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kaheart [24]
3 years ago
5

Depending upon the tip of a dropper pipet there are approximately 20 drops per milliliter of water. the experimental procedure p

art a and b indicates the addition of 5 -10 dropps of each solutions to the test tubes calculate the volume range in milliliters for the solution
Chemistry
1 answer:
Murrr4er [49]3 years ago
8 0

Answer : The volume range in milliliters for the solution is 0.25 - 0.5 mL

Explanation :

We have been given that there are approximately 20 drops per milliliter of water

This information can be used as a conversion factor as \frac{1mL}{20 drops}

We are using 5-10 drops of the solution.

Let us calculate milliliters in 5 drops.

5 drops \times\frac{1 mL}{20 drops} = 0.25 mL

Similarly, 10 drops would contain

10 drops \times\frac{1 mL}{20 drops} = 0.5 mL solution.

Therefore the volume range in milliliters for the solution is 0.25 - 0.5 mL

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A chemist wishing to do an experiment requiring 47-Ca2+ (half-life = 4.5 days) needs 5.0 μg of the nuclide. What mass of 47-CaCO
NARA [144]

Answer:

5.8μg

Explanation:

According to the rate or decay law:

N/N₀ = exp(-λt)------------------------------- (1)

Where N = Current quantity,  μg

            N₀ = Original quantity, μg

             λ= Decay constant day⁻¹

              t =  time in days

Since the half life is 4.5 days, we can calculate the  λ from (1) by  substituting N/N₀ = 0.5

0.5 = exp (-4.5λ)

ln 0.5  = -4.5λ

-0.6931 = -4.5λ

λ =   -0.6931 /-4.5

  =0.1540 day⁻¹

Substituting into (1)  we have :

N/N₀ = exp(-0.154t)----------------------------- (2)

To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:

N = 5.0 μg

N₀ = Unknown

t = 1 day

Substituting into (2) we have

[5/N₀]   = exp (-0.154 x 1)

    5/N₀        = 0.8572

N₀  =  5/0.8572

     =    5.8329μg

    ≈     5.8μg

The Chemist must order 5.8μg  of 47-CaCO3

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2 years ago
What does an atom become if it gains or loses electrons?
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2 years ago
How many moles of nitrogen gas are there in 16.8 L of this gas at STP?
sweet-ann [11.9K]
According to Avogadro's Law, same volume of any gas at standard temperature and pressure will occupy same volume. And one mole of any Ideal gas occupies 22.4 dm³ (1 dm³ = 1 L).

Data Given:

                  n = moles = ?
                  V = Volume = 16.8 L

Solution:
               
As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....

                           = ( 1 mole × 16.8 L) ÷ 22.4 L
                    
                           = 0.75 moles

Result:
           
16.8 L of Nitrogen gas will contain 0.75 moles at standard temperature and pressure.
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