What is the energy of a wave with a wavelength of 7.3E-9 m

The conditions for a standard electrochemical cell. select one or more : pressure of 1 atm temperature of 298 k solution concentrations of 1 M.

The electrochemical cell is the cell that is capable of generating the electrical energy from the chemical reactions or by the use of the electrical energy to cause the chemical reaction. The conditions for a standard electrochemical cell. select one or more : pressure of 1 atm temperature of 298 k solution concentrations of 1 M.  

There are the two types of the electrochemical cells is as follows : the galvanic called the electrolytic cells. the galvanic cell is also called as the voltaic cell.

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The balance equation for the complete combustion of pentane,C₅H₁₂ is

C₅H₁₂ (l) + 8 O₂ (g) ⇒ 5CO₂(g)  + 6H₂O(g). with the coefficient of oxygen gas after balance is 8.

The balance equation can be write as follows;

first we should  Understand the combustion of hydrocarbons. Pentane C₅H₁₂  is A hydrocarbon, or molecule with only carbon and hydrogen atoms. There are usually two predicted byproducts when we discuss the combustion of hydrocarbons: CO₂ and  6H₂O.

Remembering that "reaction with O2" refers to combustion, we may construct the following equation:

C₅H₁₂ (l) +O₂ (g) ⇒CO₂(g)  + H₂O(g)

Consequently, by counting the number of atoms in each side of the equation, we may find the equation's balance.

Right  side: 5 C atoms, 12 H atoms and 2 O atoms

Left : 1 C atom, 2 H atoms and 3 O atoms

By multiplying CO₂ by 5, we can start to create a balance:

C₅H₁₂ (l) +O₂ (g) ⇒ 5CO₂(g)  + H₂O(g)

We can now multiply  H₂O by 6 to obtain 12 H in the Left side:

C₅H₁₂ (l) +O₂ (g) ⇒ 5CO₂(g)  + 6 H₂O(g)

The LHS now has 16 O, it should be noted. Let's then multiply O₂ by 8:

C₅H₁₂ (l) + 8 O₂ (g) ⇒ 5CO₂(g)  + 6H₂O(g)

and we can see the coefficient of oxygen is 8

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If it starts off at 64 grams, there are just [tex]\frac{1}{4}[/tex] grams left after 9 days.

The half-life of a radioactive element is the amount of time it takes for it to decay to half of its original value. This means that a source's activity has a half-life when it takes a certain amount of time for it to decrease to half of what it was. These radioactive atoms release energy at a quantifiable pace known as radioactive decay to transform into new, distinct sorts of atoms.

[tex]r= \frac{1}{2}\\[/tex]

[tex]a_{1} = 64\\n=9[/tex]

[tex]a_{n}= a_{1}(r)^{n-1}[/tex]

[tex]a_{9}= 64(\frac{1}{2} )^{9-1}\\a_{9}= 64(\frac{1}{2} )^{8}\\a_{9}=64(\frac{1}{256} )\\a_{9} = \frac{64}{256} \\a_{9}= \frac{1}{4}[/tex]

[tex]a_{9} = \frac{1}{4}[/tex] grams

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Nuclear reactions are much more energetic than chemical reactions.

Nuclear reactions involve protons and neutrons in the nucleus. Chemical reactions involve electrons orbiting the nucleus. In an exothermic reaction, more energy is released than is used to break the bond of the reactants forming a bond in the product. Nuclear energy is millions of times more powerful than chemical energy.

An exothermic reaction involves an increase in the temperature of the reaction mixture. A chemical reaction that totally absorbs energy is called an endotherm. All chemical reactions require energy. However, like rust and combustion, not all chemical reactions release energy. Some chemical reactions absorb energy rather than release it.

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The stable isotopes have an infinite half-life. It's hard to get longer than infinite. There are 252 stable isotopes.

The longest known half-life is currently tellurium-128 at 7.7 × 10^24 years.

The longest directly measured half-life is presently xenon-124 with 1.8 × 10^22 years.

It is somehow possible that some isotopes that we currently think are stable might turn out to have a very long half-life, but we haven't figured it out yet.

For example, scientists used to think that bismuth-209 was stable.

In 2003 we discovered it is radioactive with a half-life of 2.01 × 10^19 years.

