ways of expressing a solution​

Answer:

US₂

Explanation:

Uranium sulfide (US₂)

Uranium atomic symbol = U

Sulfur atomic symbol = S

Uranium valency = +4

Sulfur valency = -2

So;

Uranium sulfide (US₂)

Answer:

Double replacement:

Explanation:

Chemical equation:

2AsCl  +  3H₂S     →       As₂S₃ + 6HCl

The given reaction is double displacement reaction. In this reaction arsenic trichloride and hydrogen sulfide react and produced arsenic sulfide and hydrogen chloride. In double displacement reaction both anion an cation of reactants are exchanged with each other.

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

General equation:

AB + CD → AD +CB

Answer:

an atom is the smallest part of an element that has the properties of that element

it can't be broken further into smaller substances

an element is the simplest chemical

it consists of one type of atoms

so you can consider atoms as the building block of an element

B) loses electrons and is oxidised

Its oxidation number is 0 in Co(s) and +2 in Co(C2H302)2(aq), hence it gets oxidised

In the reaction as mentioned below:-

[tex]Co(s) + Ni(C2H302)2aq -- > Co(C2H3O2)2aq + Ni(s)[/tex]

As shown in reaction cobalt losses elections so the reaction is oxidised.  Thus, option b is correct.

Oxidation is a type of reaction in which loss of electrons and gain of protons is takes place during a reaction by molecules, atom or ion. Oxidation takes place when the oxidation state of a molecule, atoms or ion is increased.

The opposite reaction is called as reduction in which gains of electron and loss of protons is takes place . In reduction reaction the oxidation state of molecules, atoms or ion is decreased.

Oxygen has been added to a compound as an older method of oxidation.  Although oxygen is typically added to a compound in compliance with the loss of electrons requirements and the oxidation state increase.

Therefore, we can say that in the reaction

[tex]Co(s) + Ni(C2H302)2aq -- > Co(C2H3O2)2aq + Ni(s)[/tex]

As shown in reaction cobalt losses elections so the reaction is oxidised.  Thus, option b is correct.

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Answer:

Cu    +    2AgNO₃  →   2Ag    +    Cu(NO₃)₂

Explanation:

The reactants are:

         Copper metal  = Cu

         Silver nitrate  = AgNO₃

The products are:

          Silver metal  = Ag

           Copper (II) nitrate  = Cu(NO₃)₂

So, the reaction equation is given as;

      Cu    +    2AgNO₃  →   2Ag    +    Cu(NO₃)₂

This is a single displacement reaction

A solid substance formed from a solution is a(n).

D. Precipitate

A solid substance formed from a solution is a(n)

a. reactant

b. equation

C. compound

d. precipitate

D. Precipitate

Answer:

3

Explanation:

Answer:

She overcame her disabilities to compete in the 1956 Summer Olympic Games, and in 1960, she became the first American woman to win three gold medals in track and field at a single Olympics. Later in life, she formed the Wilma Rudolph Foundation to promote amateur athletics.

Explanation:

Answer:

A. 1.02 moles .

Explanation:

Hello!

In this case, given the ideal gas equation, as we need to solve for moles, we divide both sides by RT to get:

[tex]n=\frac{PV}{RT}[/tex]

Thus, by plugging in the pressure and temperature at STP (1.00 atm and 273.15 K respectively) we obtain:

[tex]n=\frac{1.00atm*22.9L}{0.08206\frac{atm*L}{mol*K}*273.15K}\\\\n=1.02mol[/tex]

Therefore, the correct answer is A. 1.02 moles

Best regards!

Answer:

Here:

Explanation:

Determine the number of electrons in the atom from its atomic number. (See Below.)

Add electrons to the sublevels in the correct order of filling.

Add two electrons to each s sublevel, 6 to each p sublevel, 10 to each d sublevel, and 14 to each f sublevel.

To check your complete electron configuration, look to see whether the location of the last electron added corresponds to the element’s position on the periodic table.

Predicting the Order of Filling of the Orbitals

There are three ways to predict the order of filling of the orbitals. Probably the least reliable method is to memorize the following list (even though it shows the order of filling of all the orbitals necessary for describing the ground state electron configurations of all of the known elements).

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p

Instead of relying on memorization, you can use the memory aid shown below to remind you of the correct order of filling of the sublevels. The following steps explain how to write it and use it yourself.

