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Katyanochek1 [597]
3 years ago
10

how long would it take to administer exactly 500 mL of fluid through an IV with a drop factor of 30 gtt/mL if the drip rate is 6

0 gtt/min?
Chemistry
1 answer:
Lyrx [107]3 years ago
7 0

Answer : The time taken will be 25 min.

Explanation :

First we have to determine the amount of fluid.

As, 1 mL contains 30 gtt

So, 500 mL contains \frac{500mL}{1mL}\times 30gtt=1500gtt

Now we have to determine the time taken.

As, 60 gtt takes time = 1 min

So, 1500 gtt takes time = \frac{1500gtt}{60gtt}\times 1min=25min

Therefore, the time taken will be 25 min.

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Why doesn’t the KHP concentration have to be exactly 0.100M? Explain using complete sentences.
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Answer:

The answer is in the explanation.

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The KHP is an acid used as standard in titrations to find concentration of bases as NaOH.

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That means, at equivalence point of a titration in which titrant is NaOH, the moles of KHP = Moles of NaOH added

With the moles of KHP = Moles of NaOH and the volume used by titrant we can find the molar concentration of NaOH.

The moles of KHP are obtained from the volume and the concentration as follows:

Volume(L)*Concentration (Molarity,M) = moles of KHP

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Copper oxide, CuO, reacts with hydrochloric acid, HCI, to produce copper chloride, CuCL2 and water
spayn [35]

Explanation:

El óxido de cobre (II), también llamado antiguamente óxido cúprico ({\displaystyle {\ce {CuO}}}{\displaystyle {\ce {CuO}}}), es el óxido de cobre con mayor número de oxidación. Como mineral se conoce como tenorita.

{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

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{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

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