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What is my percent yield of titanium (II) oxide if I react 20 grams of titanium (II) oxide in excess water (that means TiS is my limiting reactant) and my actual yield of titanium (II) oxide is 13 g?

Answers

Answer 1

Answer:

Answer:

your percent yield Is

Explanation:

1.5384615385

Brainliest plz

Related Questions

Which of the following required Bohr's model of the atom to need modification ? A. Energies of electrons are quantized. B. Quantized electron energies are responsible for emission spectra lines. C. An electron's energy increases the farther it moves from the nucleus. D. Electrons do not follow circular orbits around the nucleus..
Assuming 54.81 grams of Al are consumed in the presence of excess copper II chloride dihydrate, how many grams of AlCl3 can be produced if the reaction will only produce 66.93 % yield?
Which of the following is an example of a compound? water - H2Ooxygen-O2 hydrogen - H2 helium - He
When a polar bond is formed between 2 atoms which atom receives a partial positive charge
Give an example of a substance that is MORE dense in its solid state when compared to its liquid state.

How many moles are in 12.0 grams of O2

Answers

Answer:

Moles = 0.375

Explanation:

Moles= m/M

= 12/32 = 0.375mol

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:

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

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

Learn more about Empirical Formula here:

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In the reaction 5 space B r to the power of minus space (a q )space plus space B r O subscript 3 to the power of minus space (a q )space plus space 6 space H to the power of plus space (a q )space rightwards arrow space 3 space B r subscript 2 space (a q )space plus space 3 space H subscript 2 O space (l )the rate of disappearance of Br- at some moment in time was determined to be 3.5 x 10-4 M/s. What is the rate of appearance of Br2 at that same moment

Answers

Answer:

$r_(Br_2)=2.1x10^(-4)M/s$

Explanation:

Hello,

In this case, for the reaction:

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

Thus, via the rate proportions between Br⁻ and Br₂ for which the stoichiometric coefficients are 5 and 3 respectively, we can write:

$(r_(Br^-))/(-5) =(r_(Br_2))/(3)$

Hence, the rate of appearance of Br₂ turns out:

$r_(Br_2)=(3r_(Br^-))/(-5)=(3*-3.5x10^(-4)M/s)/(-5)\n \nr_(Br_2)=2.1x10^(-4)M/s$

Take into account that the rate of disappearance is negative for reactants.

Best regards.

Redox reactions stand for oxidation/reduction reactions. True or false: an oxidation reaction is always paired with a reduction reaction.

Answers

Answer:

The answer is: true

Explanation:

In redox reactions, the half-reactions of oxidation and reduction always occur simultaneously in pair.

The oxidation half-reaction involved the lost of electrons from a reduced substance (A) to form a oxidized substance (A⁺):

A ⇒ A⁺ + e-

In contrapossition, during the reduction half-reaction the oxidized substance (B⁺) gains electrons to form the reduced subtance (B):

B⁺ + e- ⇒ B

The overall redox reaction is obtained by the addition of the two half-reactions:

A ⇒ A⁺ + e-

B⁺ + e- ⇒ B

-----------------

A + B⁺⇒ A⁺ + B

The electrons gained by B are provided by A, which lost the same number of electrons. Thus, the oxidation/reduction reactions are paired.

Final answer:

Yes, it's true that an oxidation reaction always pairs with a reduction reaction, thereby making up a redox reaction where one substance is oxidized (loses electrons) and another is reduced (gains electrons). The oxidized species is the reducing agent while the reduced one is the oxidizing agent.

Explanation:

The statement is true: an oxidation reaction is indeed always paired with a reduction reaction. This can be exemplified in the redox reactions where one substance is oxidized (loses electrons) while another is reduced (gains electrons). These reactions always occur together. The species that is oxidized is called the reducing agent, while the species that is reduced is called the oxidizing agent. Therefore, in every redox reaction, there will always be an oxidation process coupled with a reduction process.

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If the partial pressure of N2 in a scuba divers blood at the surface is 0.79 atm, what will the pressure be if he/she descends to a depth of 30 meters (4 atm) and stays there long enough to reach equilibrium (b)

Answers

Answer:

Explanation:

for every 3m that the internal pressure is lowered, it increases in an atmosphere approximately, so when the blood pressure of nitrogen decreases 30m, it will increase by approximately 10 atm, being enough there for the body to enter into equilibrium

Which description is not a property of a base? (2 points)a
pH lower than 7

b
Turns litmus paper blue

c
Bitter taste

d
Slimy feel

Answers

i think it’s A
because if the pH is lower than 7 than it is acidic

Answer:

the answer is a hope it helps.

Explanation:

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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?

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The Michael reaction is a conjugate addition process wherein a nucleophilic enolate anion (the donor) reacts with an α,β-unsaturated carbonyl compound (the acceptor). The best Michael reactions are those that take place when a particularly stable enolate anion is formed via treatment of the donor with a strong base. Alternatively, milder conditions can be used if an enamine is chosen as the donor, this variant is termed the Stork reaction. In the second step, the donor adds to the β-carbon of the acceptor in a conjugate addition, generating a new enolate. The enolate abstracts a proton from solvent or from a new donor molecule to give the conjugate addition product. Draw curved arrows to show the movement of electrons in this step of the mechanism.