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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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Describe how you might determine the m/z and relative abundance of the ions contributing to the peak at 21.876 min
aliya0001 [1]

The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.

<h3>What is relative abundance?</h3>
  • The proportion of atoms with a particular atomic mass present in an element sample taken from a naturally occurring sample is known as the relative abundance of an isotope.
  • When the relative abundances of an element's isotopes are multiplied by their atomic masses and the results are added up, the result is the element's average atomic mass, which is a weighted average.
  • Chemists often divide the number of atoms in a particular isotope by the sum of the atoms in all the isotopes of that element, then multiply the result by 100 to determine the percent abundance of each isotope in a sample of that element.

To learn more about relative abundance with the given link

brainly.com/question/1594226

#SPJ4

8 0
2 years ago
Tarnish on copper is the compound CuO. A tarnished copper plate is placed in an aluminum pan of boiling water. When enough salt
oksian1 [2.3K]

Answer:

The standard cell potential is 2.00 V

Explanation:

<u>Step 1:</u> Data given

Cu is cathode because of higher EP

Al3++3e−→Al       E∘=−1.66 V     anode

Cu2++2e−→Cu    E∘=0.340 V    cathode

<u>Step 2:</u> Balance both equations

2*(Al → Al3+-3e−)       E∘=1.66 V    

3*(Cu2++2e−→Cu)    E∘=0.340 V

<u>Step 3:</u> The netto equation

2 Al + 3Cu2+ +6e- → 2Al3+ + 3Cu -6e-

2 Al + 3Cu2+  → 2Al3+ + 3Cu

<u>Step 4:</u> Calculate the standard cell potential

E∘cell = E∘cathode - E∘anode

E∘cell = E∘ Cu2+/Cu - E∘ Al3+/Al

E∘cell =0.340 V - (-1.66) = 2.00 V

The standard cell potential is 2.00 V

4 0
3 years ago
Use Hess's Law to calculate the enthalpy change for the reaction
Marysya12 [62]

Answer:

ΔH = 125.94kJ

Explanation:

It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:

1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ

2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ

-1/2 (1):

WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ

3/2 (2):

3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ

The sum of  last both reactions:

WO3(s) + 3H2(g) → W(s) + 3H2O(g)

ΔH = 842.7kJ -716.76kJ

<h3>ΔH = 125.94kJ </h3>
3 0
3 years ago
A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol
Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the solution

c_1 = specific heat of calorimeter = 12.1J/^oC

c_2 = specific heat of water = 4.18J/g^oC

m_2 = mass of water or solution = Density\times Volume=1/mL\times 100.0mL=100.0g

\Delta T = change in temperature = T_2-T_1=(26.3-20.2)^oC=6.1^oC

Now put all the given values in the above formula, we get:

q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]

q=2623.61J

Now we have to calculate the enthalpy change for the process.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = Concentration\times Volume=1M\times 0.050L=0.050mole

\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

8 0
3 years ago
A molecule of hydrogen gas is made up of pairs of hydrogen atoms in which each atom shares its electron to form a(n) ____.
andreyandreev [35.5K]

Answer:

covalent bond

Explanation:

a covalent bond forms when electrons are shared between two nonmetals

plz mark as the brainliest

6 0
3 years ago
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