during the electrolysis of an aqueous solution of sodium nitrate, a gas forms at the anode, what gas is it?

Answer:

1120.62 kJ

Explanation:

In order to find how much heat is released for 7.70 mol, we have to compare it with the heat released from one mole.

So from the question, we have;

196.6 kJ = 1 mol

x = 5.70

x = 5.70 * 196.6 / 1

x = 1120.62 kJ

Answer:

I think the answer is " Night vision glasses detect Infrared" energy emitted from cooling objects.

Explanation:

Answer:

a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.

Answer:

Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.  

With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:  

CaO > Al2O3 > SiO2 > CO2 > P2O5 > SO3. Hence, CaO is the least acidic and SO3 is the most acidic.  

Since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:

[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]

Note the following:

Using the periodic table diagram given in the attachment below, we can rank the given oxides according to their increasing acidity.

Therefore, since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:

[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]

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Explanation:

The mass to charge ratio =136

Answer:

F, V, V , V, F

Explanation:

1 - "Os foguetes são utilizados para levar pessoas ao espaço (os astronautas), mas principalmente cargas como, por exemplo, os satélites artificiais, os telescópios espaciais, levar sondas a outros planetas etc".

2 - Tipo Meteorologia: utilizados para monitorar o tempo e o clima no planeta Terra, por exemplo, os da série Meteosat.

3 - ...

4 - ...

5 - Usam combustivel solido, liquido, hibridos (solido e liquido), iônica:

Solido:

 São sistemas simples que unem os dois propelentes envolvidos em uma massa sólida que, quando inflamada, não para de queimar até o esgotamento completo.

Liquido:

 São muito mais complexos e envolvem o bombeamento de quantidades imensas de propelentes para as câmaras de combustão dos motores.

Hibridos:

 O propelente sólido – normalmente o combustível – é distribuído ao longo do tanque de maneira homogênea. O propelente líquido ou gasoso "normalmente o oxidante" fica armazenado em tanques.

 Podem ser desligados depois de sofrerem ignição, além de permitirem um controle de queima relativamente preciso.

Iônica:

 Usando eletricidade (captada por painéis solares ou gerada por reatores atômicos) para ionizar átomos (normalmente gases nobres, como xenônio), e expulsá-los em velocidades altíssimas.

Answer:

Catalyst

Explanation:

For the reaction:

[tex]Cl_2_(_g_)~+~3F_2_(_g_)->2ClF_3_(_g_)[/tex]

We have a main observation: When platinum is added the reaction goes faster. With this in mind, we have to remember the kinetic equilibrium theory. In figure 1, we have an energy diagram. In which we have an specific energy for the reagents and the products. When the reaction takes place, the reaction has to must go through an energy peak. This energy peak is called "activation energy". When platinum is added the activation energy decreases and the reaction can go faster. Therefore, platinum is a "catalyst", a substance with the ability to reduce the activation energy.

I hope it helps!

Answer:

A.∆s>0contribute to spontaneity.

Answer:

The reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C

Explanation:

Hello,

In this case, for the given chemical reaction, we can write the law of mass action (equilibrium expression) as shown below:

[tex]Kc=\frac{[CH_3C_5H_9]}{[C_6H_{12} ]}[/tex]

Thus, since Kc < 1, we can conclude there are more moles of cyclohexane at equilibrium (denominator is greater than numerator), therefore, the reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C.

Best regards.

When a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities. The statement is True.

Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.

There are various methods of expressing the concentration of a solution.

Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.

Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.

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Answer:

heptan-2-one

Explanation:

In this case, the final product would be a ketone: heptan-2-one. To understand why this molecule is produced we have to check the reaction mechanism.

The first step is the protonation of the triple bond to produce the more stable carbocation (a secondary one) by the action of sulfuric acid [tex]H_2SO_4[/tex]. The next step is the attack of water to the carbocation to produce a new bond between C and the O, producing a positive charge in the oxygen. Then, a deprotonation step takes place to produce an enol. Finally, we will have a rearrangement (keto-enol tautomerism) to produce the final ketone.