Stable isotopes are “stable" and therefore not radioactive. An atomic nucleus needs to be “unstable” for it to be radioactive.

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Answer:

Therefore, temperature is equals to 373 divided by 273, which comes out to be 1.37 density, is given as 0.65 gram per mel. Therefore, the volume of 3 ml liquid when evaporated is equals to 3 into 0.65, divided by molar mass multiplied by 22.4 multiplied by 1.37.

a) The number of moles of KCl formed is 0.51 moles.

b) The number of molecules of KCl is 1.38 * 10^22 molecules.

c) The mass of the potassium chlorate 8.94 g.

We know that we can be able to use the stoichiometry of the reaction to be able to find the amount of the KCl that is formed. The first step is to write down the reaction equation as it has been shown;

[tex]2KClO_{3} ----- > 2KCl + 3O_{2}[/tex].

We can see that we can be able to apply the principles of stoichiometry as stated earlier.

a) Number of moles of the potassium chlorate = 62 g/122.55 g/mol

= 0.51 moles

Then we have the number of the moles of the KCl as 0.51 moles.

b) Number of moles of the potassium chlorate =  2.85 g/122.55 g/mol

= 0.023 moles

If 2 moles of potassium chlorate produces 2 moles of KCl then 0.023 moles produces 0.023 moles of KCl.

Number of molecules of KCl = 0.023 moles * 6.02 * 10^23

= 1.38 * 10^22 molecules

c) If the number of moles of oxygen is; 3.54 g/32 g/mol = 0.11 moles

If 2 moles of potassium chlorate produces 3 moles of oxygen

x moles of potassium chlorate produces  0.11 moles

x = 2 * 0.11/3

= 0.073 moles

Mass of the potassium chlorate =  0.073 moles * 122.55 g/mol

= 8.94 g

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DKA results from a lack of blood sugar entry into your cells, which occurs when your body doesn't create enough insulin to meet your needs. Instead, a procedure known as lipolysis occurs in your liver when fat is changed into acids that produce ketone bodies lead to the development of ketoacidosis.

DKA is less frequent and less severe in patients with type 2 diabetes. It is frequently caused by protracted periods of uncontrolled blood sugar, forgotten medication doses, or a significant illness or infection.

There are several types of ketoacidoses that are clinically significant, including diabetic ketoacidosis (DKA), alcoholic ketoacidosis (AKA), and starvation ketoacidosis. DKA is an untreated diabetic complication that could be lethal.

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Yes it could, but you'd have to set up the process very carefully.

I see two major challenges right away:

1).  Displacement of water would not be a wise method, since rock salt

is soluble (dissolves) in water.  So as soon as you start lowering it into

your graduated cylinder full of water, its volume would immediately start

to decrease.  If you lowered it slowly enough, you might even measure

a volume close to zero, and when you pulled the string back out of the

water, there might be nothing left on the end of it.

So you would have to choose some other fluid besides water ... one in

which rock salt doesn't dissolve.  I don't know right now what that could

be.  You'd have to shop around and find one.

2).  Whatever fluid you did choose, it would also have to be less dense

than rock salt.  If it's more dense, then the rock salt just floats in it, and

never goes all the way under.  If that happens, then you have a tough

time measuring the total volume of the lump.

So the displacement method could perhaps be used, in principle, but

it would not be easy.

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In cranberry glass, an additive used to impact red coloration is Purple of Cassius.

The purple of Cassius is a pigment which can be prepared as follows;

When gold chloride reacts with tin(II) chloride or stannous chloride, a purple pigment is formed. This pigment is called Purple of Cassius.

If gold chloride and tin chloride are present in the solution, cranberry glass is formed.

The intensity of the red color depends upon the concentration of gold present.

This method is used to test the concentration of gold.

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The pressure of gas canister at 1250° Celsius is 312 atm which is more than the total internal pressure. Hence, it will explode.

Calculating the pressure of the volume of gas after being heated to 1250° Celsius. This will be done through ideal gas equation which is as follows -

PV = nRT, where P is pressure, V is volume, n is number of moles, R is gas constant and T is temperature.

Converting temperature to Kelvin

T = 1250 + 273

T = 1523 K

P = 3.5×0.082×1523/1.4

Performing multiplication on Right Hand Side of the equation

P = 437.1/1.4

Performing division on Right Hand Side of the equation

P = 312.2 atm

The gas canister will explode as the pressure exceeds.

