what volume(in dm³) of 0.30 moldm-³ Nacl solution can be prepared from 0.060mol of solute?
​

Answer:

(a)

Explanation:

Hello,

In this case, the temperature required to boil argon, it means, transform it from liquid to gas is -197 °C. In such a way, since the temperature inside the steel sphere is -190 °C, which is greater than the boiling point, we realize argon is gaseous, therefore, the molecules will be spread inside the sphere as they will be moving based on the kinetic theory of gases.

For that reason, answer is scheme (a).

Best regards.

Answer:

A. It will shift to the left

Explanation:

In the equilibrium:

C(s) ⇄ CO2(g)2CO(g)

The system will shift to the right if any change stimulate the production of gas -LeChatelier's principle-; in the same way, if a change doesn't favors the production of gas the system will shift to the left producing less gas.

The changes that increasing the pressure of the system, doesn't favors the gas production doing the system shift to the left.

A gas that is heated expands itsellf doing the pressure increases.

In the same way, if you compress the gas, the gas increases its pressure.

Thus, both changes increase pressure of the gas doing the system shift to the left.

Answer:

[tex]n_S=1.076molS[/tex]

Explanation:

Hello,

In this case, given the undergoing chemical reaction, we can see a 4:3 mole ratio between the consumed moles of gallium and sulfur respectively, therefore, the consumed moles of sulfur, from the 100.0 g of gallium (use its atomic mass) turn out:

[tex]n_{S}=100.0gGa*\frac{1molGa}{69.72gGa}*\frac{3molS}{4molS} \\\\n_S=1.076molS[/tex]

Best regards.

Answer:

Hope this helps...

Good luck on your assignment...

Answer:

double replacement

Explanation:

Double replacement reactions are where both cations and anions of reactants switch.

For eg: AB + CD -----> AD + BC (Here we see both things are replaced)

Synthesis reactions is where two simple things make one complex thing.

So A + B --- > AB

Combustion is usually reaction where Oxygen is reactant but here we don't have that.

Single replacement only replaces either cation or anion.

So AB + C .------> AC + B

Answer:

Replacement

Explanation:

Answer:

Approximately [tex]0.180[/tex].

Explanation:

The mole fraction of a compound in a solution is:

[tex]\displaystyle \frac{\text{Number of moles of compound in question}}{\text{Number of moles of all particles in the solution}}[/tex].

In this question, the mole fraction of [tex]\rm KCl[/tex] in this solution would be:

[tex]\displaystyle X_\mathrm{KCl} = \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}[/tex].

This solution consist of only [tex]\rm KCl[/tex] and water (i.e., [tex]\rm H_2O[/tex].) Hence:

[tex]\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})}\end{aligned}[/tex].

From the question:

Apply the formula [tex]\displaystyle n = \frac{m}{M}[/tex] to find the number of moles of [tex]\rm KCl[/tex] and [tex]\rm H_2O[/tex] in this solution.

[tex]\begin{aligned}n(\mathrm{KCl}) &= \frac{m(\mathrm{KCl})}{M(\mathrm{KCl})} \\ &= \frac{195.0\; \rm g}{74.6\; \rm g \cdot mol^{-1}} \approx 2.61\; \em \rm mol\end{aligned}[/tex].

[tex]\begin{aligned}n(\mathrm{H_2O}) &= \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} \\ &= \frac{215\; \rm g}{18.0\; \rm g \cdot mol^{-1}} \approx 11.9\; \em \rm mol\end{aligned}[/tex].

The molar fraction of [tex]\rm KCl[/tex] in this solution would be:

[tex]\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})} \\ &\approx \frac{2.61 \; \rm mol}{2.61\; \rm mol + 11.9\; \rm mol} \approx 0.180\end{aligned}[/tex].

(Rounded to three significant figures.)

Answer:

the answer is monerans

Explanation:

When Carl Woese developed the modern system of classification, he broke the previous kingdom of Monera into the two kingdoms of Bacteria and Archaea.

Some biologists believed it made sense to classify prokaryotes as belonging to their own kingdom, the Monera. That served as the foundation for Richard Whittaker  and Lynn Margulis's five-kingdom proposal, which enhanced the Haeckel plan by include a kingdom of fungus.

Protists, protozoa, monera, fungi, and viruses have long been proposed as belonging to different kingdoms, but traditional evolutionists during the majority of the 20th century had given none of them any thought.

