The vapor pressure of pure water at 250C is 23.77 torr. What is the vapor pressure of water above a solution that is 1.500 m glucose, C6H12O6?

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.

Answer:

[tex]255.71~g~C_2H_5OH[/tex]

Explanation:

First, we have to check if the reaction is balanced:

[tex]C_6H_1_2O_6_(_s_)~->~2C_2H_5OH_(_l_)~+~2CO_2_(_g_)[/tex]

We have 6 carbon atoms on both sides, 12 hydrogens, and 6 oxygens. So, we can continue with the problem. If we want to calculate the mass of ethanol ([tex]C_2H_5OH[/tex]) we need to know:

1) Molar mass of glucose ([tex]C_6H_12O_6[/tex])

In this case, we have to know the atomic mass of each atom:

-) C 12 g/mol

-) O 16 g/mol

-) H 1 g/mol

With the formula we can calculate the molar mass:

(12*6) + (16*6) + (1*12) = 180 g/mol

2) Molar ratio between glucose and ethanol

In the balanced equation we have 1 mol of [tex]C_6H_12O_6[/tex] and 2 moles of [tex]C_2H_5OH[/tex]. So the molar mass is 1:2

3) Molar mass of ethanol ([tex]C_2H_5OH[/tex])

With the formula, we can calculate the molar mass:

(12*2) + (6*1) + (16*1) = 46 g/mol

Finally, we can to the calculation:

[tex]500.0g~C_6H_12O_6\frac{1mol~C_6H_12O_6}{180g~C_6H_12O_6}\frac{2mol~C_2H_5OH}{1mol~C_6H_12O_6}\frac{46g~C_2H_5OH}{1mol~C_2H_5OH}=255.71g~C_2H_5OH[/tex]

I hope it helps!

The branch of science which deals with chemicals and bonds is called chemistry.

The correct answer for the question is 255g of ethanol.

The process of digesting the glucose in absence of oxygen is called fermentation.

Before solving the question, we must check the atoms of compound and balanced it.

The molar mass of the glucose is as follows:

After balancing the equation, the mole ratio of the glucose vs ethanol is 1:2

The molar mass of the ethanol is as follows:-

[tex]C_2H_5OH[/tex] =[tex]12*2) + (6*1) + (16*1) = 46 g/mol[/tex]

After diving each molar mass of each compound with respect to its mole ratio. The mass of the ethanol produced is 255g.

Hence, the correct answer is 255g.

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Answer:

Photosynthesis

Explanation:

Photosynthesis is a process by  which photosynthetic organisms use the energy of captured sunlight to convert energy-poor molecules such as carbon (iv), CO₂ and water (H₂O) into energy-rich organic molecules sch as carbohydrates e.g. glucose.

Photosynthesis occurs in a variety of bacteria  and algae as well in vascular plants. The overall equation for the reaction of photosynthesis is as follows:

CO₂ + H₂O + light------->  (CH₂O) + O₂

It is a redox reaction in which  water donates electrons (as hydrogen) for the reduction of  CO₂ to carbohydrate, (CH₂O).

Carbohydrates are energy-rich molecules which serves as energy sources for many living organisms.

Answer:

the mass of FeSO4.7H2O  in  the 2.810 g sample was 1.5402 g

Explanation:

From the given information:

Two  moles of FeSO4.7H2O = one mole of  Fe2O3

Let recall that:

number of moles of  Fe2O3 = mass of  Fe2O3 / molar mass of  Fe2O3

Given that :

mass of Fe2O3 = 0.443 g

number of moles of  Fe2O3 = 0.443 g/ 159.69 g/mol

number of moles of  Fe2O3 = 0.00277 mol

Thus;

number of moles of FeSO4.7H2O = 2 ×  Fe2O3

number of moles of FeSO4.7H2O = 2 × 0.00277 mol

number of moles of FeSO4.7H2O = 0.00554 mol

However from the usual stoichiometry formula; the mass of a substance = number of moles  × molar mass

Now; the mass of FeSO4.7H2O = number of moles × molar mass

the mass of FeSO4.7H2O = 0.00554 mol × 278.01 g/mol

the mass of FeSO4.7H2O = 1.5402 g

Therefore; the mass of FeSO4.7H2O  in  the 2.810 g sample was 1.5402 g

Answer:

100 g of water: specific heat of water 4.18 J/g°C

Explanation:

To know the correct answer to the question, we shall determine the temperature change in each case.