Write the possible sublevels for each energy level in organized rows and columns like the image below. To do this, you need to remember that there is one sublevel on the first principal energy level, two on the second, three on the third, etc. Every principal energy level has an s orbital. The second principal energy and all higher energy levels have a p sublevel. The d sublevels start on the third principal energy level, the f sublevels start on the fourth principal energy level, etc.

Draw arrows like those you see below .

Starting with the top arrow, follow the arrows one by one in the direction they point, listing the sublevels as you pass through them.

The sublevels that are not needed for describing the known elements are enclosed in parentheses below .

Image of the memory aid that allows you to determine the order of filling for the sublevels

We can also use the block organization of the periodic table, as shown below, to remind us of the order in which sublevels are filled. To do this, we move through the elements in the order of increasing atomic number, listing new sublevels as we come to them. The type of sublevel (s, p, d, or f ) is determined from the block in which the atomic number is found. The number for the principal energy level (for example, the 3 in 3p) is determined from the row in which the element is found and the knowledge that the s sublevels start on the first principal energy level, the p sublevels start on the second principal energy level, the d sublevels start on the third principal energy level, and the f sublevels start on the fourth principal energy level.

We know that the first two electrons added to an atom go to the 1s sublevel.

Atomic numbers 3 and 4 are in the second row of the s block (look for them in the bottom half of in image below), signifying that the 3rd and 4th electrons are in the 2s sublevel.

Atomic numbers 5 through 10 are in the first row of the p block, and the p sublevels start on the second energy level. Therefore, the 5th through 10th electrons go into the 2p sublevel.

Atomic numbers 11 and 12 are in the third row of the s block, so the 11th and 12th electrons go into the 3s sublevel.

Because atomic numbers 13 through 18 are in the p block, we know they go into a p sublevel. Because the p sublevels begin on the second principal energy level and atomic numbers 13 through l8 are in the second row of the p block, the 13th through 18th electrons must go into the 3p sublevel.

The position of atomic numbers 19 and 20 in the fourth row of the s block and the position of atomic numbers 21 through 30 in the first row of the d block show that the 4s sublevel fills before the 3d sublevel.

Moving through the periodic table in this fashion produces the following order of sublevels up through 6s:

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s

Notice that atomic numbers 57 through 70 on the periodic table below are in the 4f portion of the table. It is a common mistake to forget that the 4f sublevel is filled after the 6s sublevel and before the 5d sublevel. In order to make the overall shape of the table more compact and convenient to display, scientists have adopted the convention of removing the elements with atomic number 57 through 70 and 89 through 102 (the latter being the 5f portion of the table) from their natural position between the s and d blocks and placing them at the bottom of the table. Electrons go into the 5f sublevel after the 7s sublevel and before the 6d sublevel. The second periodic table below shows how the blocks on the periodic table would fit together if the inner transition metals­—the f block—were left in their natural position.

How would you calculate the change in entropy when the pressure of a perfect gas is changed isothermally from P_i to P_f

Answer:

ΔS = nRIn(P_i/P_f)

Explanation:

Formula for entropy change is;

ΔS = nC_p•In(T_f/T_i) + nRIn(P_i/P_f)

Since the pressure of the perfect gas is changed isothermally, it means that temperature remains the same and as such T_f = T_i

Thus, we now have;

ΔS = nC_p•In(1) + nRIn(P_i/P_f)

In(1) = 0

Thus;

ΔS = nRIn(P_i/P_f)

Answer:

cup...................

Answer: [tex]CsF[/tex] > [tex]SrF_2[/tex] >  [tex]OF_2[/tex]  > [tex]F_2[/tex]

Explanation:

The percent ionic character is used to determine if a bond is more ionic or covalent. More is the percent ionic character, the more ionic the molecule is. If the electronegativity difference between two elements is higher than 1.7 it will be considered as ionic.

The electronegativity difference for CsF is (3.98-0.79) = 3.19

The electronegativity difference for [tex]OF_2[/tex] is = (3.98-3.44) = 0.54

The electronegativity difference for [tex]SrF_2[/tex] is (3.98-0.95) = 3.0

The electronegativity difference for [tex]F_2[/tex] is  0.

Thus decreasing order of  ionic character of the bonds is CsF >  [tex]SrF_2[/tex] >  [tex]OF_2[/tex]  > [tex]F_2[/tex]

The solution of manganese ion, Mn2+ is colorless. This way, the reaction involves a colour change from pink to transparent.

