Draw the Lewis structure for the polyatomic trisulfide anion. Be sure to include all resonance structures that satisfy the octet rule.

Answer:

The type of orbital hybridization around the C atom is sp3 hybridization.

Explanation:

The electron configuration of carbon is 1s² 2s² 2p². According to this configuration, carbon only has two semi-occupied orbitals that can form a covalent bond.

But a process called hybridization occurs, where the "s" and "p" orbitals at the (2s² 2p²) level combine and generate new orbitals. So this hybridization process where one "s" orbital is combined with three "p" orbitals forms four new sp³ hybrid orbitals (remember that there is 1 s orbital and 3 p orbitals for each layer or energy level). These four hybrid orbitals are half full, having only one electron and being capable of forming four covalent bonds.

So, the type of orbital hybridization around the C atom is sp³ hybridization.

Answer:

2.459 V

Explanation:

If we look at the data we have from the question closely, we will discover that aluminum has a very negative reduction potential. Recall that very negative reduction potentials are associated with strong reducing agents.

Similarly, silver has a positive reduction potential signifying that it is an oxidizing agent. Since the reducing agent is oxidized in a redox reaction and the oxidizing agent is reduced.

Hence;

E°cell= E°reduction process - E°oxidation process

E°cell= 0.799 -(-1.66)

E°cell= 2.459 V

The standard cell potential is 2.459 V

The standard cell potential is the potential difference between the cathode and anode.

The overall cell potential can be calculated by using the equation: [tex]E_0cell=E_0red-E_0oxid[/tex].

Aluminum has a very negative reduction potential. The negative reduction potentials are associated with strong reducing agents and Silver has a positive reduction potential signifying that it is an oxidizing agent. Since the reducing agent is oxidized in a redox reaction and the oxidizing agent is reduced.

Hence;

E°cell= E°reduction process - E°oxidation process

E°cell= 0.799 -(-1.66)

E°cell= 2.459 V

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Answer:

Umbilicus can't be on the hand....it is located in the abdomen....

Answer:  a)   [tex]_{88}^{234}\textrm{Ra}\rightarrow _{86}^{230}\textrm{Ra}+_{2}^{4}\textrm{He}[/tex]

b)  [tex]_{92}^{238}\textrm{Ra}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}[/tex]

c)  [tex]_{92}^{238}\textrm{Ra}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}[/tex]

 d) [tex]_{6}^{14}\textrm{C}\rightarrow _{7}^{14}\textrm{N}+_{-1}^{0}{\beta}[/tex]

e) [tex]_{12}^{24}\textrm{Mg}\rightarrow _{12}^{24}\textrm{Mg}+_{0}^{0}{\gamma}[/tex]

Explanation:

A balanced nuclear equation is one in which the atomic number and mass number remains same on both sides of the equation i.e the number of protons and neutrons remain same.

General representation of an element is given as:  

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

a) [tex]_{88}^{234}\textrm{Ra}\rightarrow _{86}^{230}\textrm{Ra}+_{2}^{4}\textrm{He}[/tex]

b)  [tex]_{92}^{238}\textrm{U}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}[/tex]

c)  [tex]_{94}^{242}\textrm{Pu}\rightarrow _{92}^{238}\textrm{Ra}+_{2}^{4}\textrm{He}[/tex]

 d)   [tex]_{6}^{14}\textrm{C}\rightarrow _{7}^{14}\textrm{N}+_{-1}^{0}{\beta}[/tex]

e)   [tex]_{12}^{24}\textrm{Mg}\rightarrow _{12}^{24}\textrm{Mg}+_{0}^{0}{\gamma}[/tex]

Answer:

Square planar.

Explanation:

Hello,

In this case, it can be demonstrated that the central atom Xenon in XeF₄ has a sp³d² hybridization which means that its geometry is likely to be octahedral.  Nevertheless Applying by applying the valence shell electron pair repulsion (VSEPR) model to in order to minimize the repulsion among lone pairs, bond pairs and lone pair-bond pairs, we we realize repulsion is actually minimized when lone pairs are anti to one another and the fluorine atoms are in equatorial position, for that reason, the corrected and properly exhibited geometry or molecular arrangement of the compound turns out square planar. You can verify it on the attached picture.

Regards.

Answer:

Explanation:

The 1000-ml graduated cylinder with marks every 100 ml has least count of 100 ml . It means it can not measure a volume less than 100 ml . It can also measure volume in the integral multiple of 100 ml like 200 ml , 300 ml , 400 ml etc perfectly .  All these have significant figure of one .

It can not measure volume like 150 , 250 , 255 , 760 etc because these measurements are not the integral multiple of 100 . 750 is not a integral multiple of 100 so it can not measure volume of 750 ml.

Answer:

45000 atoms

Explanation:

a molecule has 45 atoms

1000 molecules have how many atoms??

by mathematical cress cross calculation we get that:

45 X 1000 = 45000 atoms

Answer:

Option B. pentan-1-ol

Explanation:

We'll begin by writing the name of the compound.

