For a voltaic cell consisting of Al(s) in Al(NO3)3(aq) and Cu(s) in Cu(NO3)2(aq), what is Ecell, given [Al3 ]

Answer:

the answer is c

Explanation:

Net ionic equation is calculated by the eliminating spectator ion. The net ionic products are potassium ion and bromine ion.

Net ionic equation:

It shows the ionic products of the products that has been precipitated by the dissociation of reactant in aquas solution.

Aqueous solutions of potassium hydroxide and ammonium bromide combined to form a precipitate KBr.

The dissociation reaction of  

[tex]\bold {KBr(aq) \rightarrow K^+(aq) + Br^-(aq)}[/tex]

Aquas potassium hydroxide dissociates into potassium ion and bromine ion.

Therefore, the net ionic products are potassium ion and bromine ion.

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Answer:

1.66 moles.

Explanation:

We'll begin by calculating the number of mole in 23.2 g of nitrogen gas, N2.

This is illustrated below:

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 = 23.2 g

Mole of N2 =.?

Mole = mass /Molar mass

Mole of N2 = 23.2/28

Mole of N2 = 0.83 mole

Next, we shall determine the number of mole in 23.2 g of Hydrogen gas, H2.

This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 = 23.2 g

Mole of H2 =?

Mole = mass /Molar mass

Mole of H2 = 23.2/2

Mole of H2 = 11.6 moles

Next, the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.

From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.

Therefore, N2 is the limiting reactant.

Finally, we shall determine the maximum amount of NH3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.

The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.

Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.

Answer:

1. the study of the structure, properties, and interactions of materials and the changes they undergo

2. the study of matter and energy and how they interact

3. the systematic study and analysis of facts

4. a process to test a proposed solution to a problem

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Until the 1600s, people who investigated problems related to our understanding of the natural world were called _____.

Click ca

Explanation:

Answer:

1) If life exists in other planets

2) What type of living organisms exists on other planets

3) Predominant atoms/mixture/compound that exists on other planets

Explanation:

Earth and space science is the study of the earth, the space and other planets within the universe.

The following questions can be answered by studying earth and space science

1) If life exists in other planets

Over the years, there has been several reports suggesting this but a deep inquiry into this with scientific data can assist in putting the matter to rest.

2) What type of living organisms exists on other planets

If there is/are lives on other planets, what type of life/living organisms exist there. Are they plants? Are they animals? Or do we have a new classification for them?

3) Predominant atom/mixture/compound that exists on other planets

What is the most popular substance in each of these planets? The role of this substance in the survival of the planet or it's inhabitants also needs to be studied

Answer:

The solvent is the substance which typically determines the physical state of the solution (solid, liquid or gas). The solute is the substance which is dissolved by the solvent.

Answer:

the transfer of heat energy within the atmosphere, hydrosphere, and the earth's surface and interior occurs as a result of radiation, convection, and conduction. ocean currents play a significant role in transferring this heat toward the poles.

Answer:

True

Explanation:

Sir Joseph John Thomson OM PRS (18 December 1856 – 30 August 1940) was a British physicist and Nobel Laureate in Physics, credited with the discovery of the electron, the first subatomic particle to be discovered.

Answer:

Option C. 80.5%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Next, we shall determine the mass of C3H8 that reacted and the mass of CO2 produced from the balanced equation.

This is illustrated below:

Molar mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44 g/mol

Mass of C3H8 from the balanced equation = 1 x 44 = 44 g

Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol

Mass of CO2 from the balanced equation = 3 x 44 = 132 g

From the balanced equation above,

44 g of C3H8 reacted to produce 132 g of CO2.

Next, we shall determine the theoretical yield of CO2.

This can be obtained as shown below:

From the balanced equation above,

44 g of C3H8 reacted to produce 132 g of CO2.

Therefore, 0.295 kg (i.e 295 g) will react to produce = (295 x 132)/44 = 885 g of CO2.

Therefore, the theoretical yield of CO2 is 885 g.

Finally, we shall determine the percentage yield of CO2 as follow:

Actual yield of CO2 = 712 g

Theoretical yield of CO2 = 885 g

Percentage yield of CO2 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield of CO2 = 712/885 x 100

Percentage yield = 80.5%

Therefore, the percentage yield of CO2 is 80.5%.