There is no option with this information.

So, the best answer is (A)

Answer:

C)pink colour of the Mn+2.

Explanation:

First of all, a titration involving the change from MnO4- to Mn2+ is a redox reaction.

This redox reaction does not need an indicator because the MnO4- has a deep violet colour while Mn2+ has a light pink colour.

Hence, the change from MnO4- to Mn2+ is signaled by the appearance of the light pink colour of Mn2+ at end point.

Answer:

I would say, what helps me is really paying attention in class and asking questions, also making sure you study for upcoming test's and quizzes and completely assingments on time

Explanation:

Answer:

A or C

Explanation:

Answer: It is D.

Explanation: Edge 2021

Answer:

C. 3

Explanation:

I hope my answer will be useful

Answer:

CH2

Explanation:

there is a common factor for 9;18 which is 9 so u divide both numbers by it 9/9 and 18/9 giving 1:2 so 1 C to every 2 H so CH2

The empirical formula of C9H18 will be CH2.

The empirical formula of a compound shows the ratio of atoms present in a compound. The molecular formula of the compound shows the total number of atoms present in a compound.

Here, C9H18 if we divide the number of total atoms by 9, C9/ 9, and H18/ 9. the ratio will be in the form of 1: 2. therefore, the empirical formula will be CH2.

C9H18 is the molecular formula of the compound, while CH2 is the empirical formula of the compound. The molecular formula of butane is C4H16, then its empirical formula is CH4.  The molecular formula of compound is P4H10 then its empirical formula is P2H5.

Therefore, The empirical formula of C9H18 will be CH2.

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Answer:

To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law. The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

T

1

P

1

V

1

=

T

2

P

2

V

2

where,

P_1,V_1\text{ and }T_1P

1

,V

1

and T

1

are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2P

2

,V

2

and T

2

are the final pressure, volume and temperature of the gas

We are given:

\begin{gathered}P_1=760mmHg\\V_1=175L\\T_1=15^oC=[15+273]K=288K\\P_2=640mmHg\\V_2=198L\\T_2=?K\end{gathered}

P

1

=760mmHg

V

1

=175L

T

1

=15

o

C=[15+273]K=288K

P

2

=640mmHg

V

2

=198L

T

2

=?K

Putting values in above equation, we get:

\begin{gathered}\frac{760mmHg\times 175L}{288K}=\frac{640mmHg\times 198L}{T_2}\\\\T_2=274K\end{gathered}

288K

760mmHg×175L

=

T

2

640mmHg×198L

T

2

=274K

Hence, the temperature when the volume and pressure has changed is 274 K

Answer:

migratory distribution

Explanation: The pattern of migratory distribution of species may require collaboration between different countries for effective preservation. This is because species migration can make them enter different territories, which will need to provide an effective level of preservation for species survival

Answer:

Liquids take the shape of their containers because they are stuck together enough to not go flying off like gases but not so tightly that they stay in a stiff shape like solids do.

Explanation:

Hope this helps you! ^^

Answer:

Because of reaction between solid and liquid

Answer:

shouldn't take that long i would say from 30 to 40 minutes bc its melting in warm water its turning from a solid to a liquid.

Explanation:

Answer:

0.64 in trial 2

Explanation:

Mass (in g) of the excess reactant (P₄) left = 7.626 g

Given

Reaction

P4(s)+6Cl2(g)→4PCl3(l)

45.57 g P4 and 130.2 g Cl2

Required

mass (in g) of the excess reactant left

Solution

mol P₄ (MW=124 g/mol) :

45.57 : 124 = 0.3675

mol Cl₂(MW=71 g/mol) :

130.2 : 71 = 1.834

mol : coefficient of reactant : P₄ : Cl₂ :

= 0.3675/1 : 1.834/6

= 0.3675 : 0.306

Cl₂ as limiting reactant(smaller ratio)

Reacted mol of P₄ (as an excess reactant):

=1/6 x 1.834

= 0.306

Unreacted mol of P₄ :

= 0.3675 - 0.306

= 0.0615

Mass of P₄(left) :

= 0.0615 x 124

= 7.626 g

Answer:

C₃H₆ is an alkene

C₆H₁₂ is an alkene

C₈H₁₈ is an alkane

C₇H₁₂ is an alkyne

Explanation:

To determine which of the compound is alkane, alkene, or alkyne,we shall use the general formula of alkane, alkene, and alkyne. This is illustrated below:

General formula for alkane => CₙH₂ₙ₊₂

General formula for alkene => CₙH₂ₙ

General formula for alkyne => CₙH₂ₙ₋₂

For C₃H₆:

n = 3

Alkane => CₙH₂ₙ₊₂ => C₃H₂₍₃₎₊₂ => C₃H₈

Alkene => CₙH₂ₙ => C₃H₂₍₃₎ => C₃H₆

Alkyne => CₙH₂ₙ₋₂ => C₃H₂₍₃₎₋₂ => C₃H₄

Thus, C₃H₆ is an alkene

For C₆H₁₂:

n = 6

Alkane => CₙH₂ₙ₊₂ => C₆H₂₍₆₎₊₂ =>C₆H₁₄

Alkene => CₙH₂ₙ => C₆H₂₍₆₎ => C₆H₁₂

Alkyne => CₙH₂ₙ₋₂ => C₆H₂₍₆₎₋₂ => C₆H₁₀

Thus, C₆H₁₂ is an alkene

For C₈H₁₈:

n = 8

Alkane => CₙH₂ₙ₊₂ => C₈H₂₍₈₎₊₂ => C₈H₁₈

Alkene => CₙH₂ₙ => C₈H₂₍₈₎ => C₈H₁₆

Alkyne => CₙH₂ₙ₋₂ => C₈H₂₍₈₎₋₂ => C₈H₁₄

Thus, C₈H₁₈ is an alkane.

For C₇H₁₂:

n = 7

Alkane => CₙH₂ₙ₊₂ => C₇H₂₍₇₎₊₂ => C₇H₁₆

Alkene => CₙH₂ₙ => C₇H₂₍₇₎ => C₇H₁₄

Alkyne => CₙH₂ₙ₋₂ => C₇H₂₍₇₎₋₂ => C₇H₁₂

Thus, C₇H₁₂ is an alkyne.

SUMMARY:

C₃H₆ is an alkene

C₆H₁₂ is an alkene

C₈H₁₈ is an alkane

C₇H₁₂ is an alkyne

Answer: a) [tex]1.7\times 10^{-4}[/tex]

b) [tex]3.4\times 10^{-4}[/tex]

Explanation:

The reaction is :

[tex]2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)[/tex]

Rate = Rate of disappearance of [tex]N_2O_5[/tex] = Rate of appearance of [tex]NO_2[/tex]

Rate =  [tex]-\frac{d[N_2O_5]}{2dt}[/tex] = [tex]\frac{d[NO_2]}{4dt}[/tex]

Rate of disappearance of [tex]N_2O_5[/tex] = [tex]\frac{\text {change in concentration}}{time}[/tex] = [tex]\frac{0.100-0.066}{200-0}=1.7\times 10^{-4}[/tex]

a) Rate of disappearance of [tex]N_2O_5[/tex] = [tex]-\frac{d[N_2O_5]}{2dt}[/tex]

Rate of appearance of [tex]NO_2[/tex] = [tex]\frac{d[NO_2]}{4dt}[/tex]

b) Rate of appearance of [tex]NO_2[/tex] =  [tex]\frac{d[NO_2]}{dt}=2\times 1.7\times 10^{-4}}=3.4\times 10^{-4}[/tex]

A) Find the rate of disappearance of [tex]N_2O_5[/tex] from t = 0 s to t = 200s

[tex]Rate = \frac{1}{2}(\frac{-\delta N_2O_5}{\delta t})\\\\Rate = -\frac{1}{2}(\frac{0.066 - 0.100}{200 - 0})\\\\Rate = 8.5*10^{-5}[/tex]

B) Find the rate of appearance of [tex]NO_2[/tex] from t = 0 s to t = 200s

According to rate law,

[tex]\frac{1}{2}(\frac{-\delta N_2O_5}{\delta t}) = \frac{1}{4}(\frac{\delta NO_2}{\delta t})\\\\8.5*10^{-5} = \frac{1}{4}(\frac{\delta NO_2}{\delta t})\\\\\frac{\delta NO_2}{\delta t} = 4 * 8.5*10^{-5}\\\\Rate = 3.4*10^{-4}[/tex]

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Answer:

the moral amount of the gas.