Later, the Monera kingdom was split into Eubacteria and Archaebacteria by Carl Woese .  Moreover, he divided the five kingdoms into three domains: Eukaryotes, Archaea, and Bacteria.

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Your question is incomplete. But your complete question is as follows:

When Carl Woese developed the modern system of classification, he broke the previous kingdom of into the two kingdoms of  _____  into  Bacteria and Archaea.

Answer:

H2SO4(aq)

Explanation:

The balanced equation for the reaction is given below:

H2SO4(aq) + Mg(s) —> MgSO4(aq) + H2 (g)

An acid is a substance which dissolves in water to produce hydrogen ion, H+ as the only positive ion.

To know which of the substance is acid, let us dissolve them in water to see which will produce hydrogen ion, H+ as the only positive ion.

This is illustrated below:

H2SO4(aq) —> 2H+(aq) + SO4^2-(aq)

Mg(s) + 2H2O(l) —> Mg(OH)2(aq) + H2(g)

MgSO4(aq) —> Mg^2+(aq) + SO4^2-(aq)

H2 is insoluble in water.

From the above, only H2SO4 produces hydrogen ion H+ on dissolution in water. Therefore, H2SO4 is an acid

Answer:

D on edg 2021

Explanation:

Answer:

[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=32.23gKAl(SO_4)_2\dot \ 12H_2O[/tex]

Explanation:

Hello,

In this case, we balance the given equations as shown below:

[tex]Al(s)+KOH(aq)+3H_2O(l)\rightarrow KAl(OH)_4(aq)+\frac{3}{2} H_2(g)\\\\KAl(OH)_4(aq)+2H_2SO_4(aq)\rightarrow KAl(SO_4)_2(aq)+4H_2O(l)\\\\KAl(SO_4)_2(aq)+12H_2O\rightarrow KAl(SO_4)_2\dot\ 12H_2O(aq)[/tex]

Now, with 3.00 grams of aluminium, 50.00 mL of water and 10.00 mL of 8.00M potassium hydroxide, the first step is to identify the limiting reactant by firstly computing the moles of all of them:

[tex]n_{Al}=3.00 gAl*\frac{1molAl}{27gAl}=0.111molAl\\ \\n_{KOH}=0.010L*8.00mol/L=0.08molKOH\\\\n_{H_2O}=50.00mL*\frac{1g}{1mL} *\frac{1mol}{18g}=2.78molH_2O[/tex]

Thus, we can notice that 0.111 mol of aluminium will consume 0.11. moles of potassium hydroxide and 2.78 moles of water will consume 0.927 moles of potassium hydroxide, for that reason, we can infer that since there are only 0.08 moles of potassium hydroxide, it is the limiting reactant, therefore, we compute the yielded moles of potassium aluminium hydroxide in the first reaction:

[tex]n_{KAl(OH)_4}=0.08molKOH*\frac{1molKAl(OH)_4}{1molKOH} =0.08molKAl(OH)_4[/tex]

Next, we compute the yielded moles of potassium aluminium sulfate in the second reaction assuming sulfuric acid is in excess:

[tex]n_{KAl(SO_4)_2}=0.08molKAl(OH)_4*\frac{1molKAl(SO_4)_2}{1molKAl(OH)_4}=0.08molKAl(SO_4)_2[/tex]

Finally, in the third reaction, we compute the yielded grams of potassium aluminum sulphate dodecahydrate by using its molar mass and its mole ratio with potassium aluminium sulfate:

[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}=0.08molKAl(SO_4)_2*\frac{1molKAl(SO_4)_2\dot \ 12H_2O}{1molKAl(SO_4)_2} *\frac{474.00gKAl(SO_4)_2\dot \ 12H_2O}{1molKAl(SO_4)_2\dot \ 12H_2O} \\\\m_{KAl(SO_4)_2\dot \ 12H_2O}=37.92gKAl(SO_4)_2\dot \ 12H_2O[/tex]

Which is the theoretical yield, thus, by using the percent yield the actual yielded mass turns out:

[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=0.85*37.92gKAl(SO_4)_2\dot \ 12H_2O\\\\m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=32.23gKAl(SO_4)_2\dot \ 12H_2O[/tex]

Best regards.

Answer:

See explanation below.

Explanation:

Both carbon and silicon are members of group 4A(now group 14) i n the periodic table. Carbon is the first member of the group. CO2 is a gas while SiO2 is a solid. In SiO2, there are single bonds between silicon and oxygen and the geometry around the central atom is tetrahedral while in CO2, there are double carbon-oxygen bonds and the geometry around the central atom is linear. CO2 molecules are discrete and contain only weak vanderwaals forces.