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There are 2.66 x 10⁹ years old rocks.

Solution:

Since the formation rock U-238 gradually converts to Pb-206.

The half-life for the conversion of U-238 to Pb-206 is, t1/2 = 4.50 x 109 years

Now t1/2 = 0.693 / k = 4.50 x 10⁹ years

= k = 0.693 / 4.50 x 109 years  = 1.54 x 10-10 year-1

Given the mass of Pb-206 formed = 0.438

Hence moles of Pb-206 formed = mass / molecular mass = 0.438 / 206 = 0.0021262 mol Pb

The radioactive disintegration reaction is

U-238  ---- > Pb-206

1 mol ---------1 mol

1 moles of U-238 forms 1 mol of Pb-206.

Hence 0.0021262 mol Pb that would be formed by the moles of U-238  = 0.0021262 mol U-238

Hence mass of U-238 = (0.0021262 mol U-238) x (238 g/mol U-238) = 0.506 g U-238

Hence initial amount of U-238, N0 = 1.506 g

Current amount of U-238, Nt = 1.00 g

Now applying the integrated equation for first-order reaction

ln(N0/Nt) = kt

= t = (1/k) x ln(N0/Nt) = (1 / 1.54 x 10-10 year-1) x ln(1.506 g / 1.00 g) = 2.66 x 10⁹ years.

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The pH will fall below 7.00. Ammonia makes up 0.050 moles. In water, a polar solvent, an ionic solute dissolves because the polar water molecules draw the ions into solution, where they hydrate.

This means that equal amounts of moles of sodium hydroxide and hydrochloric acid are needed for a full neutralization to produce a neutral solution, or a solution with a pH of 7 at ambient temperature. As a result, a substance's molar mass, also known as its molecular mass or Mm, is equal to its atomic or molecular weight stated in grams. Determine the pH of a solution that contains 0.20 M ammonium chloride and 0.10 M aqueous ammonia. As you are aware, sodium hydroxide and hydrochloric acid have a mole ratio of 1:1 for neutralization.

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NO+: Nitrogen mono-oxide ion is the exception.

The octet rule refers to the characteristic  feature of any atom to have preference for possessing eight electrons in the valence shell or outer shell.

RnCl2: Radon Chloride

The central Rn atom is bound to 2 Cl atom and itself has two lone pairs. Hence the number of electrons in valance shell is: (2*2) +(2*2) =8; It fulfils Octet rule.

CF4: Carbon tetrafluoride

The central Carbon atom is bound four F atoms. Hence the number of electrons in valance shell is: (2*4) = 8: Fulfils Octet rule.

N2: Nitrogen

N atom is bound to another N atom through triple bond and the substance also have one lone pair. Hence the number of electrons in valance shell is: (3*2) + 2 =8: Fulfils octet rule.

NO+: Nitrogen mono-oxide ion

In case the ion stays in a form of N+ bound to O atom through a double bond, the nitrogen molecule here would have six molecules in valance shall and would not obey Octet rule.

But in case the substance stays as N having a triple bond with O+, both the atom having 8 electrons in valance shell would fulfil octet rule.

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Is the highest force between molecules Option a) water. hydrogen bonds are responsible for holding nucleotide bases together in DNA and RNA.

When a hydrogen atom is bound to either an oxygen, nitrogen, or fluorine atom, a peculiar type of dipole-dipole interaction takes place. In another molecule, the partially negative end of the oxygen, nitrogen, or fluorine is drawn to the partially positive end of the hydrogen atom.

A significant amount of energy is needed to break a hydrogen bond since it is a rather powerful force of attraction between molecules. This explains why substances like hydrogen fluoride (H, F) and water, H2O, have melting and boiling temperatures that are incredibly high. For instance, hydrogen bonds hold the nucleotide bases in DNA and RNA together, which is a crucial function of hydrogen bonds in biology.

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An unsusual dining experience I had was at an eatery that provided a live band as well as had wel dressed security personnel who danced to wecome their customers.

The restauarant stood out for me because of the quality of the food and the entertainment. Hence, I decised to alway visit the restaurant anytime I visist the area.

The additional tasks include:

Diners are individuals who are eating out in a restaurant or eatery.

Diners are usually out to get the best dining experience.

