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Mnenie [13.5K]
3 years ago
7

A patient needs to be given exactly 500 ml of a 5.0% (w/v) intravenous glucose solution. the stock solution is 35% (w/v). how ma

ny milliliters of the stock solution are needed to prepare the fluid for the iv bag?
Chemistry
2 answers:
Sonbull [250]3 years ago
8 0

Answer is: 71.43 milliliters of the stock solution are needed.

V(solution of glucose) = 500 mL; voume of the solution.

ω(solution) = 5.0% ÷ 100%.

ω(solution) = 0.05.

ω(stock solution) = 35% ÷ 100%.

ω(stock solution) = 0.35.

Because weight of the glucose is the same in both solutions:

0.05 · 500 mL = 0.35 · V(stock solution).

V(stock solution) = 500 mL · 0.05 ÷ 0.35.

V(stock solution) = 71.43 mL.

N76 [4]3 years ago
6 0
In this question, the <span>patient needs to be given exactly 500 ml of a 5.0%. The content of the glucose should be:
</span>weight= volume * density* concentration<span>
500ml * 1mg/ml *5%= 25mg.

The </span><span>stock solution is 35%, then the amount needed in ml would be: 
weight= volume * density* concentration
25mg= volume * 1mg/ml *35%
volume= 25/35%= 500/7= 71.43ml</span>
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Did Chelate compounds increase the rate of reaction?
Viefleur [7K]

Answer:

Chelate, any of a class of coordination or complex compounds consisting of a central metal atom attached to a large molecule, called a ligand, in a cyclic or ring structure. An example of a chelate ring occurs in the ethylenediamine-cadmium complex:

The ethylenediamine ligand has two points of attachment to the cadmium ion, thus forming a ring; it is known as a didentate ligand. (Three ethylenediamine ligands can attach to the Cd2+ ion, each one forming a ring as depicted above.) Ligands that can attach to the same metal ion at two or more points are known as polydentate ligands. All polydentate ligands are chelating agents.

Chelates are more stable than nonchelated compounds of comparable composition, and the more extensive the chelation—that is, the larger the number of ring closures to a metal atom—the more stable the compound. This phenomenon is called the chelate effect; it is generally attributed to an increase in the thermodynamic quantity called entropy that accompanies chelation. The stability of a chelate is also related to the number of atoms in the chelate ring. In general, chelates containing five- or six-membered rings are more stable than chelates with four-, seven-, or eight-membered rings.

Explanation:

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2 years ago
If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced
ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • P₄: 1 mole
  • H₂: 6 moles
  • PH₃:4 moles

Being the molar mass of the compounds:

  • P₄: 124 g/mole
  • H₂: 2 g/mole
  • PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

  • P₄: 1 mole* 124 g/mole= 124 g
  • H₂: 6 mole* 2 g/mole= 12 g
  • PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

  • P= 2 atm
  • V= 10 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 298 K

Replacing:

2 atm*10 L= n*0.082 \frac{atm*L}{mol*K} *298 K

and solving you get:

n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

  • If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3}  }{124grams of P_{4}}

moles of PH₃=0.2742

  • If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2}  }{124grams of P_{4}}

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, <u><em>the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.</em></u>

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Answer:

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