Again, silicon bonds to oxygen via its 3p orbital while carbon bonds to oxygen via a 2p orbital. As a result of this, there will be less overlap between the pi orbitals of silicon and that of oxygen. This is why tetrahedral bonds are formed with oxygen leading to a covalent network solid rather than the formation of a silicon-oxygen pi bond. A covalent network solid is known to be made up of a network of atoms of the same or different elements connected to each other continuously throughout the structure by covalent bonds.

In SiO2, each silicon atom is surrounded by four oxygen atoms. Each corner is shared with another tetrahedron. SiO2 forms an infinite three dimensional structure and melts at a very high temperature.

Answer and Explanation:

The explanation is described below:-

a. Bond that is a which is created by sp3 - sp2 orbitals

b. Bond that is b which is developed by sp2-sp orbitals

c. Bond that is c which is created by sp-sp3 orbitals

Sp2 is hybridised by a double bonded carbon, and a triple bonded carbon is hybridised. Both single bonded carbons are hybridised to sp3.

Answer:

See figure 1

Explanation:

In the structure of nylon 6,6 we have amide groups. In this functional group, We have a nitrogen bond to hydrogen, so in this bond, we will have a dipole, due to the electronegativity difference. Nitrogen has more electronegativity than hydrogen, therefore a positive dipole would be generated in the hydrogen atom. Additionally, in the carbonyl group (C=O) due to the oxygen, we will have also a dipole, in this case, a negative dipole because the oxygen atom has more electronegativity (compare with carbon).

When we put two strings of nylon 6,6 the positive dipole will interact with the negative dipole and vice-versa and we will obtain the "hydrogen bonds".

See figure 1

I hope it helps!

Answer:

The concentration of NaOH will be lower and the titration will be affected.

Explanation:

Hello,

In this case, sodium hydroxide is acknowledged as a highly hygroscopic substance, which means that is able to absorb water to its molecules. In such a way, in any measurement, if sodium hydroxide has absorbed water, the results will be wrong in terms of accuracy. More specifically, for concentration, if we have for example 30 grams of NaOH and we dissolve it a 100-mL solution, as it absorbed 30 grams of water, the total volume could be now approximated to 130 mL, thus, the concentration will change as follows:

[tex]M_1=\frac{30g/40g/mol}{0.1L}=7.5M\\ \\M_2=\frac{30g/40g/mol}{0.13L}=5.77M[/tex]

It causes the actual molarity to be decreased, it means that in a titration procedure, less acid would be used to neutralize it or more of it would be needed to neutralize a given acid.

Best regards.

We have that  the NaOH sitting on the shelf had absorbed 1 g of water for every 1 g of NaOH will affect the Molarity of NaOH and its effectiveness

With the situation of NaOH sitting on the shelf having to absorbed 1 g of water for every 1 g of NaOH.

Means that for every  g of NaOH collected 1/2g is water and 1/2g is actual NaOH.

Hence this will cause a change in the molarity of NaOH thereby causing Molarity to drop by half as well. Giveing the resultant Molarity to be somewhere around half the regular molarity.

This many cause a variation also in the results of titration

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Answer:

2. Due to the possession of resonance

Explanation:

In the benzene ring, the electrons that results in the bonds between the carbon atoms are delocalized. That is, they do not belong to a specific carbon atom. It is this unique feature that enables them to have a bond length between single and double bond lengths.

This feature is as a result of resonance.

The correct option is 2.

Answer:

[tex][Ar]4s^23d^5[/tex]

Explanation:

Hello,

In this case, we write the electron configuration of the manganese atom by noticing its atomic number is 25, so we fill the orbitals and levels up-to 25 electrons as shown below:

[tex]1s^22s^22p^63s^23p^64s^23d^5[/tex]

Moreover, for the condensed electron configuration, we consider the previous noble gas, that is argon, electron configuration which is:

[tex]1s^22s^22p^63s^23p^6[/tex]

By cause of its atomic number that is 18. In such a way, we combine argon's electron configuration with manganese's to obtain its condensed version:

[tex][Ar]4s^23d^5[/tex]

Regards.

Answer:

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10.

Explanation:

The element nitrogen is in the Group 15 in the Periodic Table, so it tends to gain 3 electrons (3 negative charges) to fill its valance shell with 8 electrons.

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10 (the original 7 plus the 3 gained). It is isoelectronic with the gas Neon, which accounts for its stability.

