of potassium present in the bloodstream. It's first and most important symptoms are tiredness, nausea and vomiting, breathing trouble and chest pain so these are the findings that will be initially noted in the client if they have hyperkalemia.

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5 0
4 years ago
manganese-56 has a half life of 2.6 hours how much manganese would have disappeared after 7.8 hours if you started with 20g of m
mamaluj [8]

Answer:

17.5 g of manganese-56

Explanation:

Now we have to use the formula;

N/No= (1/2)^t/t1/2

Where

N= mass of manganese-56 left after time (t) = ????

No= initial mass of manganese-56 present = 20 g

t= time taken for N mass of manganese-56 to remain = 7.8 hours

t1/2= half-life of manganese-56 = 2.6 hours

Then we substitute values;

N/20 = (1/2)^7.8/2.6

N/20 = (1/2)^3

N= (1/2)^3 ×20

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7 0
3 years ago
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