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What mass of glucose, C6H12O6, is needed to make 500.0 mL of a 0.200M
solution?

Answer:

48 molecules

Explanation:

From the given reaction:

[tex]\mathtt{P_4 + 6Cl_2\to 4PCl_3}[/tex]

i.e. 1 mole of Phosphorus react 6 moles of chlorine to yield 4 moles of PCl₃

This implies that for each P₄ molecule, we will require 6 molecules of Cl₂

We are provided with 15 molecules of P₄ and 42 molecules of Cl₂

Suppose we utilized the whole 15 molecules of P₄, we will require:

= 15 × 6 = 90 moles of Cl₂ which is not readily available except 42 are available.

If we utilized 42 molecules of Cl₂, we will require 42/6 = 7 molecules of P₄ which is readily available.

From this analysis, we can opine that Cl₂ serves as the limiting agent

∴

P₄ : Cl₂ : PCl₃ = 1 : 6 : 4 = 7 : 42 : 48

It implies that only 48 molecules of phosphorus trichloride will be formed.

Answer:

Equation 1 - nuclear fission

Equation 2 - nuclear fusion

Explanation:

Nuclear fission is a reaction in which a large nucleus is split into smaller nuclei when it is bombarded by neutrons. The process produces more neutrons to continue the chain reaction. This is clearly depicted in equation 1 as shown in the question.

Nuclear fusion is a reaction in which two light nuclei combine in order to form a larger nuclei. This is clearly depicted in equation 2 as shown in the question.

In the first reaction, a neutron is released, and in the second a helium atom is released. The given two equations represent nuclear fission and fusion.

A nuclear reaction is a reaction that involves the nuclei of the atom and the absorption and release of energy. In the first reaction, a big nucleus is split by the neutron bombardment into smaller nuclei.

In the second reaction the process of nuclear fusion, two nuclei combine into a single larger nucleus that is shown as:

₁¹H+ ²₁H → ³₂He

Therefore, nuclear fission and fusion are represented by each of these equations.

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Explanation:

Balanced Equation

P

2

O

3

+

3H

2

O

→

2H

3

PO

3

First use the balanced equation to determine the mole ratio between

H

3

PO

3

and

P

2

O

3

. This ratio will be used to determine the moles

P

2

O

3

required to produce

10.2

moles

H

3

PO

3

.

Mole Ratio Between

P

2

O

3

and

H

3

PO

3

from the balanced equation.

1

mol P

2

O

3

2

mol H

3

PO

3

and

2

mol H

3

PO

3

1

mol P

2

O

3

Multiply the moles

H

3

PO

3

by the molar mass that cancels

H

3

PO

3

and leaves

P

2

O

3

10.2

mol H

3

PO

3

×

1

mol P

2

O

3

2

mol H

3

PO

3

=

5.10 mol P

2

O

3

Now that the moles

P

2

O

3

required to produce

10.2 mol H

3

PO

3

are known, multiply the number of moles by its molar mass,

109.945 g/mol

. https://www.ncbi.nlm.nih.gov/pccompound?term=P2O3 This will give the mass in grams needed for

P

2

O

3

to produce

10.2 mol H

3

PO

3

.

5.10

mol P

2

O

3

×

109.945

g P

2

O

3

1

mol P

2

O

3

=

561 g P

2

O

3

rounded to three significant figures

Answer:

Br

Explanation:

because bromine is more reactive as reactivity increases on moving from left to right in p-block. hope this make sense :)

Answer:

he a real G

Explanation:

og tripple og

Explanation:

All atoms of one type were identical in mass and properties.

Answer:

M1 = 6.0M

V1 = 8.33mL

M2 = 1.0M

V2 = 50.0mL

Explanation:

It is possible to find the volume that we need to prepare a diluted solution from a stock using the equation:

M1V1 = M2V2

Where M is molar concentration and V volume of 1, stock solution and 2, diluted solution.

Replacing:

M1 = 6.0M

V1 = Our incognite

M2 = 1.0M

V2 = 50.0mL

6.0M*V1 = 1.0M*50.0mL

V1 = 1.0M * 50.0mL / 6.0M

V1 = 8.33mL of the stock solution are needed

Answer:

Mass of the ball is . 42 kg. The density of air is 1.2 kg/m3.

Explanation:

Answer:

electron shells

Explanation:

Eletrons are on the outeermost part of the atom called electron shells.

