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masya89 [10]
3 years ago
8

The decay constant of plutonium-239, a waste product from nuclear reactors, is 2.88×10−5year−1. part a what is the half-life of

239pu?
Chemistry
1 answer:
faust18 [17]3 years ago
3 0
<span>24100 years
   You can easily convert from decay constant to half-life by using the relationship:
 H = ln(2)/D
  where
 H = half life
  D = decay constant
 ln(2) = natural logarithm of 2.
So
 H = ln(2)/D
 H = 0.693147181/2.88x10^-5 1/year
  H = 2.41x10^4 year
 H = 24100 year</span>
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The equilibrium constant for the reaction
FinnZ [79.3K]

Answer: The concentrations of Cl_2 at equilibrium is 0.023 M

Explanation:

Moles of  Cl_2 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol

Volume of solution = 1 L

Initial concentration of Cl_2 = \frac{0.14mol}{1L}=0.14M

The given balanced equilibrium reaction is,

                            COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial conc.           0.14 M           0 M       0M    

At eqm. conc.     (0.14-x) M        (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}

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

4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of Cl_2 at equilibrium is 0.023 M

7 0
3 years ago
When solving a problem it is important to identify your given and needed units, but it is also important to understand the relat
Angelina_Jolie [31]

Answer:

The question has some details missing. here are the details ; Given the following ;  

1. 43.2 g of tablet with 20 cm3 of space

2. 5 cm3 of tablets weighs 10.8 g

3. 5 g of balsa wood with density 0.16 g/cm3

4. 150 g of iron. With density 79g/cm 3

5. 32 cm3 sample of gold with density 19.3 g/cm3

6. 18 ml of cooking oil with density 0.92 g/ml

Explanation:

<u>Appropriate for calculating mass</u>

32 cm3 sample of gold with density 19.3 g/cm3

18 ml of cooking oil with density 0.92 g/ml

<u>Appropriate for calculating volume</u>

5 g of balsa wood with density 0.16 g/cm3

150 g of iron. With density 79g/cm 3

<u>Appropriate for calculating density</u>

43.2 g of tablet with 20 cm3 of space

5 cm3 of tablets weighs 10.8 g

3 0
2 years ago
How can you distinguish between crystalline allotropic modifications of Sulphur from those of amorphous allotrops?​
Burka [1]

The crystalline allotropes of sulfur are very strong and have a high melting and boiling point while the amorphous allotropes of sulfur are brittle and breaks easily.

<h3>What is a crystalline substance?</h3>

A crystalline substance is one that has a definite arrangement of the atoms in the substance. An amorphous substance lacks this definite arrangement. We can see this arrangement when we conduct an X-ray crystallography of the sulfur.

Also, the crystalline allotropes of sulfur are very strong and have a high melting and boiling point while the amorphous allotropes of sulfur are brittle and breaks easily.

Learn more about sulfur:brainly.com/question/13469437

#SPJ1

4 0
1 year ago
Help me ASAP PLS
n200080 [17]

Answer:

18 g

Explanation:

We'll begin by converting 500 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:

Volume = 0.5 L

Molarity = 0.2 M

Mole of C₆H₁₂O₆ =?

Molarity = mole / Volume

0.2 = Mole of C₆H₁₂O₆ / 0.5

Cross multiply

Mole of C₆H₁₂O₆ = 0.2 × 0.5

Mole of C₆H₁₂O₆ = 0.1 mole

Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:

Mole of C₆H₁₂O₆ = 0.1 mole

Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)

= 72 + 12 + 96

= 180 g/mol

Mass of C₆H₁₂O₆ =?

Mass = mole × molar mass

Mass of C₆H₁₂O₆ = 0.1 × 180

Mass of C₆H₁₂O₆ = 18 g

Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.

6 0
2 years ago
In an exothermic process the surrounding looses heat.<br>False<br>True​
WITCHER [35]

Answer:

False

Explanation:

Message me for explanation.

3 0
3 years ago
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