10. What effect does temperature have on molecular motion? Using this explanation, explain why both pressure and volume can decrease with decreasing temperature.

Answer:

b

Explanation:

The reaction that is not a displacement reaction from all the options is [tex]C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}[/tex]

In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.

Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.

Correct option = b.

The reaction given in Option A is not a displacement reaction. In Displacement reaction functional group of one reactant is replaced by the functional group of the another reactant.

Displacement reaction:

In this reaction functional group of one reactant is replaced by the functional group of the another reactant.

[tex]\bold { Zn(s) + FeCl_2(aq) \rightarrow ZnCl_2(aq) + Fe(s).}[/tex]

In the above reaction Zinc does not any functional group to exchange with iron chloride.

Therefore, the reaction given in Option A is not a displacement reaction.

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Answer:

-55.9kJ/mol is the change in enthalpy of the reaction

Explanation:

In the reaction:

HCl(aq) + NaOH(aq) → H₂O(l) + NaCl

Some heat is released per mole of reaction.

To know how many moles reacts we need to find limiting reactant:

Moles HCl = 0.050L ₓ (1.27mol /  L) = 0.0635 moles HCl

Moles NaOH = 0.050L ₓ (1.32mol /  L) = 0.066 moles NaOH

As there are more moles of NaOH than moles of HCl, HCl is limiting reactant and moles of reaction are moles of limiting reactant, 0.0635 moles

Using the coffee-cup calorimeter equation we can find how many heat was released thus:

Q = C×m×ΔT

Where Q is heat released, C is specific heat of the solution (4.18J/g°C), m is mass of solution (100g because there are 100mL of solution -50.0mL of HCl and 50.0mL of NaOH- and density is 1g/mL) and ΔT is change in temperature (8.49°C)

Replacing:

Q = 4.18J/g°C×100g×8.49°C

Q = 3548.8J of heat are released in the reaction

Now, change in enthalpy, ΔH, is equal to change in heat (As is released heat ΔH < 0) per mole of reaction, that is:

ΔH = Heat / mol of reaction

ΔH = -3548.8J / 0.0635 moles of reaction

Negative because is released heat.

ΔH = -55887J / mol

ΔH =

The heat of reaction is  -54.7 kJ/mol.

The equation of the reaction is;

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Number of moles of HCl = 50/1000 L × 1.27 M = 0.064 moles

Number of moles of NaOH = 50/1000 L × 1.32 M = 0.066 moles

The limiting reactant is HCl

Total volume of solution = 100mL

Total mass of solution = 100 g

Temperature rise = 8.49°C

Heat capacity of solution = 4.18 J/g⋅°C

Using;

H = mcdT

m = mass of solution

c = heat capacity of solution

dT = temperature rise

H = 100 g ×  4.18 J/g⋅°C × 8.49°C = 3548.82 J

The heat of reaction = -ΔH/n = -(3.5kJ/0.064 moles)

= -54.7 kJ/mol

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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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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:

24.6g of CO₂ is theoretical yield

Explanation:

The reaction of 8.00g of octane with 38.9g of oxygen.

The reaction of octane with oxygen is:

C₈H₁₈(l) + 25/2O₂ → 9H₂O + 8CO₂

1 mole of octane reacts with 25/2 moles of oxygen to produce 8 moles of CO₂

Theoretical yield is the amount of carbon dioxide formed assuming a yield of 100%. To calculate theoretical yield, first, we need to find limiting reactant and, with the chemical reaction, we can obtain the theoretical moles of CO₂ produced and its mass to obtain theoretical yield.

Limiting reactant:

Moles octane (Molar mass: 114.23g/mol) in 8.00g:

8.00g × (1mol / 114.23g) = 0.0700 moles octane.

Moles oxygen (Molar mass: 32g/mol) in 38.9g:

38.9g × (1mol / 32g) = 1.2156 moles oxygen.

For a complete reaction of 1.2156 moles of O₂ there are necessaries:

1.2156 moles O₂ ₓ (1mol C₈H₁₈ / 25/2 moles O₂) = 0.0973 moles octane

As we have just 0.0700 moles,

Moles and mass of carbon dioxide:

As limiting reactant is octane, 0.0700 moles of C₈H₁₈ will produce:

0.0700mol C₈H₁₈ × (8 moles CO₂ / 1 mol C₈H₁₈) = 0.56 moles of CO₂ are theoretically produced. In mass (Molar mass CO₂ = 44.01g/mol):

0.56moles CO₂ × (44.01g / mol) =

-Theoretical yield because we are assuming all octane is reacting. In real life, never happens like that-

Answer:

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

Explanation:

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:

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:

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:

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:

c = 250.58 J/kg/[tex]^{0}C[/tex]

Explanation:

The specific heat of a substance is the required quantity of heat to increase or decrease the temperature of its unit mas by 1 kelvin.

Q = mcΔθ

where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is its specific heat and Δθ is the change in temperature of the substance.

Given that; m = 1.0 kg, Q = 1303 J and Δθ = 5.2 [tex]^{0}C[/tex], then;

c = Q ÷ (mΔθ)

  = 1303 ÷ (1.0 × 5.2)

  = 1303 ÷ 5.2

  = 250.58 J/kg/[tex]^{0}C[/tex]

The specific heat of the object is 250.58 J/kg/[tex]^{0}C[/tex].

