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igomit [66]
2 years ago
9

What is the product of one equivalent of phosphorus and three equivalents of hydrogen? Draw it's

Chemistry
1 answer:
yulyashka [42]2 years ago
6 0
This name is phosphine flammable , toxic gas colourless
...it's formula is PH3

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Determine the percent yield for the reaction between 6.92 g K and 4.28g of oxygen gas if 7.36 g of potassium oxide is produced.
Illusion [34]

Answer:

percentage yield = 88.25%

Explanation:

Firstly, write the chemical reaction and balance the equation.

Potassium react with oxygen to produce potassium oxide.

K + 02 → K2O

Balance the equation

4K + 02 → 2K2O

The limiting reactant is K so the yield of potassium oxide can be calculated using grams for potassium.

atomic mass of K = 39.1g/mol

grams for 4 mole of potassium =  4(39.1) = 156.4 g

grams for 2 moles of K2O = 2( 39.1 × 2 + 16) = 188.4 g

If 156.4 g of K produces 188.4 g of K2O

6.92 g of K will produce ? gram of K2O

cross multiply

grams of K2O = 6.92 × 188.4/156.4

grams of K2O = 1303.72/156.4

grams of K2O = 8.33585677749

grams of K2O = 8.34 g

percentage yield = actual yield/theoretical yield × 100

actual yield = 7.36 g

theoretical yield = 8.34 g

percentage yield = 7.36/8.34 × 100

percentage yield =         736/8.34

percentage yield = 88.2494004796%

percentage yield = 88.25%

5 0
2 years ago
A sample of an unknown compound was decomposed and found to be composed of 1.36 mol oxygen, 4.10 mol hydrogen, and 2.05 mol carb
Flura [38]

Answer:

C3H6O2

Explanation:

To find the empirical formula of the compound, we divide the amount in moles of each of the elements by the amount in mole of the element with the smallest number of mole. In this question, the element with the smallest number of moles is oxygen with 1.36 mole. Hence, we divide the number of moles of each element by this.

H = 4.10/1.36 = 3

O = 1.36/1.36 = 1

C = 2.05/1.36 = 1.5

We then multiply through by 2 to yield the compound with the empirical formula C3H6O2

7 0
2 years ago
Functional group and bond hybridization of vanillin
lana66690 [7]

Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.

See attached figure for the structure.

Vanillin have 3 functional groups:

1) aldehyde group:  R-HC=O, in which the carbon is double bonded to oxygen

2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring

3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons

Now for the hybridization we have:

The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.

The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a  <u>sp³</u>  hybridization because they are involved only in single bonds.

7 0
2 years ago
What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c
brilliants [131]

Answer:

Molecular formula → PbSO₄ → Lead sulfate

Option c.

Explanation:

The % percent composition indicates that in 100 g of compound we have:

68.3 g of Pb, 10.6 g of S and  (100 - 68.3 - 10.6) = 21.1 g of O

We divide each element by the molar mass:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

We divide each mol by the lowest value to determine, the molecular formula

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Molecular formula → PbSO₄ → Lead sulfate

8 0
2 years ago
Read 2 more answers
Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to
Delvig [45]

Answer:

H2O<en<phen

Explanation:

The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.

7 0
2 years ago
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