See figure 1

I hope it helps!

Answer:

CH3CH=NH2+>CH3CH2NH3+

Explanation:

If we look at the both species under review, we will realize that they are both amines hence they possess the polar N-H bond.

Electrons are ordinarily attracted towards the nitrogen atom hence making both compounds acidic. It is worthy of note that certain features of a compound may make it more acidic than another of close structural proximity. 'More acidic' simply means that the proton is more easily lost.

CH3CH=NH2+ contains an sp2 hybridized carbon atom which is highly electronegative and further withdraws electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ compared to CH3CH2NH3+

Answer:

(d) 2

Explanation:

Lets say that there are 2 apples or 1+1 if you count them you would do 1,2 so 2 would be the final answer

Answer:

Approximately [tex]24\; \rm g[/tex] (at most.)

Explanation:

Aluminum [tex]\rm Al[/tex] reacts with bromine [tex]\rm Br_2[/tex] at a [tex]2:3[/tex] ratio:

[tex]\rm 2\; Al\, (s) + 3\; Br_2\, (g) \to 2\; AlBr_3\, (s)[/tex].

Look up the relative atomic mass of [tex]\rm Al[/tex] and [tex]\rm Br[/tex]. From a modern periodic table:

Calculate the formula mass of the reactants and of the product:

Calculate the quantity (in number of moles of formula units) of each reactant:

Assume that [tex]\rm Al\, (s)[/tex] is the limiting reactant. From the coefficients:

[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} = 1[/tex].

Based on the assumption that [tex]\rm Al\, (s)[/tex] is the limiting reactant:

[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} \cdot n(\mathrm{Al}) \\ &=1\times 0.18528\; \rm mol \approx 0.185\; \rm mol\end{aligned}[/tex].

In other words, if [tex]\rm Al[/tex] is the limiting reactant (meaning that [tex]\rm Br_2[/tex] is in excess,) then approximately [tex]0.556\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] will be produced.

On the other hand, assume that [tex]\rm Br_2\; (g)[/tex] is the limiting reactant. Similarly, from the coefficients:

[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} = \frac{2}{3}[/tex].

Based on the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant:

[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} \cdot n(\mathrm{Br_2}) \\ &= \frac{2}{3}\times 0.13767\; \rm mol \approx 0.0918\; \rm mol\end{aligned}[/tex].

Compare the [tex]n(\mathrm{AlBr_3})[/tex] value based on the two assumptions. Only the smallest value, [tex]n(\mathrm{AlBr_3}) \approx 0.0918\; \rm mol[/tex] (under the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant,) would resemble the theoretical yield. The reason is that [tex]\rm Br_2\, (g)[/tex] would run out before all that [tex]\rm 5.0\; g[/tex] of [tex]\rm Al\, (s)[/tex] was converted to [tex]\rm AlBr_3\, (g)[/tex].

Apply the formula mass of [tex]\rm AlBr_3[/tex] to find the mass of that (approximately) [tex]0.0918\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] formula units:

[tex]\begin{aligned}m(\mathrm{AlBr_3}) &= n(\mathrm{AlBr_3}) \cdot M(\mathrm{AlBr_3}) \\ &= 0.0918\; \rm mol \times 266.698\; g \cdot mol^{-1} \approx 24\; \rm g\end{aligned}[/tex].

Answer:

Mass = 26.53 g

Explanation:

Heat of combustion = −1380.0 kJ/mol

This means 1 mol of the compound releases 1380 kJ

Molar mass = 44.53 g/mol

This means 1 mol of the compound has a mass of 44.53 g

How many grams would release 822kJ..?

First, we have to obtain the number of moles

1 mol = 1380

x = 822

x = 0.5957 moles

Moles = Mass / Molar mass

Mass = Molar mass * moles

Mass = 44.53 * 0.5957

Mass = 26.53 g

Answer:

Rate constants are temperature dependent.

