Answer:
To supply the required ions it is necessary to inject 5,6mL of 6g/30mL solution and 131,1 mL of 0,9% solution.
Explanation:
1mEq of sodium are 59mg of NaCl and 1mEq of potassium are 75mg KCl
in intravenous infusion 15 mEq of K are:
15x75mg KCl = 1,125g of KCl
And 20 mEq of Na are:
20x59mg NaCl = 1,18g of NaCl
To supply the potassium ion it is necessary to inject:
1,125g of KCl×
=<em> 5,6mL of 6g/30mL solution</em>
And, to supply the sodium ion it is necessary to inject:
1,18g of NaCl×
= <em>131,1 mL of 0,9% solution</em>
<em />
I hope it helps!
Answer:
Weather and erosion.
Explanation:
In Geography, weathering is the process of breaking down of rocks, minerals and soil as a result of contact with water, earth's atmosphere, acid, ice, plants and animals. The breaking up of the rocks by weathering weakens it and makes it susceptible to erosion.
Erosion in geography is the movement of rocks and sediments to another place by water, wind and ice.
For instance, In many parts of the Cross Timbers and Prairies ecoregion, the Brazos river has formed tall, steep cliffs in the rock along its banks through the processes of weathering and erosion.
The molecular formula for sodium chloride is NaCl. The sum of their atomic weights is (22.99 grams/mole + 35.45 grams/mole) = 58.44 grams/mole
take (17.0 grams)/(58.44 grams/mole), which equals 0.291 moles of NaCl.
Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
