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4 years ago

One isotope of Br has a half-life of 16.5 hours. How much of a 2.00 gram sample remains at the end of 1.00 day?

Chemistry

1 answer:

valina [46]4 years ago

4 0

Answer:

half life=16.5hrs

sample=2g

total time=1day=24hrs

no. of half lives=total time/half life

no. of half lives=24hrs/16.5hrs=1.45approx1.5

sample left=1/2n=1/2[1.5]=0.707gapprox

Explanation:

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motikmotik

Answer:

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Explanation:

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4 years ago

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Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to

vredina [299]

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

$H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)$

Next we will calculate how many moles of $H_2SO_4$ are present in 85.00 mL of 1.500 M sulfuric acid.

As, Molarity = $\frac{\text{moles of solute}}{\text{liters of solution
}}$

1.500 M = $\frac{n}{0.08500 L
}$

n = 0.1275 mol $H_2SO_4$

Now set up and solve a stoichiometric conversion from moles of $H_2SO_4$ to grams of $NaHCO_3$ . As, the molar mass of $NaHCO_3$ is 84.01 g/mol.

$0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})$

= 21.42 g $NaHCO_3$

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

4 0

3 years ago

What is the mass of each substance in the equation, after it is balanced

attashe74 [19]

Mass = mole x molar mass

Explanation:

The mole of the substance used in the bakanced eqn multipled by its molar mass equals to Its mass

6 0

3 years ago

The density for potassium is 0.856 g/cm3. What would be the mass of a 30 cm3 piece of potassium? A. 51.36g B. 25.68g C. 25.5g D.

CaHeK987 [17]

Density = mass / Volume 0.856 g/cm3 = x / 30 cm3(0.856)(30) = xx = 25.68 grams

4 0

4 years ago

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Express the concentration of a 0.0320 M aqueous solution of fluoride, F−, in mass percentage and in parts per million (ppm). Ass

denis23 [38]

Answer:

607 ppm

Explanation:

In this case we can start with the ppm formula:

$ppm=\frac{mg~of~solute}{Litters~of~solution}$

If we have a solution of 0.0320 M, we can say that in 1 L we have 0.032 mol of $F^-$ , because the molarity formula is:

$M=\frac{mol}{L}$

In other words:

$0.0320~M=\frac{mol}{1~L}$

$mol=0.032~M*1~L=0.032~mol$

$1~L~of~Solution=0.0320~mol~of~solute$

If we use the atomic mass of $F$ (19 g/mol) we can convert from mol to g:

$0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g$

Now we can convert from g to mg (1 g= 1000 mg), so:

$0.607~g\frac{1000~mg}{1~g}=607~mg$

Finally we can divide by 1 L to find the ppm:

$ppm=\frac{607~mg}{1~L}=~607~ppm$

We will have a concentration of 607 ppm.

I hope it helps!

4 0

4 years ago