Answer:

The amount of dissolved gases in the body. Have a good day! =)

Explanation:

Answer:

to prevent the chemical from reacting with ions present in water

Explanation:

water contains dissolved ions such as calcium from minerals and soluble rocks

Answer:

1.34x10⁻⁵ mol / L is the solubility of AgCl

Explanation:

Ksp of AgCl is defined as:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

If 0.010M of AgCl is added, some Ag⁺ and Cl⁻ will be produced until:

Ksp = [Ag⁺] [Cl⁻]

Ksp for AgCl = 1.8x10⁻¹⁰ (Taken from ALEKS Data tab):

Some Ag⁺ and Cl⁻ are produced, you can take this "some" as X:

[Ag⁺] = X

[Cl⁻] = X

Where X is the amount of AgCl that dissolvesin water. X = solubility:

Ksp = 1.8x10⁻¹⁰ = [Ag⁺] [Cl⁻]

1.8x10⁻¹⁰ = [X] [X]

1.8x10⁻¹⁰ = X²

X =

Answer:

Covalent compounds.

Explanation:

Hello,

In this case, when forming chemical bonds in order to form compounds, we say that if electrons are shared, covalent compounds are to be formed and they usually have subscripts that need prefixes to be named, for instance phosphorous pentachloride (PCl5), dichlorine heptoxide (Cl2O7), carbon tetrachloride (CCl4) and many others.

Regards.

Answer:

Cl₂ ⟶ 2Cl⁻ + 2e⁻ Oxidation

2Li + 2e⁻ ⟶ 2Li⁺ Reduction

Explanation:

The question requests to split the equation below into half equations;

Cl2+2Li⟶2LiCl

In redox chemistry, splitting into half equations simply means highlighting the reduction and oxidation reactions of the reaction.

Before proceeding, we hav to split the ionic compound; LiCl into it's component ions. So we have;

Cl₂ + 2Li ⟶ 2Li⁺Cl⁻

This leaves us with;

Cl₂ ⟶ 2Cl⁻ ............................... i

2Li ⟶ 2Li⁺  .............................. ii

Oxidation reactions can be identified by the increase in oxidation number and decrease in the case of reduction.

in reaction i, there is a decrease in oxidation number from 0 to -1. This is the reduction half equation,

in reaction ii, there is an increase in oxidation number from 0 to +1. This is the oxidation half equation

In terms of electrons, we have to even the charge;

Oxidation = Loss of electrons

Reduction = Gain of electrons

The half equations are given as;

Cl₂ ⟶ 2Cl⁻ + 2e⁻ Oxidation

2Li + 2e⁻ ⟶ 2Li⁺ Reduction

Separation of the redox reaction into its balanced component half-reactions is as follows;

Oxidation half-reaction;

Reduction half-reaction;

By definition;

Oxidation is simply characterized by an increase in oxidation number of a reacting entity.

Reduction is simply characterized by a decrease in oxidation number of a reacting entity.

A reaction in which oxidation and reduction occur simultaneously is termed a Redox reaction as in the given reaction.

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Answer:

diffraction

Explanation:

Louis de Broglie in 1924 proposed the idea of wave particle duality. In his proposition, matter could exhibit wavelike properties.

Electrons are generally regarded as particles. Electrons may also display wavelike properties such as diffraction patterns. Generally, diffraction is regarded as a property of waves.

Hence, electron diffraction effect owes to the wavelike nature of of electrons when they are passed near matter.

Answer:

the answer will be B.1 and 3 only

Explanation:1. O2- means it has in total 10 electrons as atomic no of oxygen is 8

again F- has also 10 electrons as flourine has atomic no 9

and neon  also  has 10 electrons

incase of 3rd no S2- has 18 electrons as atomic no of sulfur is 16 as its an ion so it will take 2 more electrons

argon is an inert gas so it has 18 electrons and potassium has 19 electrons eliminating 1 we will 18

# incase of the second option lithium + contains 2 electrons and Na+ contains 10 and K+ contains 18 electrons so all of them are not same

ANS is 1 and 3 only(B.i and iii) only

thank u

Answer:

c2-c

Explanation:

A mental that lost two electrons just joking I don't k ow but I think it ain't the first o e because of comment sense to the 5th power

Answer:

A and B

Explanation:

metal form ion by lossing elecron

non metal form ion by gaining electron

Answer:enamel is formed by differentiated dental epithelial cells known as ameloblasts.

