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Measurements show that unknown compound has the following composition: element mass 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. Write the empirical chemical formula of this compound?(A) Ca2PO4
(B) Ca3PO6
(C) Ca4P2O4
(D) Ca3P2O8 (or Ca3(PO4)2)
(E) CaPO4

Answers

Answer 1

Answer:

Answer:

D) empirical formula is: C₃P₂O₈

Explanation:

Given:

Mass % Calcium (Ca) = 38.7%

Mass % Phosphorus (P) = 19.9%

Mass % oxygen (O) = 41.2 %

This implies that for a 100 g sample of the unknown compound:

Mass Ca = 38.7 g

Mass P = 19.9 g

Mass O = 41.2 g

Step 1: Calculate the moles of Ca, P, O

Atomic mass Ca = 40.08 g/mol

Atomic mass P = 30.97 g/mol

Atomic mass O = 16.00 g/mol

$Moles\ Ca = (38.7g)/(40.08g/mol) =0.966\ mol\n\nMoles\ P = (19.9g)/(30.97g/mol) =0.643\ mol\n\nMoles\ O = (41.2g)/(16.00g/mol) =2.58\ mol$

Step 2: Calculate the molar ratio

$C = (0.966)/(0.643) =1.50\n\nP = (0.643)/(0.643) = 1.00\n\nO = (2.58)/(0.643) =4.00$

Step 3: Calculate the closest whole number ratio

C: P: O = 1.50 : 1.00 : 4.00

C : P : O = 3:2:8

Therefore, the empirical formula is: C₃P₂O₈

Answer 2

Answer:

Final answer:

The mass percentage composition of a compound can be used to determine its empirical formula. For a compound with 38.7% calcium (Ca), 19.9% phosphorus (P), and 41.2% oxygen (O), the empirical formula is Ca3(PO4)2.

Explanation:

To solve this problem, we're going to use the atomic mass percentages to determine the empirical formula of the compound.

We do this by assuming we have a 100g sample of the compound. Therefore:

The mass of calcium (Ca) is 38.7g.

The mass of phosphorus (P) is 19.9g.

The mass of oxygen (O) is 41.2g.

Next, we calculate how many moles we have of each element:

Ca: 38.7g / 40.08g/mol (the atomic mass of calcium) = 0.965 moles
P: 19.9g / 30.97g/mol (the atomic mass of phosphorous) = 0.643 moles
O: 41.2g / 16.00g/mol (the atomic mass of oxygen) = 2.575 moles

Then, we divide each of these numbers by the smallest number of moles, which is 0.643 (P):

Ca: 0.965/0.643 = 1.5 (~1)
P: 0.643/0.643 = 1
O: 2.575/0.643 = 4

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Answers

Answer:

Stoichiometric coefficient of hydrogen gas is 1.

Stoichiometric coefficient of palmitic acid is 1.

Explanation:

Addition of hydrogen to double bond is termed as hydrogenation reaction.

$C_(16)H_(30)O_2+H_2\rightarrow C_(16)H_(32)O_2$

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Secondary amines react with the nitrosonium ion to generate ________. n-nitrosoamines diazonium salts anilines imines oximes

Answers

Answer:
Secondary amines react with the nitrosonium ion to generate N-Nitrosoamines.

Explanation:
Nitosonium Ion is generally utilized in the formation of Diazonium Salts which are considered excellent starting Material from synthesis point of View. Diazonium salts are formed by reacting Primary Amine or Anilines with Nitrosonium Ions. In our case, the Amine given is Secondary. So, reaction of Sec. Amines with Nitrosonium Ions stops after the formation of N-Nitrosoamine as there is no Hydrogen attached to Nitrogen atom of Amine to be eliminated and form a double and eventually triple bond with the Nitrogen atom of Nitrosonium Ion.

A 14 wt% solution of cacl2 (110.98 g/mol) has a density of 1.126 g/ml. what is the mass (in milligrams) of a 20.0-ml solution of 14.0 wt% cacl2?

Answers

Density = mass / volume

Thus, Mass of the solution can be expressed as:

Mass of solution = Density of solution × volume of solution

Given- Density of solution = 1.126 g/ml

volume of solution = 20.0 ml

∴ Mass of 14 wt% solution of CaCl2 = (1.126 g/ml) × (20 ml)

= 22.52 g

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If 5.738 grams of AgNO3 is mixed with 4.115 grams of BaCl2 and allowed to react according to the balanced equation: BaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ba(NO3)2(aq) What is the limiting reagent? BaCl2AgNO3 How many grams of AgCl could be produced? grams AgCl What mass, in grams, of the excess reagent will remain? grams of excess reagent

Answers

Answer:

Limiting reagent: AgNO3

grams AgCl : 2.44 g AgCl

grams of excess reagent remain: 0.62 g BaCl2

Explanation:

1. Change grams to mol:

AgNO3:

5.738g x (1mol/169.87g) = 0.034 mol AgNO3

BaCl2:

4.115g x (1 mol/208.23g) = 0.020 mol BaCl2

2. Limiting reagent:

AgNO3:

0.034 mol AgNO3 x (1 mol BaCL2/ 2mol AgNO3) = 0.017 mol BaCl2

BaCl2:

0.020 mol BaCl2 x (2 mol AgNO3/1 mol BaCl2) = 0.04 mol AgNO3

Limiting reagent: AgNO3

3. Grams of AgCl produced:

Using the limiting reagent:

0.017 mol AgNO3 x (2mol AgCl / 2 mol AgNO3) = 0.017 mol AgCl

4. Change mol to grams:

0.017 mol AgCl x ( 143.32 g AgCl /1mol AgCl) =2.44 g AgCl

5. Grams of the excess reagent:

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0.020 mol BaCl2 - 0.017 mol BaCl2 = 0.003 mol BaCl2

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Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction 2COF2(g)⇌CO2(g)+CF4(g), Kc=4.90 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?

Answers

Answer : The concentration of $COF_2$ remains at equilibrium will be, 0.37 M

Explanation : Given,

Equilibrium constant = 4.90

Initial concentration of $COF_2$ = 2.00 M

The balanced equilibrium reaction is,

$2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)$

Initial conc. 2 M 0 0

At eqm. (2-2x) M x M x M

The expression of equilibrium constant for the reaction will be:

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Now put the values of 'x' in concentration of $COF_2$ remains at equilibrium.

Concentration of $COF_2$ remains at equilibrium = $(2-2x)M=[2-2(1.219)]M=-0.582M$

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From this we conclude that, the amount of substance can not be negative at equilibrium. So, the value of 'x' which is equal to 1.291 M is not considered.

Therefore, the concentration of $COF_2$ remains at equilibrium will be, 0.37 M

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Answers

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Answer:

True

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Hydrogen bonds can be intermolecular (occurring between separate molecules) or

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