Explanation:

the enerfy of of one photon of this light is 4.85x10^-19 J

E= 4.85x10^-19 J

Answer:

Hydrogen sulfide is used primarily to produce sulfuric acid and sulfur. It is also used to create a variety of inorganic sulfides used to create pesticides, leather, dyes, and pharmaceuticals. Hydrogen sulfide is used to produce heavy water for nuclear power plants (like CANDU reactors specifically).

Explanation:

Sana Po it's help

Answer: 19

Explanation:

In order to write the Potassium electron configuration we first need to know the number of electrons for the K atom (there are 19 electrons). When we write the configuration we'll put all 19 electrons in orbitals around the nucleus of the Potassium atom.

Answer:

8.59

⋅

10

17

Explanation:

You can start by figuring out the energy of a single photon of wavelength  

505 nm

=

505

⋅

10

−

9

m

.

To do that, use the equation

E

=

h

⋅

c

λ

Here

h

is Planck's constant, equal to  

6.626

⋅

10

−

34

.

J s

c

is the speed of light in a vacuum, usually given as  

3

⋅

10

8

.

m s

−

1

λ

is the wavelength of the photon, expressed in meters

Plug in your value to find--notice that the wavelength of the photon must be expressed in meters in order for it to work here.

E

=

6.626

⋅

10

−

34

J

s

⋅

3

⋅

10

8

m

s

−

1

505

⋅

10

−

9

m

E

=

3.936

⋅

10

−

19

J

So, you know that one photon of this wavelength has an energy of  

3.936

⋅

10

−

19

J

and that your laser pulse produces a total of  

0.338 J

of energy, so all that you need to do now is to find how many photons are needed to get the energy output given to you.

0.338

J

⋅

1 photon

3.936

⋅

10

−

19

J

=

8.59

⋅

10

17

photons

−−−−−−−−−−−−−−−−−

The answer is rounded to three sig figs.

Answer:a) organic

Explanation:

Answer:

Explanation:

25.71 × 10²³ molecules are there in 4.27 mol of tungsten (VI) oxide, [tex]WO_3[/tex].

To find the Number of molecules of substance multiply the number of moles of given substance by Avogadro number

Number of molecules = Number of moles × Avogadro's Number

Avogadro's number [tex](N_{A})[/tex] is the number of particles in one mole of substance. 6.023 × 10²³ is known as Avogadro's constant / Avogadro's number.

Avogadro's number = 6.023 × 10²³ [tex]mol^{-}[/tex]

Now,

Number of molecules in 4.27 moles of [tex]WO_{3}[/tex]

= Number of moles of [tex]WO_{3}[/tex] × Avogadro's Number

= 4.27 × 6.023 × 10²³

= 25.71 × 10²³ molecules

Thus, we can say that 25.71 × 10²³ molecules are there in 4.27 moles of tungsten (VI) oxide, [tex]WO_3[/tex].

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Answer:

b

Explanation:

because it be an the only answer there

Answer:

□In field investigations, you are usually working with much larger animals than in the lab. In field investigations, it is much more difficult to separate your control and experimental groups.

There are many ways in which a field  experiment can differ form of a classroom or a laboratory experiment,  one of which can be the environmental condition.

 A field experiment Is an experiment  which is performed in practically Outside the classroom or a laboratory in live situtations .

Environmental factors may deter the results  of a field experiment or  may alter the physical conditions of the object, For example  an anhydrous substance may become  hygroscopic, Which may  Alter the result of the field experiment.

  Hence,  field experiment can differ from  a classroom or a laboratory experiment  due to environmental conditions .

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Answer:

1. Multimeter

2. Cathode

3. Capacitor

4.  Selector switch

5. Short or Shorted

6. Resistors

7. Resistor color coding

8. Open Capacitor

9. Step-up transformer

10. Step-down transformer

Explanation:

The above-described elements are electronic components. Resistors for instance are designed to resist the flow of electric current. They are also standardized such that a deviation from the set resistance level will indicate a problem. The capacitor is another electrical component that stores energy as an electrical charge. Knowledge of these electrical components and the ways they are tested will make a person proficient in electrical electronics.

Answer:

Check the image above

explanation:

When naming organic compounds based on IUPAC; we take note of functional group, position of functional group.