The answer is 6.1*10^-3 atm.

The pictures and explanations are there.

Ideal gas law is valid only for ideal gas not for vanderwaal gas. The equation used to solve this is PV=nRT. Therefore the pressure of carbon dioxide gas is 0.005 atm.

Ideal gas equation is the mathematical expression that relates pressure volume and temperature.

Mathematically the relation between Pressure, volume and temperature can be given as

PV=nRT

where,

P = pressure of carbon dioxide gas=?

V= volume of carbon dioxide gas=4.6L

n =number of moles of carbon dioxide gas = given mass ÷Molar mass

                                                                      =0.050g÷44g/mol

                                                                     =0.001mole

T =temperature of carbon dioxide gas=303K

R = Gas constant = 0.0821 L.atm/K.mol

substituting the given values, we get

P×4.6L=0.001×0.0821×303

          =0.005 atm

Therefore the pressure of carbon dioxide gas is 0.005 atm.

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Answer:

483.27 minutes

Explanation:

using second faradays law of electrolysis

Answer:

pH = 7.29

Explanation:

Ka of butanoic acid is 1.54x10⁻⁵

To obtain the pH of the solution you must use H-H equation for butanoic acid:

pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]

Where pKa is defined as -log Ka = 4.81

Now, you need to find [C₃H₇COO⁻] and [C₃H₇COOH] concentrations (Also, you can find moles of each substance and replace them in the equation.

Butanoic acid reacts with LiOH, producing C₃H₇COO⁻, thus:

C₃H₇COOH + LiOH → C₃H₇COO⁻ + H₂O + Li⁺

Moles of both reactants, C₃H₇COOH and LiOH are:

C₃H₇COOH = 0.0300L ₓ (0.1000mol / L) = 0.003000moles of C₃H₇COOH

LiOH = 0.0299L ₓ (0.1000mol / L) = 0.00299 moles of LiOH.

That means moles of C₃H₇COO⁻ produced are 0.00299 moles.

And moles of C₃H₇COOH that remains in solution are:

0.00300 - 0.00299 = 0.00001 moles of C₃H₇COOH

Replacing in H-H equation:

pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]

pH = 4.81 + log₁₀ [0.00299moles] / [0.00001moles]

Answer:

The colour of the orange solution becomes yellow.  

Explanation:

1. Before adding NaOH

Assume the picture showed a beaker of potassium chromate and one of potassium dichromate.

Both solutions are involved in the same equilibrium:

[tex]\rm\underbrace{\hbox{2CrO$_{4}^{2-}$(aq)}}_{\text{yellow}} +2H^{+}(aq) \rightleftharpoons \, \underbrace{\hbox{Cr$_{2}$O$_{7}^{2-}$}}_{\text{orange}} + H_{2}O[/tex]

The first beaker contains mostly chromate ions with a few dichromate ions.

The position of equilibrium lies to the left and the solution is yellow.

The second beaker contains mostly dichromate ions with a few chromate ions.

The position of equilibrium lies to the right and the solution is orange.

2. After adding NaOH

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Beaker 1

If you add OH⁻ to the equilibrium solution, it removes the H⁺ (by forming water).

The system responds by having the dichromate react with water to replace the H⁺.  

At the same time, the system forms more of the yellow chromate ion.

The position of equilibrium shifts to the left.

However, the solution is already yellow, so you see no change in colour.

Beaker 2

The reaction is the same as in Beaker 1.

This time, however, as the dichromate ion disappears, do does its orange colour.

Also, the yellow chromate is being formed and its yellow colour appears .

The colour changes from orange to yellow.

Answer:

[tex]n_{CO_2}=1.93 gCO_2[/tex]

Explanation:

Hello,

In this case, considering the given chemical reaction, we can use the molar mass of octane (114.23 g/mol) and the 2:16 molar ratio with carbon dioxide to compute the emitted moles of CO2 to the atmosphere via the following stoichiometric procedure:

[tex]n_{CO2}=27.6gC_8H_{18}*\frac{1molC_8H_{18}}{114.23gC_8H_{18}} *\frac{16molCO_2}{2molC_8H_{18}} \\\\n_{CO_2}=1.93 gCO_2[/tex]

Which also corresponds to the following mass:

[tex]m_{CO_2}=1.93molCO_2*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=85.0gCO_2[/tex]

Best regards.