Explanation:

Reaction rate is used to quantify the rate of chemical reaction. There is a relationship between the reaction rate and the half-life of the reaction and the Gibbs free energy of activation, and the reaction rate is temperature dependent according to the equation.

For a reaction shown below

a A + b B ⇒ c C

The rate of reaction of the reaction is given by

[tex]r = k(T) [A]^{m}[B]^{n}[/tex]

where k(T) is the reaction constant, which is seen to be dependent on the temperature of the reaction.

Also, k(T) is numerically equal to

[tex]k(T) = Ae^{\frac{E_{a} }{RT} }[/tex]

where

r = reaction rate

A = pre exponential factor

[tex]E_{a}[/tex] = Activation energy

R = gas constant

T = temperature

and m and n are experimentally determined partial orders in [A] and [B]

Answer:

C. To determine how efficient reactions are.

D. To determine how much reactant they need.

Explanation:

When you are doing a reaction, you are hoping for a percent yield to close of 100%. You make the reaction and determine how many product you obtain. If you know the percent yield of a reaction you can calculate the amount of reactant you need to obtain a determined amount of product.

Having this in mind:

A. To balance the reaction equation.  false. To calculate percent yield you need to balance the reaction before. You don't use percent yield to balance the reaction

B. To determine how much product they will need.  false. You determine how much product you obtain after the reaction. How much product you need is independent of percent yield

C. To determine how efficient reactions are.  true. A way to determine efficience of a reaction is with percent yield. An efficient reaction has a high percent yield.

D. To determine how much reactant they need. true. If you know percent yield of a reaction you can know how many reactant you must add to obtain  the amount of product you want.

Answer:

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

3.52 g of PbCl2

3.76 g of KCl

Explanation:

The equation of the reaction is;

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

Number of moles of Pb(NO3)2 =mass/molar mass 5.06g/331.2 g/mol = 0.0153 moles

Number of moles of KCl= mass/ molar mass= 6.03g/74.5513 g/mol= 0.081 moles

Next we obtain the limiting reactant; the limiting reactant yields the least number of moles of products.

For Pb(NO3)2;

1 mole of Pb(NO3)2 yields 1 mole of PbCl2

Therefore 0.0153 moles of Pb(NO3)2 yields 0.0153 moles of PbCl2

For KCl;

2 moles of KCl yields 1 mole of PbCl2

0.081 moles of KCl yields 0.081 moles ×1/2 = 0.041 moles of PbCl2

Therefore Pb(NO3)2 is the limiting reactant.

Theoretical Mass of precipitate obtained = 0.0153 moles of PbCl2 × 278.1 g/mol = 4.25 g of PbCl2

% yield = actual yield/theoretical yield ×100

Actual yield = % yield × theoretical yield /100

Actual yield= 82.9 ×4.25/100

Actual yield = 3.52 g of PbCl2

If 1 mole of Pb(NO3) reacts with 2 moles of KCl

0.0153 moles of Pb(NO3)2 reacts with 0.0153 moles × 2 = 0.0306 moles of KCl

Amount of excess KCl= 0.081 moles - 0.0306 moles = 0.0504 moles of KCl

Mass of excess KCl = 0.0504 moles of KCl × 74.5513 g/mol = 3.76 g of KCl

Answer:

b. Al3+ + 3Br– → Al + (3/2)Br2

Explanation:

If we look at this reaction closely, aluminum was reduced while bromine was oxidized. The reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Recall that the more negative the redox potential of a chemical specie, the greater its tendency to function as a reducing agent donating electrons in an electrochemical reaction.

However, in this reaction, aluminium is found to accept rather than donate electrons. Therefore, the process is non spontaneous and can only occur in an electrolytic cell, hence the answer.

b. [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]

Electrolytic  Cell v/s Electrochemical Cell:

Out of the given reactions, [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]  is carried out in an electrolytic cell rather than in an electrochemical cell.

