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2 years ago

Assignment Directions:

Chemistry

1 answer:

Alja [10]2 years ago

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Answer:

For this assignment, I have picked three famous soccer players from around the world. I will go over their Birth, work ethic, and downfalls. I guess that most soccer players' downfalls are going to be their injuries.

The first soccer player I found was Ricardo Kakà. Ricardo Kakà was born to Simone dos Santos, his mother, and Bosco Izecson Pereira Leite, his father, on April 22, 1982. His work ethic mostly consists of "doing it". One of Ricardo Kakà's greatest downfalls is his injuries.

The second player I found was Cristiano Ronaldo. Cristiano Ronaldo was born to Maria Dolores dos Santos and José Dinis Aveiro. He was introduced to soccer by his dad because worked at a soccer field. Cristiano Ronaldo's work ethic consisted of something that Ricardo Kakà had as well. "Get it done". Cristiano Ronaldo's downfall was that his father drank too much.

I was not surprised to see that, but I thought it was going to be injuries as well.

The final player that I picked is Luis Suárez. Luis Suárez was born to Sandra Diaz and Rodolfo Suárez. He joined his local youth team that played soccer and he loved it. His work ethic is the same as the last soccer players. "Do it." His major downfall was that he was a bad-tempered teen and he got mad at stuff. That didn't help him when his parents divorced.

Overall all of the soccer players are different and very important. They have all gone through different situations and have gotten through them through soccer.

Explanation:

This is not 300 words, it is 244. You can it. Make sure to change it so it sounds like you wrote it.

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Write the formula of the conjugate acid of HCO₂⁻.

Alex

Taking into account the Brønsted-Lowry acid-base theory, the conjugate acid of HCO₂⁻ is H₂CO₂.

<h3>Brønsted-Lowry acid-base </h3>

The Brønsted-Lowry acid-base theory (or the Brønsted-Lowry theory) identifies acids and bases based on whether the species accepts or donates protons or H⁺.

According to this theory, acids are proton donors while bases are proton acceptors. That is, an acid is a species that donates an H⁺ proton while a base is a chemical species that accepts an H⁺ proton from the acid.

So, reactions between acids and bases are H⁺ proton transfer reactions.

<h3>Conjugate base and conjugate acid</h3>

Then, a conjugate base is an ion or molecule resulting from the acid that loses the proton, while a conjugate acid is an ion or molecule resulting from the base that gains the proton:

acid + base ⇄ conjugate base + conjugate acid

<h3>Conjugate acid of HCO₂⁻</h3>

Like a conjugate acid is an ion or molecule resulting from the base that gains the proton, the conjugate acid of HCO₂⁻ is H₂CO₂.

Learn more about the Brønsted-Lowry acid-base theory:

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2 years ago

A mixture containing initially 3.90 mole of NO(g) and 0.88 mole of CO2(g) was allowed to react in a flask of volume 1.00 L at a

Ymorist [56]

Answer:

Equilibrium constant Kc for the reaction will be 1.722

Explanation:

O2(g)+NO(g)→CO(g)+ NO2(g)

0.88 3.9 --- ---

0.88x 3.9-x x x

GIVEN:

0.88X-X= 0.11

⇒ X=0.77

CO2(g)+NO(g) → CO(g) + NO2(g)

0.88 3.9 --- ---

0.88-x 3.9-x x x

= 3.13 0.77 0.77

=0.11

Kc = $\frac{[CO] *[NO2]} {[CO2]*[NO]}$

= ${{0.77}×0.77÷{{0.11×3.13}}$

= 1.722

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3 years ago

What is two of same elements is called

svet-max [94.6K]

Answer:

it is a A molecule

8 0

3 years ago

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Explain the neutralization of vitamin c with sodium hydrogen carbonate?

mars1129 [50]

Scientists have known the consequences of mixing sodium bicarbonate and vitamin C ever since 1936, when a study on the subject was published in the "Journal of Nutrition." The authors of this study measured the amount of vitamin C recovered from the urine of people who drank a fixed amount of orange juice. The authors determined that the amount of vitamin C excreted was decreased by administration of sodium bicarbonate. Followup studies in the 1940s showed that this effect was due to the neutralization of the vitamin C by the sodium bicarbonate.

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3 years ago

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A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

<u>Answer:</u> The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

<u>For sulfuric acid:</u>

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

<u>For sodium sulfate:</u>

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

<u>For sulfuric acid:</u>

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

<u>For NaOH:</u>

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

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3 years ago