For 100 g of water:

Mass (M) = 100 g

Specific heat capacity (C) = 4.18 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 100 x 4.18 x ΔT

Divide both side by 100 x 4.18

ΔT = 55000/ (100 x 4.18)

ΔT = 131.6 °C

Therefore the temperature change is 131.6 °C

For 50 g of water:

Mass (M) = 50 g

Specific heat capacity (C) = 4.18 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 50 x 4.18 x ΔT

Divide both side by 50 x 4.18

ΔT = 55000/ (50 x 4.18)

ΔT = 263.2 °C

Therefore the temperature change is 263.2 °C

For 50 g of lead:

Mass (M) = 50 g

Specific heat capacity (C) = 0.128 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 50 x 0.128 x ΔT

Divide both side by 50 x 0.128

ΔT = 55000/ (50 x 0.128)

ΔT = 8593.8 °C

Therefore the temperature change is 8593.8 °C.

For 100 g of iron:

Mass (M) = 100 g

Specific heat capacity (C) = 0.449 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 100 x 0.449 x ΔT

Divide both side by 100 x 0.449

ΔT = 55000/ (100 x 0.449)

ΔT = 1224.9 °C

Therefore the temperature change is 1224.9 °C.

The table below gives the summary of the temperature change of each substance:

Mass >>> Substance >> Temp. Change

100 g >>> Water >>>>>> 131.6 °C

50 g >>>> Water >>>>>> 263.2 °C

50 g >>>> Lead >>>>>>> 8593.8 °C

100 g >>> Iron >>>>>>>> 1224.9 °C

From the table given above we can see that 100 g of water has the smallest temperature change.

Answer:

Here's what I get  

Explanation:

pKₐ of o-vanillin = 7.81; pKₐ of p-toluidine = 4.44

The higher the pKₐ, the weaker the acid.

Thus, o-vanillin is the weaker acid and has a stronger conjugate base.

The conjugate acid of p-toluidine is the stronger and has the weaker conjugate base.

The equation for the equilibrium is

H-OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₂ ⇌ ⁻OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₃⁺

    weaker acid              weaker base          stronger  base        stronger acid

The reaction between the stronger acid and the stronger base pushes the position of equilibrium to the left.

Thus, o-vanillin does not fully protonate p-toluidine.

O-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base; hence, o-vanillin does not fully protonate p-toluidine.

The pKa is defined as the negative logarithm of Ka. The dissociation constant of an acid Ka shows the extent of dissociation of an acid in solution. The higher the pKa, the lower the Ka and the weaker the acid.

The pKₐ of o-vanillin is 7.81 while the pKₐ of p-toluidine is 4.44. This means that  o-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base. Hence, o-vanillin does not fully protonate p-toluidine.

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Answer:

Combination of H2N(CH2)8NH2 and HOOC(CH2)6COOH leads to the loss of water molecules at each linkage position.

Explanation:

A condensation polymer is a polymer formed when two monomers combine with the elimination of a small molecule such as water. The removal of the small molecule occurs at the point where the two monomers are joined to each other.

Nylon is known to form condensation polymers. This is because it involves the linkage of an -OH group to an -NH2 group. Water is eliminated in the process.

In the case of H2N(CH2)8NH2 and HOOC(CH2)6COOH, linkage of the both monomers at the 8 position of each chain leads to the formation of nylon- 8,8 with loss of water molecules at each linkage position. This stepwise loss of water molecules at each linkage makes it a condensation polymer.

Answer:

The answers to your questions are given below

Explanation:

The following data were obtained from the question:

CH3COOH → CH3C00^- + H^+

Equilibrium constant, Ka = 1.8 x 10^-5

AgCl → Ag^+ + Cl^-

Equilibrium constant, Ksp = 1.6 x 10^-10

Al(OH)3 → Al^3+ + 3OH^-

Equilibrium constant, Ksp = 3.7 x 10^-15

A+B → C

Equilibrium constant, K = 4.9 x 10^3

When the value of the equilibrium constant is grater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.

When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.

When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.

Now, we shall the question given above as follow:

A. CH3COOH → CH3C00^- + H^+

Equilibrium constant, Ka = 1.8 x 10^-5

Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.

B. AgCl → Ag^+ + Cl^-

Equilibrium constant, Ksp = 1.6 x 10^-10

Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.

C. Al(OH)3 → Al^3+ + 3OH^-

Equilibrium constant, Ksp = 3.7 x 10^-15

Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.

D. A+B → C

Equilibrium constant, K = 4.9 x 10^3

Since the value of the equilibrium constant is greater than 1, it means that the forward reaction is favored.