The name of the compound above is pentylpropanoate.

From the name of the compound and the structure, we can suggest the following equation:

CH3CH2COOH + HOCH2CH2CH2CH2CH3 —> CH3CH2COOCH2CH2CH2CH2CH3 + H2O

Thus,

Propanoic acid + pentan-1-ol —> pentylpropanoate + water.

Therefore, the alcohol used in the reaction is pentan-1-ol

Answer:

They produce ions when dissolved in water.

Explanation:

Acids and bases have the characteristic in common to each other. Both of them have the property of reacting and dissolving in the water. Both acids and bases lead to the production of the ions when they are placed in a water solution. Acids produce Hydrogen ions when they are dissolved in water. Bases produce hydroxide ion when they are dissolved in water.

Answer:

A

Explanation:

Answer:

[tex]Ksp=3.39x10^{-14}[/tex]

Explanation:

Hello,

In this case, since the dissociation of gold (V) oxalate is:

[tex]Au_2(C_2O_4)_5(s)\rightleftharpoons 2Au^{5+}(aq)+5(C_2O_4)^{2-}(aq)[/tex]

In such a way, the equilibrium expression is:

[tex]Ksp=[Au^{5+}]^2[(C_2O_4)^{2-}]^5[/tex]

Thus, since the molar solubility of the gold (V) oxalate is computed by considering its molar mass (834 g/mol):

[tex][Au_2(C_2O_4)_5]=2.58\frac{g}{L} *\frac{1mol}{834g} =3.09M[/tex]

In such a way, since gold (V) is in a 2:1 molar ratio with the salt and the oxalate in a 5:1 in the chemical reaction, the corresponding concentrations at equilibrium are:

[tex][Au^{5+}]=3.09x10^{-3}\frac{mol}{L} *\frac{2molAu^{5+}}{1mol} =6.19x10^{-3}M[/tex]

[tex][(C_2O_4)^{2-}]=3.09x10^{-3}\frac{mol}{L} *\frac{5mol(C_2O_4)^{2-}}{1mol} =0.0155M[/tex]

Therefore, the solubility product turns out:

[tex]Ksp=(6.19x10^{-3})^2*(0.0155)^5\\\\Ksp=3.39x10^{-14}[/tex]

Regards.

Answer:

the overall cell potential

Explanation:

We must bear in mind that the standard hydrogen electrode is a reference electrode whose electrode potential has been arbitrarily set at 0 V.

The standard hydrogen electrode consists of hydrogen ion solution and hydrogen gas together with a platinum electrode.

The overall cell potential is the reduction potential of the substance being determined using the standard hydrogen electrode as a reference electrode since its electrode potential is set at zero volts.

Answer:

The correct answer is A.

Explanation:

Helium's Atomic Model has 2 Protons and 2 Neutrons, and on the outside it has 2 electrons circling around it.

Answer:

is A

Explanation:

Answer:

Rate = k [OCl] [I]

Explanation:

OCI+r → or +CI

Experiment [OCI] M I(-M) Rate (M/s)2

1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3

2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3

3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3

4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3

The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.

In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.

In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.

The rate law is given as;

Rate = k [OCl] [I]

Answer:

True

Explanation:

Significant digits include non-zero digits (unless the zero is between two non-zero numbers)

In 12,785.000, there are 5 non-zero digits:

1 2 7 8 5

ANSWER: the answer is true.

Answer:

a. NaNO3 and CuCl2

Explanation:

The other 3 react as follows:

Ba(OH)2 and HNO3 ---> BaNO3 + H2O

CaCl2  + Na3PO4 --->  CaPO4 + NaCl

AgNO3 + HCl --->  AgCl +  HNO3

Answer:

Benzophenone

Explanation:

Given that:

an unknown compound’s semicarbazone melts at 162° -165°

The compound does not give a silver mirror in the Tollens’ test.

On reaction with KI/I2, NaOH/H2O no straw yellow precipitate forms.

The objective is to identify the unknown compound.

Since the unknown compound result to a semicarbazone, we can deduce that the unknown compound is a carbonyl compound. A carbonyl compound is either an aldehyde or ketone in nature. Also, the absence of  silver mirror in the Tollens’ test carried out in the reaction confirms that the compound is a ketone because  ketones will  never give a silver mirror in Tollens’ test.

Similarly, on reaction with KI/I2, NaOH/H2O no straw yellow precipitate forms,  that is an iodoform test. That implies that a keto-methyl group is absent in the unknown compound.

Finally, since the unknown compound melting point is between 162° -165° and Benzophenone semicarbazone  melting point is 164°.

We can conclude that the unknown  compound is Benzophenone semicarbazone  

Answer:

Sample Response: Both passive and active transport move substances into and out of the cell across the cell membrane. However, passive transport does not require the use of energy because molecules move from an area of high concentration to an area of low concentration. Active transport does require the use of energy because molecules move from an area of low concentration to an area of high concentration.

Explanation:

Active and passive transport are the two main biological processes which play an important role in providing the nutrients, oxygen, water and other essential molecules to the cells along with removing waste materials.

