What is a negatively charged subatomic particle?

Please help!! A compound with a molar mass of 544.0 g/mol is made up of 26.5 grams Carbon, 2.94 grams

lukranit [14]

Answer:

1. Empirical formula = C₃H₄O₆

2. Molecular formula = C₁₂H₁₆O₂₄

Explanation:

From the question given above, the following data were obtained:

Molar mass of compound = 544 g/mol

Mass of Carbon (C) = 26.5 g

Mass of Hydrogen (H) = 2.94 g

Mass of oxygen (O) = 70.6 g

Empirical formula =?

Molecular formula =?

1. Determination of the empirical formula of the compound.

C = 26.5 g

H = 2.94 g

O = 70.6 g

Divide by their molar mass

C = 26.5 / 12 = 2.208

H = 2.94 / 1 = 2.94

O = 70.6 / 16 = 4.4125

Divide by the smallest

C = 2.208 / 2.208 = 1

H = 2.94 / 2.208 = 1.33

O = 4.4125 / 2.208 = 2

Muitiply through by 3 to express in whole number.

C = 1 × 3 = 3

H = 1.33 × 3 = 4

O = 2 × 3 = 6

Empirical formula = C₃H₄O₆

2. Determination of the molecular formula of the compound.

Molar mass of compound = 544 g/mol

Empirical formula = C₃H₄O₆

Molecular formula = [C₃H₄O₆]ₙ

[C₃H₄O₆]ₙ = 544

[(12×3) + (4×1) + (16×6)]n = 544

[36 + 4 + 96]n = 544

136n = 544

Divide both side by 136

n = 544 / 136

n = 4

Molecular formula = [C₃H₄O₆]ₙ

Molecular formula = [C₃H₄O₆]₄

Molecular formula = C₁₂H₁₆O₂₄

6 0

3 years ago

Read 2 more answers

A chemist makes of nickel(II) chloride working solution by adding distilled water to of a stock solution of nickel(II) chloride

Lady_Fox [76]

Answer:

0.0900 mol/L

Explanation:

<em>A chemist makes 330. mL of nickel(II) chloride working solution by adding distilled water to 220. mL of a 0.135 mol/L stock solution of nickel(II) chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.</em>

Step 1: Given data

Initial concentration (C₁): 0.135 mol/L

Initial volume (V₁): 220. mL

Final concentration (C₂): ?

Final volume (V₂): 330. mL

Step 2: Calculate the concentration of the final solution

We prepare a dilute solution from a concentrated one. We can calculate the concentration of the working solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂

C₂ = 0.135 mol/L × 220. mL/330. mL = 0.0900 mol/L

3 0

3 years ago

When looking at Newman projections are there instances where the staggered confirmation is not the lowest in energy? Explain you

stiks02 [169]

Answer:

let's summarize the caution as we are given with human projection. We have to tell them stances where staggered confirmation is not the lowest in energy. Okay, so let's see that. Are there any instances that which conforms tiger is not the lowest energy. Okay, so the lowest energy is what then happened? The lowest in energy already happens on living. Honestly, when, when it happens the lowest in energy only happens when the bulky groups are at 1 80 degree with each other. When it happens when bulky, bulky, bulky groups are at 1 80 degree. Okay, When they are, for example, when the bulky groups are in this way At 180°, then it is Louis Why? Because repulsion are very less repulsion very less. Got it. So we have to tell when it is not the lowest so far not the lowest instance. These bulky groups should be together. So if they either make of angle of 60° or 300° together, suppose this is a bulky group one and this is a bulky group two and okay, so if it makes a 60 degree or 300 degree then this um have the highest energy Or has the energy is not lowest in it. Why? It is not lowest in it Because at 300° or 60°. Why? Because the pulsations are very high repulsion. Are you ready? Hi. So by looking at the humans projection we can conclude that when the if the bulky group parrot 1, 80 degree be considered as the lowest energy. And if they are at 300 or 60 degree or close to each other, republicans are higher and they are not in lowest energy.

7 0

3 years ago

Iodine pentafluoride gas reacts with iodine fluoride gas producing iodine heptafluoride gas and iodine gas. What is the maximum

dybincka [34]

Answer:

63.45g of I₂ can be produced

Explanation:

IF₅ reacts with IF to produce IF₇ and I₂. The reaction is:

IF₅ + 2 IF → IF₇ + I₂

Moles of 10.0g of IF₅ (221.89g/mol):

10.0g IF₅ × (1mol / 221.89g) = <em>0.0451 moles of IF₅</em>

Using PV / RT = n, it is possible to find moles of 11.20L of IF, thus:

1atm×11.20L / 0.082atmL/molK × 273K = 0.500 moles of IF.

<em>At STP, pressure is 1atm, temperature is 273K and gas constant R is 0.082atmL/molK</em>

For a complete reaction of IF₅ you need:

0.0451 moles of IF₅ × (2 moles IF / 1 mole IF₅) = 0.902 moles of IF. As you have just 0.500 moles of IF, the IF is the limiting reactant.

2 moles of IF produce 1 mole of I₂. 0.500 moles of IF produce:

0.500mol IF ₓ ( 1 mol I₂ / 2 mol IF) = 0.250 mol I₂

As molar mass of I₂ is 253.81g/mol, mass of 0.250 mol I₂ are:

0.250mol I₂ ₓ (253.81g / mol) =

<h3>63.45g of I₂ can be produced</h3>

3 0

4 years ago

Energy that is made from the power of moving water

motikmotik

Hydroelectric energy is produced.

4 0

3 years ago