The reaction conditions are:

A. The reverse reaction is favored.

B. The reverse reaction is favored.

C. The reverse reaction is favored.

D. The forward reaction is favored.

Chemical reaction:

A. [tex]CH_3COOH[/tex] → [tex]CH_3COO^- + H^+[/tex]

Equilibrium constant, Ka = [tex]1.8 * 10^{-5}[/tex]

B. [tex]AgCl[/tex] → [tex]Ag^+ + Cl^-[/tex]

Equilibrium constant, Ksp = [tex]1.6 * 10^{-10}[/tex]

C. [tex]Al(OH)_3[/tex] → [tex]Al^{3+} + 3OH^-[/tex]

Equilibrium constant, Ksp = [tex]3.7 * 10^{-15}[/tex]

D. A+B → C

Equilibrium constant, K = [tex]4.9 * 10^3[/tex]

When the value of the equilibrium constant is greater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.

When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.

When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.

Thus, the reactions will be:

A. The reverse reaction is favored.

B. The reverse reaction is favored.

C. The reverse reaction is favored.

D. The forward reaction is favored.

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Answer:

The correct answer is 16.61 grams methanol and 57.38 grams water.

Explanation:

The mole fraction (X) of methanol can be determined by using the formula,  

X₁ = mole number of methanol (n₁) / Total mole number (n₁ + n₂)

X₁ = n₁/n₁ + n₂ = 0.14

n₁ / n₁ + n₂ = 0.14 ---------(i)

n₁ mole CH₃OH = n₁ mol × 32.042 gram/mol (The molecular mass of CH₃OH is 32.042 grams per mole)

n₁ mole CH₃OH = 32.042 n₁ g

n₂ mole H2O = n₂ mole × 18.015 g/mol  

n₂ mole H2O = 18.015 n₂ g

Thus, total mole number is,  

32.042 n₁ + 18.015 n₂ = 74 ------------(ii)

From equation (i)

n₁/n₁ + n₂ = 0.14

n₁ = 0.14 n₁ + 0.14 n₂

n₁ - 0.14 n₁ = 0.14 n₂

n₁ = 0.14 n₂ / 1-0.14

n₁ = 0.14 n₂/0.86 ----------(iii)

From eq (ii) and (iii) we get,  

32.042 × 0.14/0.86 n₂ + 18.015 n₂ = 74

n₂ (32.042 × 0.14/0.86 + 18.015) = 74

n₂ = 74 / (32.042 × 0.14/0.86 + 18.0.15)

n₂ = 3.1854 mol

From equation (iii),  

n₁ = 0.14/0.86 n₂

n₁ = 0.14/0.86 × 3.1854

n₁ = 0.5185 mol

Now, presence of water in the mixture is,  

= 3.1854 mole × 18.015 gram per mole  

= 57.38 grams

Methanol present in the mixture is,  

= 0.5185 mol × 32.042 gram per mole

= 16.61 grams

Answer:

The correct answer is 51.2 degree C.

Explanation:

The standard enthalpy for H₂SO₄ (l) is -814 kJ/mole and the standard enthalpy for H₂SO₄ (aq) is -909.3 kJ/mole.  

Now the dHreaction = dHf (product) - dHf (reactant)  

= -909.3 - (-814)

dHreaction or q = -95.3 kJ of energy will be used for dissociating one mole of H₂SO₄.  

The heat change in calorimetry can be determined by using the formula,  

q = mass * specific heat capacity * change in temperature -----------(i)

Based on the given information, the density of H₂SO₄ is 1.060 g/ml

The volume of H₂SO₄ is 1 Liter

Therefore, the mass of H₂SO₄ will be, density/Volume = 1.060 g/ml / 1 × 10⁻³ ml = 1060 grams

The initial temperature given is 25.2 degrees C, or 273+25.2 = 298.2 K, let us consider the final temperature to be T₂.  

ΔT = T₂ -T₁ = T₂ - 298.2 K

Now putting the values in equation (i) we get,  

95.3 kJ = 1060 grams × 3.458 j/gK (T₂ - 298.2 K) (the specific heat capacity of the final solution is 3.458 J/gK)

(T₂ - 298.2 K) = 95300 J / 1060 × 3.458 = 26 K

T₂ = 298.2 K + 26 K

T₂ = 324.2 K or 324.2 - 273 = 51.2 degree C.  

Answer:

Explanation:

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)

In this reaction we see that 3 moles of O₂ reacts with one mole of C₂H₄ .

