4. A gas at 400 K is placed in a 4.0 L moveable piston. The gas

Model describe how you would show j.j. thomson's model of the atom using small beads and clay

faltersainse [42]

Make a ball of clay and embed small beads throughout it. The plum pudding model.

5 0

3 years ago

using the equation you wrote determine how many moles of butane c4h10 are needed to react with 5.5 moles of oxygen

GaryK [48]

Answer:

0.846 moles.

Explanation:

This is a stichiometric problem.

The balanced equation of complete combustion of butane is:

C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O

It is clear from the stichiometry of the balanced equation that complete combustion of 1.0 mole of butane needs 6.5 moles of O₂ to produce 4 moles of CO₂ and 5 moles of H₂O.

<u><em>Using cross multiplication:</em></u>

1.0 mole of C₄H₁₀ reacts with → 6.5 moles of O₂

??? moles of C₄H₁₀ are needed to react with → 5.5 moles of O₂

The number of moles of C₄H₁₀ that are needed to react with 5.5 moles of O₂ = (1.0 x 5.5 moles of O₂) / (6.5 moles of O₂) = 0.846 moles.

3 0

3 years ago

write a generalization correlating solubility with ionization. start with: "as compounds become more ionized,..."

Tema [17]

Answer:

the solubility increases

Explanation:

The solubility of any ionic compound is due to the ionization of the compound and then the strong ion-dipole interactions acting between the ions and the solvent.

Thus, solubility also depends on the extent of the ionization of the salt.

The more the salt ionizes, the more there is ion-dipole interaction between the ions the solvent and more is the solubility.

4 0

3 years ago

A graduated cylinder is filled to the 40.00-mL mark with a mineral oil. The masses of the cylinder before and after the addition

castortr0y [4]

Answer:

Firstly, Let's experiment this !

Experiment 1 :

159.446g - 124.966g = 34.48g

34.48g = The mass of Mineral oil.

The density of the mineral oil = M/V = 34.48g/40mL = 0.862g/cm³.

Experiment 2 :

124.966 + 18.173 = 143.139 = The mass of solid + cylinder.

124.966 + 50.952 = 175.918 = The mass of solid + cylinder + Mineral water.

175.918 - 143.139 = 32.779 = The mass of added mineral oil.

Explanation:

Now we have to find the volume of the added mineral oil using the density from experiment 1.

V = 32.779g/0.862g/cm³ = 38.02668213mL

Since we found the volume of the solid, we then have to subtract the added mineral oil volume from the total volume from experiment 1.

Volume of solid = 40-38.02668213 = 1.97331787mL

Density of solid = 18.713g/1.97331787mL = 9.483013499g/cm^3

1.97331787 = (4/3)(3.14)r³

1.97331787*(3/4)(3.14) = .4713338861

.4713338861 = r ³

r = 0.7782328425158433

r = 0.78

Now that's our final answer ! r = 0.78

5 0

3 years ago

If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically

Phantasy [73]

548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = $\frac{mass}{atomic mass of 1 mole}$

putting the values in the equation

NaOH = $\frac{842}{39.9}$

= 21.1 MOLES OF NaOH

Al = $\frac{750}{26.9}$

= 27.8 moles

from the equation

from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

$\frac{1}{3}$ = $\frac{x}{21.1}$

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

= 548.55 grams

theoretical yield from the given data is 548.55 grams

3 0

3 years ago