In 2,3,4-trimethyl-3-heptanol, the functional group is hydroxyl group ( OH ). It is on position 3 (2,3,4-trimethyl-3-heptanol. Then we put it on the third carbon. Another functional group is methyl group, with three positions, 2, 3, and 4.

In 4-ethyl-4-octanol, the functional group is hydroxyl group ( OH ) which is in position 4 on the fourth carbon. Another functional group is ethyl group in position 4 on the fourth carbon. In this case, the functional groups that have same position, are put on that same carbon.

Answer:

A. 1. Strong acid (sa): Hydrobromic acid: HBr (aq)

2. Strong acid (sa); Hydrochloric acid: HCl (aq)

3. Weak acid (wa); Carbonic acid: H₂CO₃ (aq)

B. H+ (aq) + OH- (aq) ----> H₂O (l)

Explanation:

Strong acids are which ionize completely in aqueous solution into hydrogen ions and the corresponding anion. Examples of strong acids include hydrobromic acid, hydrochloric acid, tetraoxosulfate (vi) acid.

The ionization of hdyrobromic and hydrochloric acids in aqueous solution is given below:

1. Hydrobromic acid: HBr (aq) ----> H+ (aq) + Br- (aq)

Hydrobromic acid in aqueous solution ionizes completely into hydrogen ions and bromide ions

2. Hydrochloric acid: HCl (aq) ----> H+ (aq) + Cl- (aq)

Hydrochloric acid in aqueous solution ionizes completely into hydrogen ions and chloride ions

Weak acids are acids which ionizes only partially in aqueous solutions to hydrogen ions and the corresponding anions. Examples of weak acids are carbonic acid and ethanoic acid. The ionization of carbonic acid in aqueous solution is shown below:

3. Carbonic acid: H₂CO₃ (aq) ⇄ 2 H+ (aq) + CO₃²- (aq)

Carbonic acid ionizes partially only to give hydrogen ions and trioxocarbonate (iv) ions. The unionized acid exists in equilibrium with the ions produced by the partial ionization of the acid.

Part B:

The reaction between hydrochloric acid and barium hydroxide is a neutralization reaction producing barium chloride salt and water.

The net ionic equation of the neutralization reaction is given below :

H+ (aq) + OH- (aq) ----> H₂O (l)

Answer:

When a wine bottle is uncorked, its contents are exposed to air, causing them to oxidize. ... Bacteria naturally present in grapes can turn either the sugars in grape juice or the alcohol in wine into acetic acid, giving it a vinegar taste (and eventually producing a wine vinegar).

Explanation:

Answer:

P4 + 4NaOH + 2H2O → 2PH3 + 2Na2HPO3

Explanation:

A chemical equation is said to be balanced if the quantity of each type of atom in the reaction is the same on both the reactant and product sides. In a balanced chemical equation, the mass and the charge are both equal.

A chemical equation must balance according to the rule of conservation of mass. According to the rule, mass cannot be generated or removed during a chemical process.

Chemical equations must be balanced, which means that the atom types and numbers on both sides of the reaction arrow must match. Coefficients are the values added in front of formulas to balance equations; they multiply each atom in a formula.

Here the given equation is balanced as:

P₄ + 4NaOH + 2H₂O——> 2PH₃ + 2Na₂HPO₃

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Answer:

b. changing the compound changes the absorbance behavior.

Explanation:

Answer:

did it give you answers to in blanks

Answer:

D

Explanation:

Answer:

Emerald is a compound made of Beryllium, Aluminium, Silicon and Oxygen with the formula, Be3Al2(SiO3)6

Explanation:

Answer:

A pure substance because its a compound

Explanation:

Answer:

∑Q(TTL) = 347 Kj. = 82.8 Kcal. (3 sig. figs.)

Explanation:

Draw a typical heating curve for water (see figure below) and label each section with data needed to calculate the amount of heat flow in the specified section.  

Super-imposing given data onto a trace of a typical heat flow chart for water finds there are three segments of the heating curve that can utilize the given data to calculate heat flow for each segment. With all three heat flow quantities, the total heat quantity associated with this problem is marked in orange. Calculate the heat quantity associated with the designated segments and add to obtain total heat flow for the transitions listed.  

Point E to D => Q₄ = m∙ΔHᵥ = (115g)(540 cal/g) = 62,100 cals.

Point D to C => Q₃ = m∙c∙ΔT = (115g)(1 cal/g∙˚C)(100˚C) = 11,500 cals.

