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Step2247 [10]
3 years ago
12

if you administering an oral acetaminophen dose of 320.mg from a bottle of 32.0 mg/ ml, how many mililitters of the solution do

you need
Chemistry
1 answer:
Yuri [45]3 years ago
7 0
10 milliliters are needed
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The gaseous product of a reaction is collected in a 25.0L container at 27.0 C. The pressure in the container is 3.0atm and the g
NeX [460]

Answer: The molar mass of the gas is 31.6 g/mol

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 3.0 atm

V = Volume of gas = 25.0 L

n = number of moles  = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =27.0^0C=(27.0+273)K=300K

n=\frac{PV}{RT}

n=\frac{3.0atm\times 25.0L}{0.0821 L atm/K mol\times 300K}=3.04moles

Moles =\frac{\text {given mass}}{\text {Molar mass}}

3.04=\frac{96.0g}{\text {Molar mass}}

{\text {Molar mass}}=31.6g/mol

The molar mass of the gas is 31.6 g/mol

4 0
3 years ago
If the length, width, and height of a box are 8.00 cm, 6.75 cm and 3.50 cm, respectively, what is the volume of the box in units
svp [43]
I think it is 1620 (lxwxh) x 10 to get to millimeters
5 0
4 years ago
Read 2 more answers
The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat
Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

 Half life of lipase t_{1/2} = 8 min x 60 s/min

                                       = 480 s

Rate constant for first order reaction is as follows.

         k_{d} = \frac{0.6932}{480}

                        = 1.44 \times 10^{-3}s^{-1}


Initial fat concentration S_{o} = 45 mol/m^{3}

                                                = 45 mmol/L

Rate of hydrolysis V_m_{o} = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = S_{o} (1 - X)

                                      = 45 (1 - 0.80)

                                      = 9 mol/m^{3}

or,                                  = 9 mmol/L

It is given that K_{m} = 5mmol/L

Therefore, time taken will be calculated as follows.

                    t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]

Now, putting the given values into the above formula as follows.

            t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]  

             = -\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]

              = 1642.83 s \times \frac{1 min}{60 sec}

              = 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

6 0
3 years ago
26.6 mL of 2.50 M stock solution of sucrose is diluted to 50.0 mL. A 16.0 mL sample of the resulting solution is then diluted to
frozen [14]

Answer:

In the final solution, the concentration of sucrose is 0.126 M

Explanation:

Hi there!

The number of moles of solute in the volume taken from the more concentrated solution will be equal to the number of moles of solute in the diluted solution. Then, the concentration of the first solution can be calculated using the following equation:

Ci · Vi = Cf · Vf

Where:

Ci = concentration of the original solution

Vi = volume of the solution taken to prepare the more diluted solution.

Cf = concentration of the more diluted solution.

Vf = volume of the more diluted solution.

For the first dillution:

26.6 ml · 2.50 M = 50.0 ml · Cf

Cf = 26.6 ml · 2.50 M / 50.0 ml

Cf = 1.33 M

For the second dilution:

16.0 ml · 1.33 M = 45.0 ml · Cf

Cf = 16.0 ml · 1.33 M / 45.0 ml

Cf = 0.473 M

For the third dilution:

20.0 ml · 0.473 M = 75.0 ml · Cf

Cf = 20.0 ml · 0.473 M / 75.0 ml

Cf = 0.126 M

In the final solution, the concentration of sucrose is 0.126 M

7 0
3 years ago
Hydrogen (H), oxygen (O), and Nitrogen (N) are all examples of which of the following?
Llana [10]

Answer:

All are correct

Explanation:

This might be a little deceptive.  The question shows H, O and N which would designate ELEMENTS.

However, all of these can also be considered compounds and molecules as well.  BUT, they are H₂, O₂ and N₂ when they are molecules and compounds.

So it depends on the whether the question is being literal or not.

8 0
3 years ago
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