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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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A candle slowly burns until the last bit of wax and wick are gone. The two reactants in this chemical reaction are the wick and
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The component of the candle burning in the surrounding has been the oxygen in the air.

The burning of candle wax and wick has been the chemical reaction. It has been based on the reaction of wick with the atmospheric oxygen, resulting in the formulation of the wax burning.

<h3>Chemical reaction of burning of wax</h3>

The wax has been vaporizes by the heat of the flame, that has been resulted by the burning. The wick has been able to react with the oxygen and form the byproducts that helps in flame burning.

The end products have been wick and oxygen as the wax has been consumed in the reaction. The air in the surrounding has oxygen as the part of the system, as it has been involved in the reaction.

Learn more about candle burning, here:

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6 0
3 years ago
Question 5(Multiple Choice Worth 5 points)
mars1129 [50]

Answer:

Density

Explanation:

8 0
3 years ago
Read 2 more answers
Consider the following data showing the initial rate of a reaction (A→products) at several different concentrations of A.
OlgaM077 [116]

Answer:

(1) order = 2

(2) R = K [A]²

Explanation:

Given the reaction:

A--------->Product

The rate constant relation for the reaction is given as:

R(i) = K [A]............(*)

Where R(I) is rate constant at different concentration of A.

Taking the rate constant as R1, R2 and R3 for the different concentrations respectively. Then the following equations results

0.011 = K [0.15] ⁿ.........(1)

0.044 = K [0.30]ⁿ .......(2)

0.177 = K [0.60]ⁿ .........(3)

Dividing (2) by (1) and (3) by (1)

Gives:

0.044/0.011 = [0.3/0.15]ⁿ

4 = 2ⁿ; 2² = 2ⁿ; n = 2

Similarly

0.177/0.011 = [0.60/0.15]ⁿ

16.09 = 4ⁿ

16.09 = 16 (approximately)

4² = 4ⁿ ; n = 2

Hence the order of the reaction is 2.

The rate law is R = K [A]²

4 0
3 years ago
Positron Emission Tomography (PET) is an advanced medical scanning technique. Fluorine-18 is a commonly used isotope for PET sca
TiliK225 [7]

Answer:

Option D. equation D

Explanation:

Positron is simply a beta plus decay or superscript 0 e subscription 1 (0 1 e).

Now, let us consider the options given:

Option A is emitting alpha decay

Option B is emitting beta minus decay.

Option C is emitting beta minus decay.

Option D is emitting beta plus decay also known as Positron.

3 0
3 years ago
A student titrates a 20.00 mL sample of an aqueous borax solution with 1.03 M H2SO4. If 2.07 mL of acid are needed to reach the
RSB [31]

Answer:

The concentration of the borax solution is 0.1066 M

Explanation:

Step 1: Dtaa given

Volume of a sample of aqueous borax solution = 20.00 mL = 0.020 L

Molarity of H2SO4  = 1.03 M

Volume of the H2SO4 = 2.07 mL = 0.00207 L

Step 2: The balanced equation

Na2B4O7*10H2O(borax) + H2SO4 ⇆ Na2SO4 + 4 H3BO3 + 5 H2O

Step 3: Calculate molarity of borax solution

b*Ca*Va = a * Cb*Vb

⇒with B = the coefficient of H2SO4 = 1

⇒with Ca = the concentration of borax = TO BE DETERMINED

⇒with Va = the volume of borax = 0.020 L

⇒with a = the coefficient of borax = 1

⇒with Cb = the concentration of H2SO4 = 1.03 M

⇒with Vb = the volume of H2SO4 = 0.00207 L

Ca*0.020 L = 1.03 M * 0.00207 L

Ca = (1.03 * 0.00207) / 0.020

Ca = 0.1066 M

The concentration of the borax solution is 0.1066 M

6 0
4 years ago
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