Point C to B => Q₂ = m∙ΔHₓ = (115g)(80cal/g) = 9,200 cals.

Total Heat Flow (∑Qₙ) = Q₄ + Q₃ + Q₂ = 62,100 cals.  +  11,500 cals.  +  9,200 cals. = 82,800 cals.

= (82,800 cals.)(4.184 joules/cal.) = 346,435 joules = 346.435 Kj ~ 347 Kj (3 sig. figs.) _____________________________________________________

m = mass = 115g

c = specific heat of liquid water = 1 calorie/gram·°C = 4.184 j/g·°C

ΔT = temperature change in degrees Celsius

ΔHᵥ = heat of vaporization = 540 cals./gram

ΔHₓ = heat of crystallization = 80 cal./gram

Qₙ = heat flow quantity per specific segment (calories or joules)

∑Qₙ = total heat flow

Heating curve for water:

Note in the diagram that only two formulas are used.

Q = m·c·ΔT => heating or cooling the pure condensed state. The segments demonstrate temperature change.

Q = m·ΔHₙ => heat flow during phase change. Note, in these segments of the heating curve two phases are in contact. That is solid/liquid or liquid/gas phase substances. Also note, in these segments, when two phases are in contact no temperature change occurs. Examples, ice water remains at a constant temperature until all ice is melted or all of the liquid water is frozen depending upon the direction of heat flow. The same is true for boiling water in that when two phases are in contact (liquid/gas), temperature remains constant. The portions of the heating curve designating phase transitions are horizontal and are defined by the equation Q = m·ΔH,  while the curve segments that are only one phase demonstrate temperature change and are defined by the equation containing temperature change, Q = m·c·ΔT.  

The total energy released during this whole process was 346.4 kiloJoules.

Explanation:

Given:

115 g of steam at 100°C condenses, cools, and falls as snow at 0.0°C.

To find:

The energy released during this whole process.

Solution:

In the given three stages will appear:

1) [tex]H_2O(g)\rightarrow H_2O(l) (at 100^oC)[/tex]

Phase change will occur at 100°C, that is from gas to liquid.

Energy released during the phase change of water from gas to liquid =[tex]Q_1[/tex]

Latent heat of vaporization of water =[tex]\Delta H_{vap}=2.260 kJ/g[/tex]

Latent heat of condensation of water =[tex]\Delta H_{con}=- \Delta H_{vap}=2.260 kJ/g=-2.260 kJ/g[/tex]

Mass of steam = 115 g

[tex]Q_1=115 g\times \Delta H_{con}\\=115g\times -2.260 kJ/g=-259.9 kJ[/tex]

2) [tex]H_2O(l)(at 100^oC)\rightarrow H_2O(l) (at 0.0^oC)[/tex]

Mass of water = m = 115 g ( steam converted into water)

Initial temperature of water = [tex]T_i=100^oC[/tex]

Final temperature of water =[tex]T_f=0.0^oC[/tex]

The specific heat of water = c = 8.186J/g°C

Energy released during this stage = [tex]Q_2[/tex]

[tex]Q_2=m\times c\times (T_f-T_i)\\=115g\times 4.186 J/g^oC\times (0.0^oC-100^oC)\\=-48,139 J= -48.139 kJ\\(1 J=0.001kJ)[/tex]

3)[tex]H_2O(l)(at 0.0^oC)\rightarrow H_2O(s) (at 0.0^oC)[/tex]

Water freezes at 0°C ,phase change will occur at 0.0°C, that is from liquid to solid.

Energy released during the phase change of water from liquid to solid=[tex]Q_3[/tex]

Latent heat of fusion of water =[tex]\Delta H_{fus}=-334J/g[/tex]

Mass of water = 115 g

Energy released during this stage:

[tex]Q_3=115g\times \Delta H_{fus}\\115\times -334 J/g=-38,410 J\\=-38,410 J=-38.410 kJ\\(1 J=0.001kJ)[/tex]

Total energy released during this whole process = Q

[tex]Q = Q_1+Q_2+Q_3\\Q=(-259.9 kJ)+( -48.139 kJ)+(-38.410 kJ)\\=-346.449 kJ\approx -346.4 kJ[/tex]

(negative sign indicates that heat energy is released))

The total energy released during this whole process was 346.4 kiloJoules.

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Answer:

C.

Explanation:

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