Answer:

0.25

Explanation:

Answer:

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

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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#SPJ2

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:

[tex]\boxed{Heptene}[/tex]

Explanation:

Double Bond => An Alkene molecule

So, the suffix will be "-ene"

7 Carbons => So, we'll use the prefix "Hept-"

Combining the suffix and prefix, we get:

=> Heptene

Answer:

[tex]\boxed{\mathrm{Heptene}}[/tex]

Explanation:

Alkenes have double bonds. The molecule has one double bond.

Suffix ⇒ ene

The molecule has 7 carbon atoms and 14 hydrogen atoms.

Prefix ⇒ Hept (7 carbons)

The molecule is Heptene.

[tex]\mathrm{C_7H_{14}}[/tex]

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:

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:

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: 9.08 L

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles[/tex]

[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]

According to stoichiometry :

4 moles of [tex]Al[/tex] require  = 3 moles of [tex]O_2[/tex]

Thus 0.54 moles of [tex]Al[/tex] will require=[tex]\frac{3}{4}\times 0.54=0.405moles[/tex]  of [tex]O_2[/tex]

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atm respectively.

According to the ideal gas equation:

[tex]PV=nRT[/tex]

P = Pressure of the gas = 1 atm

V= Volume of the gas = ?

T= Temperature of the gas = 273 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 0.405

[tex]V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L[/tex]

Thus 9.08 L of [tex]O_2[/tex] at STP would be required

Considering the reaction stoichiometry and STP conditions, 9.072 L of O₂ at STP would be required.  

The balanced reaction is:

4 Al + 3 O₂ → 2 Al₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Being 27 g/mole the molar mass of Al, this is the amount of mass that a substance contains in one mole, then if 14.6 grams Al are reacted,   the number of moles of Al that react is calculated as:

[tex]14.6 gramsx\frac{1 mole}{27 grams}= 0.54 moles[/tex]

Then you can apply the following rule of three: if by stoichiometry 4 moles of Al react with 3 moles of O₂, 0.54 moles of Al react with how many moles of O₂?

[tex]amount of moles of O_{2} =\frac{0.54 moles of Alx3 moles of O_{2} }{4 moles of Al}[/tex]

amount of moles of O₂= 0.405 moles

On the other side, the STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

Then you can apply the following rule of three: if by definition of STP 1 mole of O₂ occupies 22.4 L, 0.405 moles of O₂, how much volume does it occupy?

[tex]volume=\frac{0.405 moles of O_{2}x22.4 L }{1 mole of O_{2} }[/tex]

volume= 9.072 L

Finally, 9.072 L of O₂ at STP would be required.  

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Explanation:

The mass to charge ratio =136

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:

Oxygen and Chlorine

Explanation:

Covalent bonds involve the sharing of electrons between nonmetals.

Answer:

oxygen (O) and chlorine (Cl)

Explanation:

cuz i said so

Answer:

1. False

2. True

3. True

4. True

5. True

6. True

7. True

8. False

9. True

Explanation:

Density is a physical property since its measurement does not involve any chemical process.

Since transition elements exhibit variable oxidation states, the actual oxidation state of the transition element must be specified in the compound.

Due to the fact that neutron has no charge, it was discovered by Chadwick long after the electron and proton were discovered.

The balancing of chemical reaction equations is a demonstration that atoms are neither created no destroyed. It also shows that mass is neither created nor destroyed in chemical reactions.

When a gas is heated, it expands. Its volume and its kinetic energy increases. Since volume and pressure are inversely proportional (Boyle's law) the pressure decreases.

Enthalpy is said to be an extensive property. This implies that the magnitude of change in enthalpy is known to depend on the amount of reactants that is actually reacted.

The combined gas law is applicable to all ideal gas systems irrespective of their individual chemical formulas.

10g of CO2 contains 0.227 moles of CO2 while 10g of NO contains 0.33 moles of NO hence 10.0 g of NO will contain more molecules than 10.0g of CO2.

If a sample is not at absolute zero, the particles are known to possess kinetic energy which decreases continuously until absolute zero is attained.

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:

(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:

[tex]M=0.763\frac{mol}{L}=0.763M[/tex]

Explanation:

Hello,

In this case, as the osmotic pressure (π) is widely known as a colligative property, we can see that the solution in this case is formed by water and tree sap, that is mathematically defined by:

[tex]\pi =iMRT[/tex]

Thus, since tree sap is a covalent substance that is nonionizing, we can infer its van't Hoff factor to be 1, therefore, for the given osmotic pressure and temperature, we can compute the molar concentration (in molar units mol/L) as follows:

[tex]M=\frac{\pi }{RT} =\frac{19.1atm}{0.082\frac{atm*L}{mol*K}*(32+273.15)K} \\\\M=0.763\frac{mol}{L}=0.763M[/tex]

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.

Learn more about Concentrations, here:

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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:

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:

5.643 kJ

Explanation:

The quantity of heat released or absorbed by a substance (Q) is given by the equation:

Q = mcΔT

Where m is the mass of the substance, c is the specific heat of substance and ΔT is the difference between the final temperature and the initial temperature.

Given that:

m = 45 g, Final temperature = 48°C, Initial temperature = 18°C, c = specific heat of water = 4.18 J/g°C

ΔT = Final temperature - Initial temperature = 48°C - 18°C = 30°C

The quantity of heat is:

Q = mcΔT = 45 g × 4.18 J/g°C × 30°C = 5643 J

Q = 5.643 kJ