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1 year ago

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1. Which trait is most likely to be passed on to its organism’s offspring?

1 answer:

mart [117]1 year ago

4 0

Answer:Evolution

Evolution is the process by which populations of organisms change over generations. … If a trait is advantageous and helps the individual survive and reproduce, the genetic variation is more likely to be passed to the next generation (a process known as natural selection).

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Given that equation Na+(s)O2–> Na2O. Find the mass of sodium needed to produce 12.5 grams of sodium oxide.

Norma-Jean [14]

Answer: 9.27 g of Na

Explanation:Please see attachment for explanation

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3 years ago

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emmasim [6.3K]

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3 years ago

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What is the principle feature of a metallic bond?

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3 years ago

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Calculate the pH of a solution that is 0.235M benzoic acid and 0.130M sodium benzoate, a salt whose anion is the conjugate base

lesantik [10]

Hello!

We have the following data:

ps: we apply Ka in benzoic acid to the solution.

[acid] = 0.235 M (mol/L)

[salt] = 0.130 M (mol/L)

pKa (acetic acid buffer) =?

pH of a buffer =?

Let us first find pKa of benzoic acid, knowing that Ka (benzoic acid) = $6.20*10^{-5}$

So:

$pKa = - log\:(Ka)$

$pKa = - log\:(6.20*10^{-5})$

$pKa = 5 - log\:6.20$

$pKa = 5 - 0.79$

$\boxed{pKa = 4.21}$

Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:

$pH = pKa + log\:\dfrac{[salt]}{[acid]}$

$pH = 4.21 + log\:\dfrac{0.130}{0.235}$

$pH = 4.21 + log\:0.55$

$pH = 4.21 + (-0.26)$

$pH = 4.21 - 0.26$

$\boxed{\boxed{pH = 3.95}}\end{array}}\qquad\checkmark$

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR! =)

8 0

2 years ago

g The decomposition reaction of A to B is a first-order reaction with a half-life of 2.42×103 seconds: A → 2B If the initial con

Tanzania [10]

Answer:

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Explanation:

The equation used to calculate the constant for first order kinetics:

$t_{1/2}=\frac{0.693}{k}}$ .....(1)

Rate law expression for first order kinetics is given by the equation:

$t=\frac{2.303}{k}\log\frac{[A_o]}{[A]}$ ......(2)

where,

k = rate constant

$t_{1/2}$ =Half life of the reaction = $2.42\times 10^3 s$

t = time taken for decay process = ?

$[A_o]$ = initial amount of the reactant = 0.163 M

[A] = amount left after time t = 66.8% of $[A_o]$

[A]= $\frac{66.8}{100}\times 0.163 M=0.108884 M$

$k=\frac{0.693}{2.42\times 10^3 s}$

$t=\frac{2.303}{\frac{0.693}{2.42\times 10^3 s}}\log\frac{0.163 M}{0.108884 M}$

t = 1,409.19 s

1 minute = 60 sec

$t=\frac{1,409.19 }{60} min=23.49 min$

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

6 0

2 years ago