In recent times,  Diners are not only interested in a good meal these days; they often want to enjoy entertainment as well.

Hence, eateries or restuarants are now seeking out new ways to provide entertainment to their customers.

They do so by providing the best of meals in great environments as well providing an accompanying entertainment.

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90 g CH₃NH₃Cl should be dissolved in the solution to turn it into a buffer with pH.

90 g CH₃NH₃Cl should be used by the student to dissolve in the solution and to turn methylamine into a buffer with pH.

A 1.70 M of methylamine solution, weighing 450 mL, is delivered to a graduate student in chemistry.

Methylamine (weak base) has a Kb value of 4.4 x 104.

To create a buffer solution with a pH of 10.40 out of the methylamine solution, we should know how much CH₃NH₃Cl should the student dissolve in it.

We will use the Henderson-Hasselbach equation:

pH = pKa + log

So first we use Kb to calculate Ka and then calculate pKa:

Ka = Kw/Kb

⇒ Ka = 1x10⁻¹⁴/4.4x10⁻⁴ = 2.27x10⁻¹¹

pKa = -log(Ka) = 10.64

Now we can calculate the concentration of the salt, CH₃NH₃Cl:

pH = pKa + log (methylamine/salt)

10.40 = 10.64 + log (1.7/salt)

-0.24 = log (1.7/salt)

10⁻²⁴ = log (1.7/salt)

[Salt] = 2.95 M

Now, using the final volume and CH₃NH₃Cl's molecular weight, we can determine its mass and moles by,

450 mL ⇒ 450/1000 = 0.450 L

2.95 M × 0.450 L = 1.3275 mol CH₃NH₃Cl

1.3275 mol CH₃NH₃Cl × 67.45 g/mol = 89.54 g CH₃NH₃Cl

After rounding off to 2 significant digits it becomes 90 g CH₃NH₃Cl

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(2R, 3R, 4S) - 2,3,4 - trichloro pentane and (2R,3R,4R)- 2,3,4- trichloro heptane both pairs are Enantiomers which exactly have the same connectivity byt opposite three- dimensional space.

The Enantiomers are non-superimposable mirror images of each other. In a mirror view absolute configuration R becomes S and S becomes R. This can be observed in the names of the two given isomers. Also, all other positions and basic skeleton structure of the compound is same in both the isomers. Two compounds with the exact same connectivity, that are mirror images of each other but that are not identical to each other. Enantiomers contain no mirror planes. Enantiomers do not contain two equal and opposite halves. Enantiomers are another example of a type of stereoisomers. Two enantiomers have identical physical properties, except for optical rotation.

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So in this question, we have to explain two ways through which food is changed as it passes through the digestive system.

So first of all, food is taken inside our body with the help of our mouth. Once the food enters our mouth, the food gets dissolved in our saliva. Once the food is dissolved in our saliva, it enters the esophagus (which is also called the food pipe ) . From the esophagus, it goes into the stomach. In our stomach, dilute hydrochloric acid is present, which kills all the bacteria present in our food. The food stays in our stomach for 2-3 hours.It then goes into the small intestine which absorbs all the vital and important nutrients like protein, vitamins, and carbohydrates from the food. It then enters the large intestine and after the large intestine, it reaches the anus from where it is excreted.

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478 cal of heat is applied to a glass plate with a mass of 90 g, and the temperature of the plate increases will be 26.6 g⁻¹.°C⁻¹

The amount of heat applied = 478 cal

The specific heat capacity of glass = 0.2 cal g⁻¹.°C⁻¹

Mass of glass = 90 g

Temperature increased ΔT = ?

We will calculate the change in temperature by using the following equation

    Q = mcΔT

Rearrange the equation for ΔT

    ΔT = Q / mc

    ΔT = 478 cal / 90 * 0.2 cal g⁻¹.°C⁻¹

   ΔT =  478 cal / 18 g⁻¹.°C⁻¹

   ΔT = 26.6 g⁻¹.°C⁻¹

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If 4 moles of zinc hydroxide react, The reaction consumes2 moles of sulfuric acid The reaction produces _2_ moles of zinc sulfate and _2__ moles of water

Zinc sulfate is the inorganic compound with the components ZnSO4 and traditionally recognised as "white vitriol".

