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3 years ago

Show the calculation for determining the % composition of glycerin which is the major

Chemistry

1 answer:

Shkiper50 [21]3 years ago

6 0

Answer:

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A mineral forms from water at the edge of a lake.

irga5000 [103]

Answer:C

Explanation:The mineral is only formed when hot water cooler

8 0

3 years ago

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A certain drug has a half-life in the body of 3.5h. Suppose a patient takes one 200.Mg pill at :500PM and another identical pill

tekilochka [14]

Answer:

The amount of drug left in his body at 7:00 pm is 315.7 mg.

Explanation:

First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:

$N_{t} = N_{0}e^{-\lambda t}$

Where:

λ: is the decay constant = $ln(2)/t_{1/2}$

$t_{1/2}$ : is the half-life of the drug = 3.5 h

N(t): is the quantity of the drug at time t

N₀: is the initial quantity

After 90 min and before he takes the other 200 mg pill, we have:

$N_{t} = 200e^{-\frac{ln(2)}{3.5 h}*90 min*\frac{1 h}{60 min}} = 148.6 mg$

Now, at 7:00 pm we have:

$t = 7:00 pm - (5:00 pm + 90 min) = 30 min$

$N_{t} = (200 + 148.6)e^{-\frac{ln(2)}{3.5 h}*30 min*\frac{1 h}{60 min}} = 315.7 mg$

Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).

I hope it helps you!

3 0

2 years ago

A mixture of three noble gases has a total pressure of 1.25 atm. The individual pressures exerted by neon and argon are 0.68

FinnZ [79.3K]

First of all, as you seen the gases are noble which means that will not react with each other and in this case each gas create individual pressure.

P $_{T}$ = total pressure

P $_{Ne}$ = pressure of neon

P $_{Ar}$ = pressure of argon

P $_{He}$ = pressure of helium {which is required}

P $_{T}$ = P $_{Ne}$ + P $_{Ar}$ + P $_{He}$

1.25 = 0.68 + 0.35 + P $_{He}$

P $_{He}$ = 1.25 - [0.68 + 0.35] = 0.22 atm

4 0

3 years ago

Sugar cannot be separated from sugar solution by filtering explain why?

alexdok [17]

Answer:

The sugar particles are so small that they have dissolved in the water, and can easily pass through the filter.

5 0

2 years ago

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What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)

blsea [12.9K]

<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine, $(C_{2}H_{5})_{3}N$ , the reaction will be
$(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}$

and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value = 5.3 x 10^-4

And $kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}$

thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

=0.00516 M

Thus, pOH = 2.30
pH = 14 - pOH = 11.7

6 0

3 years ago