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1 answer:
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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>
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I hope it helps!
The mass percent of sulfurous acid in the new solution : 38.9%
<h3>Further explanation</h3><em>In a container you have 800 g of a 35% by mass solution of sulfurous acid, from which 80 ml of water evaporates. What is the mass percent of sulfurous acid in the new solution? data: density of water is 1g / ml.</em>
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solution 1
composition :
- 35% acid :
- water :
solution 2(new solution)
composition :
- water
- Total mass of new solution after water evaporated
- %mass of acid in a new solution
Answer:
Explanation:
First, we find in the tables the ΔH of formation of each compound. As you can see in the (image 1)
Then we solve the ecuation for ΔH°reaction
ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(Reactants)
ΔH°reaction= (-2* 393.5 - 2*285.8) - (52.4 + 0) kJ/mol
ΔH°reaction = -1.41 *10^3 kJ/mol
