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3 years ago

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1.36g H2 is allowed to react with 10.1g N2 producing 2.05g NH3 What is the theoretical yield in grams for this reaction under th

e given conditions?
Express your answer to three significant figures and include the appropriate units.

1 answer:

vladimir2022 [97]3 years ago

6 0

here's the answer to your question

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Write the chemical formula for sodium hypobromite.

irga5000 [103]

Answer:

NaOBr (or) Na⁺ ⁻OBr

Explanation:

The Oxo-Acids of Bromine are as follow,

Hypobromous Acid = HOBr

Bromous Acid = HOBrO

Bromic Acid = HBrO₃

Perbromic Acid = HBrO₄

When these acids are converted to their conjugate bases their names are as follow,

Hypobromite = ⁻OBr

Bromite = ⁻OBrO

Bromate = ⁻OBrO₂

Perbromate = ⁻OBrO₃

According to rules, the positive part of ionic compound is named first and the negative part is named second. So, Sodium Hypobromite has a chemical formula of Na⁺ ⁻OBr or NaOBr.

6 0

3 years ago

⚠️NO LINKS⚠️(it will get reported)

brilliants [131]

Answer:

I'm thinking cooper but not sure

6 0

3 years ago

Choose the thermochemical equation that illustrates ΔH°f for Li2SO4. Choose the thermochemical equation that illustrates ΔH°f fo

Deffense [45]

Answer:

2Li(s) + ⅛S₈(s, rhombic) + 2O₂(g) → Li₂SO₄(s)

Explanation:

A thermochemical equation must show the formation of 1 mol of a substance from its elements in their most stable state,.

The only equation that meets those conditions is the last one.

A and B are wrong , because they show Li₂SO₄ as a reactant, not a product.

C is wrong because Li⁺ and SO₄²⁻ are not elements.

D is wrong because it shows the formation of 8 mol of Li₂SO₄.

8 0

3 years ago

You make an experiment to determine the percent mass of carbon in sugar. You find out that there is 46.63g of carbon in 75.00g o

Aleksandr [31]

Taking into account the definition of percentage composition, the percent composition of carbon in this sample is 62.17%.

<h3>Percentage composition</h3>

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

Then, the percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

$percentageby mass= \frac{mass of solute}{mass of solution}x100$

<h3>This case</h3>

In this case, you know:

mass of solute= 46.63 g
mass of solution= 75 g

Replacing:

$percentageby mass= \frac{46.63 g}{75 g}x100$

Solving:

<u><em>percentage by mass= 62.17 %</em></u>

Finally, the percent composition of carbon in this sample is 62.17%.

Learn more about percent composition:

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3 0

2 years ago

If 45 mL of water are added to 250 mL of a 0.75 M K2SO4 solution, what will the molarity of the diluted solution be?

krok68 [10]

Answer:

$\large\boxed{\large\boxed{0.64M}}$

Explanation:

When you form a <em>diluted solution</em> from a mother (concentrated) solution, the moles of solute are determined by the mother solution.

The main equation is:

$Molarity=\dfrac{\text{moles of solute}}{\text{volume of the solution in liters}}$

Then, since the moles of solute is the same for both the mother solution and the diluted solution:

$\text{Molarity mother solution }\times\text{ volume mother solution}=\\\\=\text{Molarity diluted solution }\times\text{ volume diluted solution}$

Substitute and solve for the molarity of the diluted solution:

$250mL\times 0.75M=(45mL+250mL)\times M\\\\\\M=\dfrac{250mL\times 0.75M}{295mL}=0.64M$

8 0

3 years ago