As we know, In electrolytic cells, like galvanic cells, are composed of two half-cells

In the given reaction, aluminum is being reduced while bromine gets oxidized and the value of reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Therefore, the process is non spontaneous and can only occur in an electrolytic cell.

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Answer:

0.312g

Explanation:

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of sulphur also contains 6.02x10^23 molecules

1mole of sulphur = 32g

If 1 mole(i.e 32g) of sulphur contains 6.02x10^23 molecules

Then, Xg of sulphur will contain 5.87x10^21 molecules i.e

Xg of sulphur = (32x5.87x10^21)/6.02x10^23 = 0.312g

just saying this is not my work, thank Eduard22sly

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Answer:

The answer to your question is given below.

Explanation:

Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:

Al (13) —› 1s² 2s²2p⁶ 3s²3p¹

The orbital diagram is shown on the attached photo.

Answer: screen shot

Explanation:

The name of the molecule which is given below is 2-pentene.

Alkenes are the organic compounds which are composed of carbon and hydrogen atoms, in which double bond is present.

In the given diagram:

So the compound is 2 pentene.

Hence, 2 pentene is the name of the compound.

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Answer:

Explanation:

The given formula is empirical formula

Let the molecular formula of first be

[tex]( CCl )_n[/tex]

molecular weight = n x ( 12 + 35.5 )

= 47.5 n

Given molecular weight = 189.83 so

47.5 n = 189.83

n = 3.99 or 4 approx

Molecular formula =

[tex]( CCl )_4[/tex]

= C₄ Cl₄

Let the molecular formula of second compound  be

[tex]( C_3H_2N)_n[/tex]

molecular weight = n x ( 3 x 12 +2+14 )

= 52 n

Given molecular weight = 156.23  so

52 n = 156.23

n = 3.0044 or 3 approx

Molecular formula =

[tex]( C_3H_2N )_3[/tex]

= C₉H₆ N₃

Answer:

Approximately [tex]10.88[/tex].

Explanation:

[tex]\rm OBr^{-}[/tex] can act as a weak Bronsted-Lowry base:

[tex]\rm OBr^{-}\; (aq) + H_2O\; (l) \rightleftharpoons HOBr\; (l) + OH^{-}\; (aq)[/tex].

(Side note: the state symbol of [tex]\rm HOBr[/tex] in this equation is [tex]\rm (l)[/tex] (meaning liquid) because [tex]\rm HOBr[/tex] is a weak acid.)

However, the equilibrium constant of this reaction, [tex]K_\text{eq}[/tex], isn't directly given. The idea is to find [tex]K_\text{eq}[/tex] using the [tex]\rm pH[/tex] value at the half-equivalence point. Keep in mind that this system is at equilibrium all the time during the titration. If temperature stays the same, then the same [tex]K_\text{eq}[/tex] value could also be used to find the [tex]\rm pH[/tex] of the solution before the acid was added.

At equilibrium:

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At the half-equivalence point of this titration, exactly half of the base, [tex]\rm OBr^{-}[/tex], has been converted to its conjugate acid, [tex]\rm HOBr[/tex]. Therefore, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should both be equal to one-half the initial concentration of [tex]\rm OBr^{-}[/tex].

As a result, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should be the same. The expression for [tex]K_\text{eq}[/tex] can thus be simplified:

[tex]\begin{aligned}& K_\text{eq} \\&= \frac{\left(\text{half-equivalence $[\rm HOBr\; (l)]$}\right)\cdot \left(\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\right)}{\text{half-equivalence $[\rm OBr^{-}\; (l)]$}}\\ &=\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\end{aligned}[/tex].

