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Juliette [100K]
3 years ago
6

Help please with this tough question

Chemistry
1 answer:
Alla [95]3 years ago
5 0

Answer:

  1. option B
  2. option A

Explanation:

please mark me as brainliest ❤️

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the first option: lowering the control rods

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The half-life of bismuth is 5 days. How much of a 100 gram sample will remain after 20 days?
sesenic [268]
Half-life is the duration needed for a compound to decay into half of its initial weight. In this case, the initial weight is 100gram. The calculation for the final weight should be:

final weight= initial weight * (1/2)^(duration/half-life)
final weight= 100 grams * (1/2)^(20days/5 days)
final weight= 100 grams * (1/2)^(4)
final weight= 100 * 1/16= 100/16 gram= 6.25 grams

8 0
3 years ago
Find mass percent of each element URGENT PLS HELP!!!
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Answer: Think and you will get it right!

Explanation:

To find the mass percent composition of an element, divide the mass contribution of the element by the total molecular mass. This number must then be multiplied by 100% to be expressed as a percent.

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3 years ago
Which element is being represented by the electron configuration below? [Ar] 4s2 3d3
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3 years ago
Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →
Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0
3 years ago
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