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Rasek [7]
3 years ago
8

An order is given to administer methylprednisolone, an anti-inflammatory drug, by IV at a rate of 39 mg every 30 min. The IV bag

contains 125 mg of methylprednisolone in every 2 mL. What should the flow rate be in milliliters per minute (mL/min)? flow rate=_______ mL/min
Chemistry
1 answer:
Korvikt [17]3 years ago
3 0

Answer:

The Flow rate = 0.0208 mL/min

Explanation:

Data provided:

Rate of dose = 39 mg every 30 min = (39/30) mg/min  = 1.3 mg/min

also,

125mg of methylprednisolone is present in every 2 mL

thus,

concentration = (125/2) mg/mL = 62.5 mg/mL

Now,

The flow rate is given as:

Flow rate = Rate / Concentration

on substituting the respective values, we get

Flow rate = (1.3 mg/min) / (62.5mg/mL)

or

The Flow rate = 0.0208 mL/min

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Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer
bogdanovich [222]

Answer:

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

VA− = 9.125 mL

VHA = 15.875 mL

The Henderson-Hasselbalch equation is:

pH = pKa + log [A−][HA]

We have a pH 4.5

solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

[A−][HA]=10

pH − pKa = 104.5 − 4.74 = 0.5754

Now, if the total concentration is

0.10 M , then:

[HA] + [A−] 0.5754

[HA] = 0.10 M

⇒[HA] = 0.10 M 1.0000 +0.5754

= 0.0635 M

−−−−−−−−

⇒[A−] = 0.0365 M

−−−−−−−−

and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

−−−−−−−−−−−−

nHA = 0.0635 molL × 0.050L =

0.003175 mols

−−−−−−−−−−−−

So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

6 0
3 years ago
Choose a likely identity for X, Y, and Z in these structures. A central X atom is bonded to two chlorine atoms through single bo
11111nata11111 [884]

Answer:

X= Be

Y= B

Z=O

Explanation:

From the description of the compound XCl2, among the options listed only beryllium can form such compound with three lone pairs in the two chlorine atoms and no lone pair on the central atom X.

From the description of YCl3, only Boron among the options listed can form such a compound with no lone pair on the central atom and three lone pairs on each of the chlorine atoms.

From the description of ZCl2, only oxygen forms the compound OCl2 among the elements listed where oxygen possesses two lone pairs and each chlorine atom possesses three lone pairs each.

8 0
3 years ago
What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.
AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

pH=pK_{a} + log(\frac{[Base]}{[Acid]})

According to the given conditions, the equation will become as follow

pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

pH=6.86 + log(\frac{0.058}{0.042})

pH=7.00

5 0
3 years ago
What volume of nitrogen gas is equal too 3.42x10^22 molecules of this substance
Bas_tet [7]

if i am correct the volume of nitrogen gas has to equal to molecules density, making the substance 1.27 liters :)

8 0
2 years ago
Which of the following is an empirical formula?<br> OP205<br> O C2H602<br> ON3F6<br> OC8H18
jarptica [38.1K]

Answer:

None are empirical formulas

Explanation:

All are actual compounds. An example of an empirical formula could be CH2O, the empirical formula for carbohydrates like glucose (C6H12O6).

7 0
3 years ago
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