Answer:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, we can write the law of mass action for the undergoing chemical reaction, based on the rates and the stoichiometric coefficients:

[tex]\frac{1}{-2}r_{NH_3} =\frac{1}{1} r_{N_2}=\frac{1}{3}r_{H_2}[/tex]

In such a way, knowing the rate of formation hydrogen (H₂), we can know the  rate of change of ammonia, that must be negative for consumption:

[tex]r_{NH_3} =\frac{-2}{3}r_{H_2}=\frac{-2}{3}*9.00\frac{M}{s} \\\\r_{NH_3} =-6.00\frac{M}{s}[/tex]

Nevertheless, the absolute magnitude will be positive:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Best regards.

Answer:

Option C. Will always.

Explanation:

A spontaneous reaction is a reaction that occurs without an external supply of heat.

This implies that spontaneous reaction will always occur as no external supply of heat is needed.

Answer:

An ionic compound is formed between a metal and a nonmetal (or polyatomic ions), and is held together through the attraction of opposite charges. An example is KCl. A molecular compound is usually formed between two or more nonmetals, and is held together through the sharing of electrons between atoms. An example is SO2.

Explanation:

When we talk about ionic bonds, the first thing that must come to mind is the electrostatic attraction between oppositely charged ions. Hence, we know that metals form cations and nonmetals form anions, thus metals could transfer electrons to nonmetals to facilitate the formation of ionic bonds. Ionic bonds could also be formed by the combination of metals with polyatomic ions such as CaCO3. Always keep it in mind that ionic bonds are characterized by electrostatic attraction between any pair of oppositely charged ions.

Molecular compounds are formed by sharing of electrons between nonmetals. We find covalent or polar covalent bonds in molecular compounds such as SO2.

Answer:

1. nickel(II) carbonate is MORE soluble than barium sulfate, nickel hydroxide and silver chromate because its Ksp is higher than those of such compounds, it means more ions will be dissolved.

2. nickel(II) carbonate is LESS soluble than calcium sulfite because its Ksp is lower than it of such compound, it means, less ions will be dissolved.

Explanation:

Hello,

In this case, for calcium sulfite, barium sulfate, nickel hydroxide and silver chromate Kps is 6.13x10-5, 1.53x10-9, 1.63x10-16 and 9.03x10-12 respectively. Now, since Ksp for nickel (II) carbonate is 1.43 x10-7, we can notice that:

1. nickel(II) carbonate is MORE soluble than barium sulfate, nickel hydroxide and silver chromate because its Ksp is higher than those of such compounds, it means more ions will be dissolved.

2. nickel(II) carbonate is LESS soluble than calcium sulfite because its Ksp is lower than it of such compound, it means, less ions will be dissolved.

Best regards.

Answer:

b) increases by a factor of about 2.

Explanation:

Ignore the nitrogen and oxygen. Each gas acts independently of the others.

You have 0.02 mol of CO₂ gas at some pressure in equilibrium with the CO₂ in solution.

According to Graham's Law,

S  = kp

That is, the solubility of a gas in a liquid is directly proportional to its partial pressure above the liquid.

If you add another 0.02 mol of CO₂, you have doubled the number of moles.

According to Avogadro's Law, doubling the number of moles doubles the pressure.

According to Graham's Law, doubling the pressure doubles the solubility.

The solubility of CO₂ increases by a factor of two.

Answer:

H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)

Explanation:

In a redox reaction, one half-reaction is the oxidation (where the atom loss electrons) whereas the other reaction is the reduction (Where the atom is gaining electrons.

In the reactions:

H₂(g) → 2H⁺(aq) + 2e⁻ oxidation

Here, the reaction is written as the oxidation because the hydrogen H₂ is in oxidation state 0 and H⁺ in +1. That means each atom is loosing one electron.

Cu²⁺(aq) + 2e⁻ → Cu(s) reduction

And here, the Cu²⁺ is in +2 oxidation state and after the reaction is in Cu(s) 0 state. Thus, each atom is gaining 2 electrons.

The sum of both reactions is:

H₂(g) + Cu²⁺(aq) + 2e⁻ → 2H⁺(aq) + 2e⁻ + Cu(s)

Subtracting the electrons in both sides of the reaction:

Answer:

The answer to your question is given below

Explanation:

The balanced equation for the reaction is given below:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

1. Writing an expression for the equilibrium constant, K.

The equilibrium constant, K for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Thus, we can write the equilibrium constant, K for the reaction as follow:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

K = [CaCO3] [H2]⁴ / [CaO] [CH4] [H2O]²

2. Based on the value of K, more products will be in the equilibrium mixture since the value of K is a positive large number.