Explanation:

Answer:

Final temperature = 100°C

Explanation:

The general equation of a coffee cup calorimeter is:

Q = m×C×ΔT

Where Q is the heat added to the system (3.6kJ = 3600J), m is the mass of the substance (200.0g), C is specific heat (0.240J /g°C) and ΔT is change in temperature).

Computing the values:

3600J = 200.0g×0.240J/(g°C)×ΔT

75°C = ΔT

ΔT is the difference in temperature between final and initial temperature. As intital temperature was 25°C:

75°C = Final T - 25°C

Answer:

0.64 M

Explanation:

Step 1: Given data

Step 2: Calculate the final concentration of the dextrose solution

We want to prepare a diluted solution from a concentrated one. We can find the final concentration using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 1.0 M × 16 mL / 25 mL

C₂ = 0.64 M

Answer:

[tex]m_{Au_{31}}=6106.07 g[/tex]

Explanation:

Hello,

In this case, since one gold atom has a mass of 196.97 g, for the given gold cluster that has 31 gold atoms, the mass will be:

[tex]m_{Au_{31}}=31*196.97g\\\\m_{Au_{31}}=6106.07 g[/tex]

Whereas we consider the mass of all the 31 gold atoms to compute the overall mass.

Best regards.

Answer:

Explanation:

Atomic mass of Au = 197u

So, mass of 1 Au31 cluster (in u)

[tex]= 197 * 31\\\\= 6107u[/tex]

So, mass of 1 Au31 cluster (in gram)

[tex]= 6107 * 1.66*10^{-24}\\\\= 1.014*10^{-20}grams[/tex]

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Answer:

[tex]m_{MgCO_3}=0.0202molMgCO_3[/tex]

Explanation:

Hello,

In this case, for this purpose we first have to write the undergoing chemical reaction:

[tex]Na_2CO_3+Mg(NO_3)_2\rightarrow MgCO_3+2NaNO_3[/tex]

Thus, since the mole ratio between the reactants is 1:1, we next identify the limiting reactant by computing the available moles of sodium carbonate and those moles of the same reactant consumed by the magnesium nitrate considering the given solutions:

[tex]n_{Na_2CO_3}=0.010L*0.300\frac{molNa_2CO_3}{1L}=0.003molNa_2CO_3 \\\\n_{Na_2CO_3}^{consumed}=0.006L*0.0400\frac{molMg(NO_3)_2}{1L}*\frac{1molNa_2CO_3}{1molMg(NO_3)_2} =0.00024molNa_2CO_3[/tex]

In such a way, since less moles are consumed, we can say that the sodium carbonate is excess whereas the magnesium nitrate is the limiting one, therefore, the yielded mass of magnesium carbonate turns out:

[tex]m_{MgCO_3}=0.00024molMg(NO_3)_2*\frac{1molMgCO_3}{1molMg(NO_3)_2}*\frac{84.31gMgCO_3}{1molMgCO_3} \\\\m_{MgCO_3}=0.0202molMgCO_3[/tex]

Regards.

Answer:

B, adding an indicator to a solution to test its acidity

Explanation:

Answer:

B - Adding an indicator to a solution to test its acidity

Explanation:

Took the test

Answer:

Pressure of hydrogen gas = 695.2 mmHg

Explanation:

Given:

Water temperature = 22°C

Pressure inside the tube = 715 mmHg

Find:

Pressure of hydrogen gas

Computation:

Using vapor pressure of water table

Water pressure at 22°C = 19.8 mmHg

Pressure inside the tube = Pressure of hydrogen gas + Water pressure at 22°C

715 = Pressure of hydrogen gas + 19.8

Pressure of hydrogen gas = 715 - 19.8

Pressure of hydrogen gas = 695.2 mmHg

Answer:

The correct answer is : 0.025.

Explanation:

The precipitation reaction is as follows in this procedure:

NiSO4.6H20 (aq) + Na2CO3.10H2O (aq)⇒ NiCO3 (s) + Na2SO4 (aq) + 16H2O

So, the number of moles of reactants can be found by

number of moles : mass used/ molar mass

n (NiSO4×6H2O) = (6.57 g)/(262.84 g/mol) = 0.025 mol (for Ni)

n (Na2CO3×10H2O) = (7.15 g)/(286.14 g/mol) = 0.0250 mol (for CO3).

Thus, The maximum number of moles of NiCO3 that can form is 0.025 mol.

Answer:

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Explanation:

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False

Propyne can be converted to

propene by using H2+Lindlar's

catalyst ?

True

or False