The active transport involves the movement of molecules across a membrane against a concentration gradient which requires energy. The passive transport is the movement of molecules or ions across the cell membrane without using energy.

The differences between active and passive transport are: Active transport is a rapid process, passive transport is a slow process. Active transport transpires in one direction whereas passive one transpires bidirectionally. Active transport requires carrier proteins and energy whereas passive one does not require both.

The similarities among them are: both transport system require concentration gradient, both helps in homeostatic regulation and allow entry and exit of ions.

To know more about passive and active transport, visit;

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hypothesis

need 20 characters hope this helps

Answer:

Follow safety measures and safety rules to avoid any accident.

Explanation:

1. Follow all safety measures is needed from adults when they conducting any experiment at home in order to prevent any damage occurs due to that experiment.

2. Safety rules should always be followed when conducting an experiment at any place such as at home, class room and laboratory etc in order to avoid any accident.

3. Some chemicals are very dangerous because they are flammable and catch fire when allow it to the heat source. Some chemicals also cause suffocation by absorbing all the oxygen from the environment and as a result death occurs.

4. Science is a systematic study of physical objects through observation and experiments. Science is a knowledge gain from experiments.

Answer:

10 cm (Approx)

Explanation:

Given:

8.20 cm + 2 cm

Find:

Round of the following number

Computation:

⇒ 8.20 cm + 2 cm

⇒ 10.20 cm

We know that 0.20 < 0.50

So,

⇒ 10.20 cm = 10 cm (Approx)

Answer:

No effect.

Explanation:

Hello,

In this case, considering the widely studied Le Chatelier's principle, we can realize that the factors affecting equilibrium are concentration, temperature and pressure and volume if the reaction is in gaseous phase and with non-zero change in the number of moles. In such a way, by adding a catalyst to given reaction will have no effect on the equilibrium direction.

Best regards.

Answer:

separating funnel method

Answer and Explanation: Ksp is the Solubility Product Constant and is the equilibrium constant that happens when a solid is dissolved in an aqueous solution.

The dissolution of chromium (III) hydroxide:

[tex]Cr(OH)_{3}_{(s)}[/tex] ⇄ [tex]Cr^{3+}_{(aq)} + 3OH^{-}_{(aq0)}[/tex]

Every equilibrium constant is of the form:

[tex]K = \frac{[products]^{coefficient}}{[reagents]^{coefficient}}[/tex]

Then,

[tex]K_{sp} = [Cr^{3+}_{(aq)}][OH^{-}_{aq}]^{3}[/tex]

The reagent is not included because solids don't take part in euqilibrium constants.

So, Ksp of chromium (III) hydroxide is [tex]K_{sp} = [Cr^{3+}_{(aq)}][OH^{-}_{aq}]^{3}[/tex]

Answer: chlorine 17Cl35 has 18 neutrons, and 17Cl37 has 20 neutrons

Explanation: 17 is the atomic number of chlorine which tells us what is the number of protons in chlorine nucleus. 35 and 37 is the mass number of two isotopes of chlorine, it tells us what is the total number of protons and neutrons in atomic nucleus of two chlorine isotopes .

Now to know how many neutrons are in both isotopes we have to substract the number of protons from the mass number of each isotope

For 17Cl35 it will be 35- 17 what makes 18 and for 17 Cl 37 it will be 20 neutrons

Answer:

pH = 7.0

Explanation:

The HCl reaction with KOH as follows:

HCl + KOH → H₂O + KCl

To find pH first we need to know how many moles of each reactant reacts as follows:

HCl = 0.1000L * (0.18 mol / L) = 0.018 moles HCl

KOH = 0.06667L * (0.27mol / L) = 0.018 moles KOH

That means all HCl reacts with all KOH and you will have in solution just water. The equilibrium of water is:

H₂O(l) ⇆ H⁺(aq) + OH⁻(aq)

K = 1x10⁻¹⁴ = [H⁺] [OH⁻]

As the amount of H⁺ = OH⁻:

1x10⁻¹⁴ = [H⁺]²

[H⁺] = 1x10⁻⁷M

as pH = -log [H⁺]

Answer:

X

Anode

Loses mass

Electrons in the wire flow away from it

Anions from the salt bridge flow toward it

Y

Cathode

Gains mass

Electrons in the wire flow toward it

Cations from the salt bridge flow toward it

Explanation:

A galvanic cell consists of an anode, a cathode and a salt bridge. The cathode is the negative electrode. The cathode increases in mass due to mass deposition. The cathode becomes increasingly negative as the cell reaction progresses, hence cations from the salt bridge flow towards it.

The anode is the positive electrode. It becomes more positive as the cell reaction progresses hence anions from the salt bridge flow towards it. The anode is the electrode where oxidation reaction occurs hence electrons are lost here and flow towards the cathode. The anode looses mass as the cell reaction progresses due to oxidation.

Answer:

5ml

Explanation:not sure why

Answer:

Solvent = Water

Solute = Salt