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Umnica [9.8K]
3 years ago
5

How do i salve 50,200 mL to L

Chemistry
1 answer:
sveta [45]3 years ago
3 0
You will have divided 50200 by 1000000 which 0.0502
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C2F4 effuses through a barrier at a rate of 4.6x10-6 mol/hour, while an unknown gas effuses at a rate of 5.8x10-6 mol/hour. What
umka21 [38]
The  molar mass  of  the unknown  compound  is   calculated   as   follows

let the unknown  gas be represented by   letter  Y

Rate of C2F4/  rate of  Y  = sqrt of   molar  mass of gas Y/ molar mass of  C2F4

 =  (4.6  x10^-6/ 5.8  x10^-6)  = sqrt  of  Y/ 100

remove  the  square  root  sign  by  squaring  in both  side

(4.6  x  10^-6 / 5.8  x10^-6)^2 =  Y/100

= 0.629 =Y/100

multiply  both side  by  100

Y=  62.9 is  the molar  mass of unknown  gas



5 0
3 years ago
A physician has ordered 0.50 mg of atropine, intramuscularly. If atropine were available as 0.25 mg/mL of solution, how many mil
sladkih [1.3K]
If 0.25mg of atropine is in 1mL
                  so
    0.50mg of atropine is in x

x=\frac{0.5mg*1mL}{0.25mg}=2mL
6 0
3 years ago
A standard drink of beer is _ ounces.
kiruha [24]

A standard drink of beer is 12 ounces


4 0
3 years ago
Read 2 more answers
I WILL GIVE 15 POINTS PLS I NEED THIS BEFORE THE CLASS FINISHES IN 30 MINN
bulgar [2K]

Answer:

likely be the same

Explanation:

this is because we have one color that both atoms share (green). both sample 1 and sample 2 have green and another color. yet, since they share one color, they are likely similar

7 0
3 years ago
A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m
kogti [31]

Answer:

The molar mass of the unknown gas is \mathbf{ 51.865 \  g/mol}

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

R \  \alpha  \ \dfrac{1}{\sqrt{d}}

R = \dfrac{k}{d}  where K = constant

If we compare the rate o diffusion of two gases;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}

Since the density of a gas d is proportional to its relative molecular mass M. Then;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}

Rate is the reciprocal of time ; i.e

R = \dfrac{1}{t}

Thus; replacing the value of R into the above previous equation;we have:

\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}

We can equally say:

{\dfrac{t_2}{t_1}}=  {\sqrt{\dfrac{M_2}{M_1}}

{\dfrac{6.34}{4.98}}=  {\sqrt{\dfrac{M_2}{32}}

M_2 = 32 \times ( \dfrac{6.34}{4.98})^2

M_2 = 32 \times ( 1.273092369)^2

M_2 = 32 \times 1.62076418

\mathbf{M_2 = 51.865 \  g/mol}

7 0
3 years ago
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