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weqwewe [10]
3 years ago
9

A client presents with fatigue, nausea, vomiting, muscle weakness, and leg cramps. The laboratory values are as follows: sodium

147 mEq/L (147 mmol/L) potassium 3.0 mEq/L (3.0 mmol/L) chloride 112 mEq/L (112 mmol/L) Magnesium 2.3 mg/dL (0.95 mmol/L) What laboratory value is consistent with the client?
Chemistry
1 answer:
Ganezh [65]3 years ago
6 0

Answer: The laboratory value of potassium (3.0 mmol / L) is consistent with the client's symptoms of hypokalemia.

Explanation:

Hello!

Let's solve this!

Hypokalemia is a disorder in the body's electrolyte balance, when the decrease in blood potassium (K) ion levels is below 3.5 mmol / L. Potassium losses can occur through the digestive tract: such as vomiting and

diarrhea The most frequent symptoms of potassium loss include: tiredness, muscle weakness and cramping.

In conclusion, the laboratory value of potassium (3.0 mmol / L) is consistent with the client's symptoms of hypokalemia.

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its because fibrinogen is a chemical substance that builds up at the wound and gets hard by the action of air to prevent bleeding

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which of the following elements are in this unbalanced chemical reaction: li(s) + H2O(l) lioh(aq)+h2(g)​
Arada [10]

Answer:

Li and H

Explanation:

2Li(s)+2H2O(i)→2LiOH(aq)+H2(g) is full balanced

5 0
2 years ago
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Consider a solution that contains 0.274 M potassium chloride and 0.155 M magnesium chloride.
solong [7]

Answer:

Concentration of chloride ions = 0.584M

Explanation:

The step by step calculations is shown as attached below.

3 0
3 years ago
Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C
ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O  (1)  

The enthalpy of reaction (1) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}   (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

  • 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
  • 7O₂: 7 moles of 1 O=O bond  
  • 4CO₂: 4 moles of 2 C=O bonds  
  • 6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

\Delta H = \Delta H_{r} - \Delta H_{p}

\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})      

\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})  

\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol  

\Delta H = -2860 kJ/mol          

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

Read more here:

brainly.com/question/11753370?referrer=searchResults  

I hope it helps you!        

7 0
2 years ago
Janice bought 40 shares of stock at $31.82 per share. She received dividends of $1.11 per share for 1 year. (Do not use commas,
Alborosie

Answer:

  • $30.71 per share

Explanation:

The <em>purchase price</em> is what Janice invested for every share.

Since the stock was priced at $31.82 per share and she received a $1.11 dividend per share, her investment was:

  • $31.82 - $1.11 = $30.71 per share ← answer

This price is the cost for Janice, over which she shall calculate their returns (gains or losses) on the future, when she sells the shares, for instance.

The total investment of Janice was the number of shares multipled by the purchase price:

  • 40 shares × ($31.82 -  $1.11)/ share
  • 40 shares × ($30.71) / share = $1,228.40 (total investment)
5 0
3 years ago
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