It is at the World Health Organization's List of Essential Medicines, a listing of the maximum vital remedy wanted in a simple fitness system.neutralization reactionThe given response is a neutralization response. Zinc oxide (base) reacts with sulfuric acid (acid) to shape zinc sulfate (salt) and water.

When sulfuric acid reacts with zinc hydroxide, zinc sulfate and water are produced. The balanced equation for this response is: H2SO4(aq)+Zn(OH)2(s)=ZnSO4(aq)+2H2O(1) If three moles of zinc hydroxide react, The response consumes _____ moles of sulfuric acid.

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Answer:

an article published by a university or a government agency

and

an article published in 2015 and updated a week ago

Explanation:

Both of these meet the criteria for a credible source. Cites with Advertisements, blogposts, and anonymous cites do not

The empirical formula is obtained by compositons: Cl2Zn for the empircal formula of a compound, given that the compound is found to be 47.9% zinc and 52.1% chlorine by mass

Calculate the molar mass of each element: Cl=35.453, Zn=65.409

Cl=1.469551236848, Zn=0.73231512482992 when expressed in moles.

Identify the least mole value: 0.73231512482992

Multiplying all elements by their lowest value Cl=2.0067197672451, Zn=1

rounding to the nearest whole number Cl=2, Zn=1

Cl[tex]_{2}[/tex]Zn for the empircal formula

The empirical formula, which is defined as the ratio of compositons of the least whole number of the components included in the formula, is the simplest formula for a compound compositons. The simplest formula is another name for it.

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Answer:

ammonium carbonate option e

Potassium hydroxide will have the greatest total ion concentration at 25°C. The correct option is (C)

The total ion concentration is the sum of the molar concentration of all the ions in a solution. A solution of a compound will produce ions when dissolved in water, and the sum of the molar concentration of all the ions is known as the total ion concentration.

Potassium hydroxide (KOH) is a strong base that fully dissociates in water to produce potassium ions (K+) and hydroxide ions (OH-).So, if we add 1.0 mole of KOH to 1.0 L of water, we will obtain a 1.0 molar solution of KOH in which the molar concentration of K+ and OH- will both be 1.0 M.

On the other hand, ammonium carbonate, calcium phosphate, potassium chloride, and iron (II) nitrate will produce fewer ions in solution since they do not fully dissociate, so their total ion concentration will be lower.

Hence, we can say that, when 1.0 mole of the following compounds is added to 1.0 L of water, potassium hydroxide will have the greatest total ion concentration at 25°C.

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The Ground state electronic configuration of  Cr2+:[Ar] 3d4, Cu2+: [Ar] 3d9, Co3+: [Ar] 3d6

Electronic configuration, sometimes referred to as electronic structure or electron configuration, is the arrangement of electrons in orbitals that surround an atomic nucleus.

1. Cr2+

Atomic no.of Chromium(Cr)=24  

a ground state Chromium's electronic structure is as follows: [Ar] 4s1 3d5 = 1s2 2s2 2p6 3s2 3p6

(Atomic no.of [Ar](argon) = 18)

so,ground state electronic configuration for Cr2+:

[Ar] 4s0 3d4 =[Ar] 3d4

so, ground state electronic configuration of Cr2+:

[Ar] 3d4

2. Cu2+

Atomic number of Copper(Cu)= 29

Ground state electronic configuration of Cu:

[Ar] 4s1 3d10

so, ground state electronic configuration of Cu2+:

[Ar] 4s0 3d9 = [Ar] 3d9

so,electronic configuration of Cu2+:

[Ar] 3d9

3.Co3+

Atomic number of Cobalt(Co)=27

Ground state electronic configuration of Co:

[Ar] 4s2 3d7

so,electronic configuration of Co3+:

[Ar] 4s0 3d6 = [Ar] 3d6

So,electronic configuration of Co3+:

[Ar] 3d6

Therefore the Ground state electronic configuration of

Cr2+:[Ar] 3d4, Cu2+: [Ar] 3d9, Co3+: [Ar] 3d6

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The formula for pH at a weak acid-strong base titration's equivalence point must be used to determine the pH at each stage of the titration:

[tex]pH = pK_a + log \frac{[A^-]}{[HA]}[/tex]

where pK a is the weak acid's acid dissociation constant, [A-] is the concentration of the acid's conjugate base, and [HA] is the acid's concentration.