In other words, the [tex]K_\text{eq}[/tex] of this system is equal to the [tex]\rm OH^{-}[/tex] concentration at the half-equivalence point. Assume that [tex]\rm p\mathnormal{K}_\text{w}[/tex] the self-ionization constant of water, is [tex]14[/tex]. The concentration of [tex]\rm OH^{-}[/tex] can be found from the [tex]\rm pH[/tex] value:

[tex]\begin{aligned}& \text{half-equivalence $[\rm OH^{-}\; (aq)]$} \\ &= 10^{\rm pH - p\mathnormal{K}_\text{w}}\;\rm mol \cdot L^{-1} \\ &= 10^{7.75 - 14}\; \rm mol \cdot L^{-1}\\ &= 10^{-6.25}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

Therefore, [tex]\begin{aligned} K_\text{eq} &= 10^{-6.25}\end{aligned}[/tex].

Again, since [tex]\rm KOBr[/tex] is a soluble salt, all that [tex]0.200\; \rm M[/tex] of [tex]\rm KOBr[/tex] in this solution will be in the form of [tex]\rm K^{+}[/tex] and [tex]\rm OBr^{-}[/tex] ions. Before any hydrolysis takes place, the concentration of [tex]\rm OBr^{-}[/tex] should be equal to that of [tex]\rm KOBr[/tex]. Therefore:

[tex]\text{$[\rm OBr^{-}\; (aq)]$ before hydrolysis} = 0.200\; \rm M[/tex].

Let the equilibrium concentration of [tex][\rm OH^{-}\; (aq)][/tex] be [tex]x\; \rm M[/tex]. Create a RICE table for this reversible reaction:

[tex]\begin{array}{c|ccccccc} & \rm OBr^{-}\; (aq) &+&\rm H_2O\; (l)& \rightleftharpoons & \rm HOBr\; (l)& + & \rm OH^{-}\; (aq) \\ \textbf{I}& 0.200\; \rm M & & & & 0 \; \rm M & & 0\; \rm M \\ \textbf{C} & -x\; \rm M & & & & +x \; \rm M & & +x\; \rm M \\ \textbf{E}& (0.200 + x)\; \rm M & & & & x \; \rm M & & x\; \rm M \end{array}[/tex].

Assume that external factors (such as temperature) stays the same. The [tex]K_\text{eq}[/tex] found at the half-equivalence point should apply here, as well.

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At equilibrium:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x}[/tex].

Assume that [tex]x[/tex] is much smaller than [tex]0.200[/tex], such that the denominator is approximately the same as [tex]0.200[/tex]:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x} \approx \frac{x^2}{0.200}[/tex].

That should be equal to the equilibrium constant, [tex]K_\text{eq}[/tex]. In other words:

[tex]\displaystyle \frac{x^2}{0.200} \approx K_\text{eq} = 10^{-6.25}[/tex].

Solve for [tex]x[/tex]:

[tex]x \approx 3.35\times 10^{-4}[/tex].

In other words, the [tex]\rm OH^{-}[/tex] before acid was added was approximately [tex]3.35\times 10^{-4}\; \rm M[/tex], which is the same as [tex]3.35\times 10^{-4}\; \rm mol \cdot L^{-1}[/tex]. Again, assume that [tex]\rm p\mathnormal{K}_\text{w} = 14[/tex]. Calculate the [tex]\rm pH[/tex] of that solution:

[tex]\begin{aligned}\rm pH &= \rm p\mathnormal{K}_\text{w} + \log [\mathrm{OH^{-}}] \approx 10.88\end{aligned}[/tex].

(Rounded to two decimal places.)

Answer:

A 50:50 mixture of ethylene glycol and water is effective both summer and winter extremes in temperature because of high boiling point of 106°C and low freezing point of about -37°C. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.

Explanation:

Ethylene glycol or antifreeze is an organic compound which is used in automobile engines as a coolant and also as an anti-freezing agent, however it does not conduct heat effectively as water due to its lower heat capacity. It has a freezing point of -12.9°C and boiling point of 197.3°C.

