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stira [4]
2 years ago
15

When a person in the hospital is given fluids intravenously, the fluid is typically a saline (salt) solution with about the same

water concentration as human body tissues. Explain how the use of distilled water (water without salt) in place of this saline solution would possibly upset the patients homeostasis.
Chemistry
1 answer:
Sati [7]2 years ago
5 0

Answer:

Swelling of the red blood cells occurs.

Explanation:

  • <u>Distilled water makes the blood hypotonic ,. that is a less concentrated solution, to the body  tissues including  the red blood cells. </u>
  • Therefore the water will enter the red blood cells and may cause them to lyse or swell.
  • <em>Additionally , water will enter the tissue spaces and cause sweling. </em>
  • But in the case of a hypertonic solution , ( a salt or saline solution ) , <em>This swelling does not happen because the concentration of the saline will be equal or more than the blood. </em>
  • Thus the water will flow only from the blood to the solution. <u>This will not cause swelling.</u>
  • But too much of salt is also not recommended.
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The answer is 4. It can hold 4 sublevels as s.p,d,f and it holds 32 electrons
8 0
3 years ago
density of a sample of tin is 8 g/'ll when the actual density was 7.28 g/ml. what was the percentage error
riadik2000 [5.3K]

9.9%

|Approximate Value − Exact Value| divided by

|Exact Value| X 100

7 0
2 years ago
A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C
Mamont248 [21]

Answer:

T_{eq}=23.85^oC

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

\Delta H_{Cu}=-\Delta H_{H_2O}

Therefore the equilibrium temperature shows up as:

m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC

Best regards.

6 0
3 years ago
Read 2 more answers
Group 18 noble gases are relatively inert because *
choli [55]

Answer:I believe its A. There noble gasses so there happy with there electron count!

Explanation:

3 0
3 years ago
Sort each of the following elements by effective nuclear charge, Zeff, from smallest to largest:
krek1111 [17]

Answer:

Rb = +1 , Sr = +2, In= +3,  Sn = +4, Sb= +5

Explanation:

Formula:

Zeff = Z - S

Z = atomic number

S = number of core shell or inner shell electrons

For Sn:

Electronic configuration:

Sn₅₀ = [Kr] 4d¹⁰ 5s² 5p²

Zeff = Z - S

Zeff = 50 - 46

Zeff = +4

For Rb:

Electronic configuration:

Rb₃₇ = [Kr] 5s¹

Zeff = Z - S

Zeff = 37 - 36

Zeff = +1

For Sb:

Electronic configuration:

Sb₅₁ = [Kr] 4d¹⁰ 5s² 5p³

Zeff = Z - S

Zeff = 51 - 46

Zeff = +5

For In:

Electronic configuration:

In₄₉ = [Kr] 4d¹⁰ 5s² 5p¹

Zeff = Z - S

Zeff = 49 - 46

Zeff = +3

For Sr:

Electronic configuration:

Sr₃₈= [Kr]  5s²

Zeff = Z - S

Zeff = 38 - 36

Zeff = +2

8 0
3 years ago
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