The weak acid (formic acid) is at a concentration of 0.100 M at the start of the titration (0 mL of additional base), and the conjugate base (formate ion) is at a concentration of 0.00 M, resulting in a pH of:

[tex]pH = pK_a + log \frac{[A^-]}{[HA]} [/tex]

[tex]= pK_a + log \frac{0}{0.100} [/tex]

= pK_a + log(0)

We are unable to determine the pH at this time since the logarithm of 0 is undefinable.

The weak acid concentration falls as the titration goes along and more base is added, while the conjugate base concentration rises. The concentration of the weak acid is 15.0 mL when 15.0 mL of additional base is used.

[tex]0.100 M - \frac{0.200 M \times15 mL }{50 mL} [/tex]

= 0.080 M,

and the concentration of the conjugate base is

[tex] \frac{0.200 M \times 15 mL}{50 mL} [/tex]

= 0.060 M. The pH at this point is:

[tex]pH = pK_a + log \frac{[A^-]}{[HA]}[/tex]

[tex]= pK_a + log \frac{0.060}{0.080} [/tex]

= pK_a + log(0.75)

= pK_a + (-0.122)

You may use the same formula and the appropriate numbers for [A-] and [HA] to determine the pH at the other points in the titration.

It should be noted that the pH at the equivalence point (when the ratio of acid to base is equal) will be the same as the weak acid's pK a. Since the concentration of the weak acid is zero at the equivalence point, the pH is:

[tex]pH = pK_a + log \frac{[A^-]}{[HA]}[/tex]

[tex]= pK_a + log \frac{[A^-]}{0}[/tex]

[tex]= pK_a + log([A^-])[/tex]

The pH at the equivalence point will be lower than the pK a of the weak acid since the logarithm of a number less than 1 has a negative value. Beyond the equivalence point, adding more base will cause the pH to continue to drop.

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The empirical formula for the given compound is Al₂Se₃.

Step 1: Write the data given

Mass of the sample of Aluminium = 7.300 grams

Mass of the definite compound = 39.35 grams

Molar mass of the Aluminium = 26.98 g/mol

Molar mass of the Selenium = 78.96 g/mol

Step 2: Calculate the mass of selenium

Mass of selenium = mass of compound - mass of aluminium

Mass of selenium = 39.35 - 7.3 = 32.05 grams

Step 3: Calculate the number moles of Al

Moles Al = Mass Al/ Molar mass Al

Moles Al = 7.300 grams / 26.98 g/mol

Moles Al = 0.2706 moles

Step 4: Calculate the number moles of Se

Moles Se = Mass Se / Molar mass Se

Moles Se =32.05 g / 78.96 g/mol

Moles Se = 0.4059 moles

Step 5: Divide through the smallest amount of moles

Aluminium: 0.2706 / 0.2706 = 1

Selenium: 0.4059/0.2706 = 1.5

This means for each moles of aluminium, we have 1.5 moles of selenium.

For each 2 moles of aluminium, we have 3 moles of selenium

The empirical formula is Al₂Se₃  

Hence, the compound is aluminium(III) selenide.

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The periodic table would element x be found is 1.

Atomic number or number of protons in an atom. The number of neutrons and the number of electrons is often the same as the number of protons but varies from atom to atom.

The current periodic table has 18 vertical columns called groups, arranged from left to right, and 7 horizontal rows called periods. Elements of this group form salts. They are noble gases and inert under normal conditions. It is called the periodic table because of the arrangement of the elements. You'll notice they are in rows and columns.

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Ag (aq) → Ag⁺ (aq) + 1 e⁻ at the anode and NO₃ (aq) + 1e⁻ → NO₃⁻ at the cathode.

The half cell reaction are as follows:

In the anode oxidation always occurs, during the oxidation the reducing agent loses electrons, and its oxidation number increases.

NO₃ (aq) + 1e⁻ → NO₃⁻

In the cathode reduction always occurs, during the reduction the oxidizing agent gains electrons, and its oxidation number decreases.

Ag (aq) → Ag⁺ (aq) + 1 e⁻

Hence, Ag (aq) → Ag⁺ (aq) + 1 e⁻ occurs at the anode and NO₃ (aq) + 1e⁻ → NO₃⁻ occurs at the cathode.

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