Water is also used as a coolant in automobile engine but it has a limited range due to its boiling point of 100°C. It is also not a good anti-freezing agent due to it high freezing point of 0.°C

However, when ethylene glycol is mixed with water in a ratio of  50:50, the property of the mixture is enhanced to both serve as a coolant and as an antifreeze. The boiling point is elevated to about 106°C while its freezing point is lowered to about -37°C.

This temperature range is effective for both summer and winter temperatures. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.

Answer:

11445.8years

Explanation:

Half-life of carbon-14 = 5720 years

First we have to calculate the rate constant, we use the formula :

Answer:

C. 1.07 M.

Explanation:

Hello,

In this case, we can define the molarity of the bleach as shown below:

[tex]M=\frac{moles_{NaClO}}{V_{solution}}[/tex]

In such a way, given the mass of bleach in a 1-L solution, we can compute the density:

[tex]\rho = \frac{1100g}{1L}=1100g/L =1.1 kg/L=1.1g/mL[/tex]

In such a way, we can use the previously computed density to compute the volume of the solution, assuming a 100-g solution given the by-mass percent:

[tex]V_{solution}=100g*\frac{1mL}{1.1g} *\frac{1L}{1000mL} =0.091L[/tex]

Afterwards, using the by-mass percent of bleach we compute the mass:

[tex]m_{NaClO}=100g*0.0725=7.25g[/tex]

And the moles:

[tex]n_{NaClO}=7.25g*\frac{1mol}{74.44g} =0.097mol[/tex]

Therefore, the molarity turns out:

[tex]M=\frac{0.097mol}{0.091L}\\ \\M=1.07M[/tex]

Thus, answer is C. 1.07 M.

Regards.

Answer:

(B)

Explanation:

edg 2020

The four seasons in earth is originating from the earth's revolution around the sun. Therefore, the most suitable model for Amanda is  Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season.

Seasons in earth is originating from the difference in the distance from the sun over each time period. Hence, revolution of earth around sun make these seasons.

The time period at which earth comes closer to sun more hot will be earth's surface and we experience summer season. When we far from sun winter season occurs.

Therefore, different season are coming based on the distance of earth from sun at each revolution point. This is also affected by the tilt of earth's in its own axis.

Hence, different earth models with different distance from sun is most suitable model here for Amanda. Thus option B is correct.

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Answer:

Option B. 30 KJ.

Explanation:

The following data were obtained from the question:

Temperature (T) = 5000 K

Enthalpy change (ΔH) = – 220 kJ/mol

Change in entropy (ΔS) = – 0.05 KJ/mol•K

Gibbs free energy (ΔG) =...?

The Gibbs free energy, ΔG can be obtained by using the following equation as illustrated below:

ΔG = ΔH – TΔS

ΔG = – 220 – (5000 x – 0.05)

ΔG = – 220 – (– 250)

ΔG = – 220 + 250

ΔG = 30 KJ

Therefore, the Gibbs free energy, ΔG is 30 KJ.

Answer:

58.45g is the answer

Explanation:

took the test

The mass of chromium that can be deposited is equal to 58.45 g. Therefore, option B is correct.

For a steady flow of charge through a conductor, the current can be determined with the following equation:

[tex]{\displaystyle I={Q \over t}[/tex]

where Q is the electric charge while time t. If Q and t are measured in coulombs (C) and seconds then I will be in amperes.

Electric charge flows by electrons, from lower potential to higher electrical potential. Any stream of charged objects can constitute an electric current.

Given, the amount of electric current flowing through the solution:

I = 45.2 A

The time for which the current flows, t = 2 hrs = 2 × 60 ×60 = 7200 sec

The charge flowing through the solution, Q = I × t

Q = 45.2 × 7200

Q = 325440 C

The number of moles of electrons in 325440 C charge = 325440/96500 = 3.37 mol

We know Cr³⁺ + 3e⁻ →  Cr (s)

3 moles of electrons deposit of chromium  = 1 mol

3.37 mol of electrons deposit of chromium  = 3.37/3 = 1.12 mol

The mass of chromium in 1.12 mol = 1.12 × 52 = 58.45 g

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