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Kisachek [45]
3 years ago
15

Amning At Home -Chemisu

Chemistry
1 answer:
levacccp [35]3 years ago
4 0

Answer:first one: 3.0g second one: 10g H2O(I)

Explanation:

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The half life of plutonium-239 (239pu, pu-239) is 24,100 years. How much plutonium will remain after 1000 years if the initial a
Anestetic [448]

The mass of plutonium that will remain after 1000 years if the initial amount is 5 g when the half life of plutonium-239 (239pu, pu-239) is 24,100 years is 2.5 g

The equation is Mr=Mi(1/2)^n

where n is the number of half-lives

Mr is the mass remaining after n half lives

Mi is the initial mass of the sample

To find n, the number of half-lives, divide the total time 1000 by the time of the half-life(24,100)

n=1000/24100=0.0414

So Mr=5x(1/2)^1=2.5 g

The mass remaining is 2.5 g

  • The half life is the time in which the concentration of a substance decreases to half of the initial value.

Learn more about half life at:

brainly.com/question/24710827

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8 0
1 year ago
Two compounds are standing at the same temperature. Compound "A" is evaporating more slowly than compound "B." According to the
frosja888 [35]

Answer:

Option D) Compound B may have a lower molecular weight.

Explanation:

Compound A and B are standing at the same temperature yet compound A is evaporating more slowly than compound B.

This simply indicates that compound B have a lower molecular weight than compound A.

This can further be seen when gasoline and kerosene are placed under same temperature. The gasoline will evaporate faster than kerosene because the molecular weight of the gasoline is low when compared to that of the kerosene.

4 0
3 years ago
A buffer contains the weak acid HA and its conjugate base A-. the weak acid has a p Ka of 4.82 and the buffer has a pH of 4.25.
klasskru [66]
<span>I think the correct answer is A. A buffer is a substance that resists small change in the acidity of a solution when an acid or base is added to the solution. Usually, a buffer involves a weak acid or a weak alkali and one of its salt.</span>
3 0
3 years ago
For each trial, compute the mol of titrant; (molarity x L) and keep the number of significant figures to 4.
MrMuchimi

Answer:

Trial     Number of moles

           

  1          0.001249mol

  2         0.001232mol

  3          0.001187 mol

Explanation:

To calculate the <em>number of moles of tritant</em> you need its<em> molarity</em>.

Since the<em> molarity</em> is not reported, I will use 0.1000M (four significant figures), which is used in other similar problems.

<em>Molarity</em> is the concentration of the solution in number of moles of solute per liter of solution.

In this case the solute is <em>NaOH</em>.

The formula is:

          Molarity=\dfrac{\text{Number of moles of solute}}{\text{Volume of solution in liters}}

Solve for the <em>number of moles:</em>

          \text{Number of moles}=Molarity\times Volume\text{ }in\text{ }liters

Then, using the molarity of 0.1000M and the volumes for each trial you can calculate the number of moles of tritant.

Trial    mL           liters          Number of moles

           

1          12.49       0.01249        0.01249liters × 0.1000M = 0.001249mol

2         12.32      0.01232         0.01232liters × 0.1000M = 0.001232mol

3          11.87       0.01187         0.01187liters × 0.1000M = 0.001187 mol

3 0
3 years ago
Read 2 more answers
A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles
alexgriva [62]

Answer : The pH of the solution is, 3.41

Explanation :

First we have to calculate the moles of HF.

\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}

\text{Moles of HF}=0.250M\times 1.50L=0.375mol

Now we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (6.8\times 10^{-4})

pK_a=4-\log (6.8)

pK_a=3.17

The reaction will be:

                             HF+OH^-\rightleftharpoons F^-+H_2O

Initial moles     0.375     0.100   0.375

At eqm.   (0.375-0.100)      0     (0.375+0.100)

                     = 0.275                    = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[F^-]}{[HF]}

Now put all the given values in this expression, we get:

pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]

pH=3.41

Thus, the pH of the solution is, 3.41

8 0
3 years ago
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