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3 years ago

8

Using the values from Step 1, predict the pattern for the decay of 100 atoms over the course of eight half-life cycles. Round to

the nearest whole number of atoms. Record in the appropriate blanks. A = B = C = D = E =

1 answer:

ratelena [41]3 years ago

8 0

Answer:

A=50

B = 13

C = 3

D = 2

E = 0

Explanation:

First guy got it wrong^^^

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The answer would be letter A

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3 years ago

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Why do we use balanced equations ?

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Answer:

Balanced equations are used to allow chemists to calculate how much product they will produce from their reactants. The co-efficients in balanced equations represent the number of moles reacting and being produced, it is rare that this exact ratio will be used in a reaction.

Explanation:

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3 years ago

Which statement best explains how mitochrondria help get a cell material it needs

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Answer:

By breaking down glucose into an energy molecule (ATP), so by producing energy which is necessary for the cell's survival.

Explanation:

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3 years ago

When the enthalpy of the reaction is provided, it is always in the context of moles; i.e., ΔHrxn=+252.8 kJ when two moles of CH3

Morgarella [4.7K]

<u>Answer:</u> The balanced chemical equation is written below.

<u>Explanation:</u>

There are 2 types of chemical reaction classified on the basis of heat change:

<u>Endothermic reactions </u>: They are the reactions in which energy of products is more than the energy of the reactants. For these reactions, energy is absorbed by the system. The $\Delta H$ comes out to be positive and is written on the reactants side.
<u>Exothermic reactions:</u> They are the reactions in which energy of reactants is more than the energy of the products. For these reactions, energy is released by the system. The $\Delta H$ comes out to be negative and is written on the product side.

We are given:

Moles of methanol = 2 moles

Moles of methane = 2 moles

Moles of oxygen gas = 1 mole

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Hence, the balanced chemical equation is written above.

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3 years ago

An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an

Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that $pH = -log[H^{+}]$ , and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

$pH = pKa - log \frac{[HA]}{[A^{-}]}$

Where [HA] is the concentration of the acid, and $[A^{-}]$ is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of $OH^-$ will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of $OH^-$ reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of $OH^-$ , which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

$6.72 = pKa - log \frac{0.01}{0.0025}$

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

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4 years ago