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Alex787 [66]
1 year ago
7

One of the waste products produced by kidneys is creatinine. A healthy range ofcreatinine in human blood is between 0.50 mg/dL a

nd 1.1 mg/dL. Elevated levels of thiscompound could indicate a problem with kidney function.The concentration of creatinine in patient’s blood was reported in g/L: 0.0082 g/L.Is this value within the normal range? If it is not, is this concentration low or high?Confirm your answer with your calculations and show all the work.The concentration of creatinine = _____________mg/dL. It is ______________the normal range
Chemistry
1 answer:
Anna35 [415]1 year ago
8 0

The normal range of creatinine in human blood is between 0.50 mg/dL and 1.1 mg/dL. The patient's blood has a concentration of 0.0082 g/L. Let's convert that value into mg/dL.

We kwnot that there are 1000 mg in 1 g. And there are 10 dL in 1 L. We have to use those conversions.

1000 mg = 1 g 10 dL = 1 L

0.0082 g/L = 0.0082 g/L * 1000 mg/g = 8.2 mg/L * 1 L/ (10 dL) = 0.82 mg/dL

0.0082 g/L = 0.82 mg/dL

0.50 mg/dL < 0.82 mg/dL < 1.1 mg/dL

Answer: The concentration of creatinine = 0.82 mg/dL. It is in the normal range.

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Single and double replacement reactions practice WS CuCl2 +<br> F2 »<br> CuF2<br> +<br> Cl2
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3 years ago
Read 2 more answers
There is a 50 g sample of ra-229. It has a half-life of 4 minutes.how much will be left after 12 minutes? a. 3.13 g b. 6.25 g c.
polet [3.4K]

It's B. 6.25

first divide 50/2

then divide 25/2

after that divide 12.5/2

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Hope it helps!

3 0
2 years ago
If the concentration of the HCl used in your titration was 1.00 M and you used 23.68 mL to reach the endpoint, calculate the con
drek231 [11]

Answer: The concentration of the OH-, CB = 0.473 M.

Explanation:

The balanced equation of reaction is:

2HCl + Ca(OH)2 ===> CaCl2 + 2H2O

Using titration equation of formula

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)

NB is the number of mole of base = 1 (from the balanced equation of reaction)

CA is the concentration of acid = 1M

CB is the concentration of base = to be calculated

VA is the volume of acid = 23.65 ml

VB is the volume of base = 25mL

Substituting

1×23.65/CB×25 = 2/1

Therefore CB =1×23.65×1/25×2

CB = 0.473 M.

6 0
3 years ago
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