Definition: This is an acid that does not dissociate 100% into its ions. HA<-> H+ + A-

yKpoI14uk [10]

It is in a large elliptical shape!

3 0

2 years ago

If electrons were used in the two slit experiment instead of light, what change would need to be made to the slit spacing in ord

irina [24]

The change that would need to be made to the slit spacing in order to see a diffraction pattern is bending, because in understanding why light behaves like a wave, it is the interference and diffraction were the phenomena distinguish waves from particles but waves are the only one can interfere and diffract while particles do not. The light bends around obstacles or cylinder like waves do, then it is bending which cause and resulted in the single slit diffraction pattern.

5 0

2 years ago

In a mixture of hydrogen and nitrogen gases, the mole fraction of nitrogen is 0.333. If the partial pressure of hydrogen in the

notka56 [123]

Answer:

$P_T=112.4torr$

Explanation:

Hello there!

In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:

$P_{H_2}=x_{H_2}*P_T$

And can be solved for the total pressure as follows:

$P_T=\frac{P_{H_2}}{x_{H_2}}$

However, we first calculate the mole fraction of hydrogen by subtracting that of nitrogen to 1 due to:

$x_{H_2}+x_{N_2}=1\\\\x_{H_2}=1-0.333=0.667$

Then, we can plug in to obtain the total pressure:

$P_T=\frac{75.0torr}{0.667}\\\\P_T=112.4torr$

Regards!

4 0

2 years ago

Convert each into decimal form. a) 1.56× 10^3 b) 0.56×10-4

Lera25 [3.4K]

Answer: A = 1560

B = 1.6

Explanation: brainlest please

3 0

2 years ago

A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solut

Irina-Kira [14]

Answer:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

Explanation:

It is a stichiometry problem.

We should write the balance equation of the mentioned chemical reaction:

2Al + 3CuCl₂ → 3Cu + 2AlCl₃.

It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.

Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:

n = mass / molar mass

The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.

The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³ mol.

From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.

∴ 3.85 x 10⁻³ mol of Al foil reacts completely with 5.578 x 10⁻³ mol of CuCl₂ with (2:3) ratio and CuCl₂ is the limiting reactant while Al foil is in excess.

From the stichiometry 3.0 moles of CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
Finally, we can calculate the mass of copper produced using:

mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³ mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.

So, the answer is:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

5 0

3 years ago