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Furkat [3]
3 years ago
14

0.265g of an organic compound produced on evaporation 102cm cube of vapour at 373K and 775mmHg. Percentage composition of the co

nstituent elements are 92.24% C and 7.76% H. Find the molecular mass and molecular formula of the composition.​
Chemistry
1 answer:
grigory [225]3 years ago
5 0

Answer:

No results found for 0.265g of an organic compound produced on evaporation 102cm cube of vapour at 373K and 775mmHg. Percentage composition of the constituent elements are 92.24% C and 7.76% H. Find the molecular mass and molecular formula of the composition..

Results for 0.265g of an organic compound produced on evaporation 102cm cube of vapour at 373K and 775mmHg Percentage composition of the constituent elements are 92.24 C and 7.76 H Find the molecular mass and molecular formula of the composition

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Different between organic fertilizer and inorganic fertilizer ​
Svetradugi [14.3K]

Organic fertilizers are made from mined rock minerals, and natural plant and animal materials. They include ingredients like manure, guano, dried and powdered blood, ground bone, crushed shells, finely pulverized fish, phosphate rock, and wood.

Inorganic fertilizer is synthetic, comprised of minerals and synthetic chemicals. Most of the minerals in inorganic fertilizer are mined from the earth, and balanced inorganic fertilizers are high in all three macronutrients and can contain ammonium sulfate, magnesium sulfate, and potassium chloride.

mark me brainliesttt :))

5 0
3 years ago
Read 2 more answers
What is the mass of solute dissolved in 50,0 g of 12.5% saline solution? A) 7.148 B) 6.258 C) 43.88 D) 4.00 g
Marizza181 [45]

Answer:

6.25 grams is the mass of solute dissolved.

Explanation:

w/w % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

Mass of the solution = 50.0 g

Mass of the solvent = x

w/w % = 12.5%

12.5\%=\frac{x}{50.0 g}\times 100

x = 6.25 g

6.25 grams is the mass of solute dissolved.

6 0
3 years ago
A 17.9 mL sample of a 0.445 M aqueous hypochlorous acid solution is titrated with a 0.374 M aqueous sodium hydroxide solution. W
In-s [12.5K]

Answer:

the pH at the start of the titration is 3

4 0
4 years ago
Select the correct answer​
Vladimir79 [104]
C. 14 protons and 14 electrons
6 0
3 years ago
How many moles of H2 are produced from 5.8 moles of NH3
vodomira [7]

Answer:

1. 8.7moles of H2

2. 2.25moles of O2

Explanation:

1. 2NH3 —> N2 + 3H2

From the equation,

2moles of NH3 produce 3 moles of H2.

Therefore, 5.8moles of NH3 will produce Xmol of H2 i.e

Xmol of H2 = (5.8x3)/2 = 8.7moles

2. C3H8 + 5O2 —> 3CO2 + 4H2O

From the equation,

5moles of O2 produced 4moles of H2O.

Therefore, Xmol of O2 will produce 1.8mol of H2O i.e

Xmol of O2 = (5x1.8)/4 = 2.25moles

4 0
4 years ago
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