All categories

3 years ago

What is the maximum amount in moles of P2O5 that can theoretically be made from 176 g of O2 and excess phosphorus?

Chemistry

1 answer:

Strike441 [17]3 years ago

5 0

Answer:

2.60 mol2

Explanation:

2.60mol2 is ur answer

You might be interested in

How do you work out question 1a?

Sliva [168]

Answer:

-125 kJ

Explanation:

You calculate the energy required to break all the bonds in the reactants. Then you subtract the energy to break all the bonds in the products.

H₂C=CH₂ + H₂ ⟶ H₃C-CH₃

Bonds: 4C-H + 1C=C 1H-H 6C-H + 1C-C

D/kJ·mol⁻¹: 413 612 436 413 347

The formula relating ΔHrxn and bond dissociation energies (D) is

ΔHrxn = Σ(Dreactants) – Σ(Dproducts)

(Note: This is an exception to the rule. All other thermochemical reactions are “products – reactants”. With bond energies, it’s “reactants – products”. The reason comes from the way we define bond energies.)

For the reactants:

Σ(Dreactants) = 4 × 413 + 1 × 612 + 1 × 436 = 2700 kJ

For the products:

Σ(Dproducts) = 6 × 413 + 1 × 347 = 2825 kJ

For the system :

ΔHrxn = 2700 - 2825 = -125 kJ

4 0

3 years ago

Match the element with number of valence electrons

Alla [95]

Answer:

Oxygen-O = 6

Carbon C = 4

Gallium G = 3

Astatine A = 7

Arsenic As = 5

Potassium K = 1

Hydrogen H = 1

Helium H = 2

Explanation:

6 0

2 years ago

The nonvolatile, nonelectrolyte TNT (trinitrotoluene), C7H5N3O6 (227.1 g/mol), is soluble in benzene C6H6. How many grams of TNT

Alexeev081 [22]

Answer:

Explanation:

formula of osmotic pressure is as follows

p= n RT

n is mole of solute per unit volume

If m be the grams of solute needed

m gram = m / 227.1 moles

m / 227.1 moles dissolved in .279 litres

n = m / (227.1 x .279 )

= m / 63.36

substituting the values in the osmotic pressure formula

5.14 = (m / 63.36) x .082 x 298

m / 63.36 = .21

m = 13.32 grams .

7 0

2 years ago

A metal, M , of atomic mass 56 amu reacts with chlorine to form a salt that can be represented as MClx. A boiling point elevatio

nordsb [41]

Answer:

Formula for the salt: MCl₃

Explanation:

MClₓ → M⁺ + xCl⁻

We apply the colligative property of boiliing point elevation.

We convert the boiling T° to °C

375.93 K - 273K = 102.93°C

ΔT = Kb . m . i

where ΔT means the difference of temperature, Keb, the ebulloscopic constant for water, m the molality of solution (mol of solute/kg of solvent) and i, the Van't Hoff factor (numbers of ions dissolved)

ΔT = 102.93°C - 100°C = 2.93°C

Kb = 0.512 °C/m

We replace data: 2.93°C = 0.512 °C/m . m . i

i = x + 1 (according to the equation)

22.9 g / (56g/m + 35.45x) = moles of salt / 0.1kg = molality

We have calculated the moles of salt in order to determine the molar mass, cause we do not have the data. We replace

2.93°C = 0.512 °C/m . [22.9 g / (56g/m + 35.45x)] / 0.1kg . (x+1)

2.93°C / 0.512 m/°C = [22.9 g / (56g/m + 35.45x)] / 0.1kg . (x+1)

5.72 m = [22.9 g / (56g/m + 35.45x)]/ 0.1 (x+1)

5.72 . 0.1 / [22.9 g / (56g/m + 35.45x)] = x+1

0.572 / (22.9 g / (56g/m + 35.45x) = x+1

0.572 (56 + 35.45x) / 22.9 = x+1

0.572 (56 + 35.45x) = 22.9x + 22.9

32.03 + 20.27x = 22.9x + 22.9

9.13 = 2.62x

x = 3.48 ≅ 3

6 0

3 years ago

Suppose that for the same 10.0 ml sample described in q3, the mass of the crucible with precipitate was 17.550 g, and the mass o

marusya05 [52]

The weight of the precipitate was 17.550g - 17.410g = 0.140g. From a 10ml sample, this means the concentration was 0.140g / 10ml, which is equal to 0.014g/ml.

6 0

3 years ago