Which set contains no ionic species?

What phase of matter is oxygen at room temperature

Vedmedyk [2.9K]

Gas

good luck mate :)

8 0

2 years ago

Aryan want to deposit silver on an iron spoon using silver nitrate as electrolyte. Which terminal of the battery he should conne

34kurt

Silver from the anode gets dissolved to reach the cathode, where the spoon will be plated.

<h3>What is electroplating?</h3>

Electroplating is a way of electrolysis, where a thin layer of metal is used to plate a determined object. It is a kinda process to purify a material you want.

The anode contains the metal you want to plate with, in this case, the Ag.

The anode contains the half-reaction of oxidation so:

Ag(s) → Ag ⁺ (aq) + e⁻

In the cathode, you have the spoon, which it takes place the half-reaction of reduction:

Ag ⁺ (aq) + e⁻ → Ag(s)

The electrolytic cell, where the redox reaction takes place, must be filled with a AgNO₃ solution.

Silver from the anode gets dissolved to reach the cathode, where the spoon will be plated.

Learn more about the electroplating here:

brainly.com/question/20112817

#SPJ1

8 0

1 year ago

Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO

sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

CaO : CaCO₃

1 : 1

0.26 : 0.26

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ = 26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

7 0

3 years ago

How many atoms of carbon are combined to with 2.81 moles of hydrogen atoms in a sample of the compound ethane, C2H6?

puteri [66]

Answer:

5.64×10²³ atoms C

Explanation:

Convert moles of H to moles of C:

2.81 mol H × (2 mol C / 6 mol H) = 0.937 mol C

Convert moles of C to atoms of C:

0.937 mol C × (6.02×10²³ atoms C / mol C) = 5.64×10²³ atoms C

5 0

3 years ago

Suppose you have been given the task of distilling a mixture of hexane + toluene. Pure hexane has a refractive index of 1.375 an

miv72 [106K]

Explanation:

The given data is as follows.

Refractive index of mixture = 1.456

Refractive index of hexane = 1.375

Refractive index of toulene = 1.497

Let mole fraction of hexane = $x_{1}$

and, mole fraction of toulene = $x_{2}$

Also, $x_{1} + x_{2} = 1$

or, $x_{1} = 1 - x_{2}$

Hence, calculate the mole fraction of hexane as follows.

refractive index mixture= mole fraction hexane × ref index hexane + mole fraction toluene × ref index toluene.

1.456 = $(1 - x_{2}) \times 1.375 + x_{2} \times 1.497$

1.456 = $1.375 - 1.375x_{2} + 1.497x_{2}$

0.081 = $0.122x_{2}$

$x_{2}$ = $\frac{0.081}{0.122}$

= 0.66

Since, $x_{1} = 1 - x_{2}$

= 1 - 0.66

= 0.34

Thus, we can conclude that mole fraction of hexane in your sample is 0.34.

3 0

3 years ago