Hence rate of disappearance of O₂ is 3 times faster .

- d [O₂] / dt = - 3 d [ C₂ H₄ ] / dt

Putting the given value

.23 = 3 d [ C₂ H₄ ] / dt

d [ C₂ H₄ ] / dt  = .23 / 3

= .077 M / s

Hence the rate of disappearance of C₂ H₄ is .077 moles / s .

The given question is incomplete, the complete question is:

Identify the solutions that will form a precipitate when mixed with aqueous barium chloride, BaCl2 (aq). Select all that apply. potassium carbonate, K2CO3 (aq) silver nitrate, AgNO3(aq) sulfuric acid, H2SO4 (aq) sodium hydroxide, NaOH(aq) sodium chloride, NaCl (aq) copper(II) nitrate, Cu(NO3)2 (aq)

Answer:

The aqueous solution of BaCl2 form precipitate with potassium carbonate, silver nitrate, sulfuric acid and sodium hydroxide.

Explanation:

The formation of an insoluble salt, which takes place when two solutions comprising soluble salts are reacted with each other, the reaction is termed as precipitation reaction. The insoluble salt produced in the given reaction is termed as the precipitate.  

BaCl₂ reacts with K₂CO₃ to produce white precipitate in the form of BaCO₃,  

BaCl₂ (aq) + K₂CO₃ (aq) ⇒ BaCO₃ (s) + 2KCl (aq)

BaCl₂ reacts with H₂SO₄ to produce white precipitate in the form of BaSO₄,  

BaCl₂ (aq) + H₂SO₄ (aq) ⇒ BaSO₄ (s) + 2HCl (aq)

BaCl₂ (aq) reacts with NaOH to produce white precipitate in the form of Ba(OH)2,  

BaCl₂ (aq) + 2NaOH (aq) ⇒ Ba(OH)2 (s) + 2NaCl (aq)

BaCl₂ reacts with AgNO₃ to produce white precipitate in the form of 2AgCl (s),  

BaCl₂ (aq) + 2AgNO₃ (aq) ⇒ Ba(NO₃)₂ (aq) + 2AgCl (s)

Upon reaction of BaCl₂ with Cu(NO₃)₂ no formation of precipitate takes place, and BaCl₂ does not react with NaCl.  

Answer:

d. increases PFK activity, decreases FBPase activity

Explanation:

Fructose-2,6-bisphophate is formed by the phosphorylation of fructose-6-phosphate catalyzed by phosphofructokinase-2, PFK-2.

Fructose-2,6-bisphophate functions as an allosteric effector of the enzymes phosphofructokinase-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase.

Fructose-2,6-bisphophate has opposite effects on the enzymes, PFK-1 and FBPase. When it binds to the allosteric site of the enzyme, PFK-1, it increases the enzymes's activity by increasing its affinity for its substrate fructose-6-phosphate and reduces its affinity for its allosteric inhibitors ATP and citrate. However, when it binds to FBPase, it reduces its activity by reducing its affinity for glucose, its substrate

Answer:

El volumen del gas era 12.95 L

Explanation:

Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:

“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”

La ley de Boyle se expresa matemáticamente como:  P*V=k

Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:  

[tex]\frac{V}{T}=k[/tex]

Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:

[tex]\frac{P}{T}=k[/tex]

Combinado las mencionadas tres leyes se obtiene:

[tex]\frac{P*V}{T} =k[/tex]

Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:

[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]

Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:

Reemplazando:

[tex]\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}[/tex]

Resolviendo:

[tex]V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}[/tex]

V2= 12.95 L

El volumen del gas era 12.95 L

Answer:

Explanation:

a ) one mole = 6.02 x 10²³ atoms

no of moles in given no of atoms

= 8 x 10⁹ / 6.02 x 10²³

= 1.329 x 10⁻¹⁴ moles .

b ) one mole of calcium = 40 gram

102 .5 g calcium

= 102 .5 / 40 moles

= 2.5625 moles

c )

no of moles in 5.04 g lead = 5.04 / 207

= 2.4347 x 10⁻² moles

= 2.4347 x 10⁻²x 6.02 x 10²³ no of atoms of lead

= 14.6568 x 10²¹ no of atoms .

d)  

one mole = 6.02 x 10²³ atoms

2.85 mole = 17.157 x 10²³ atoms .

e )

moles of fluorine gas = 1.08 x 10³ / 6.02 x 10²³

= .1794 x 10⁻²⁰ moles

mass in grams =  .1794 x 10⁻²⁰ x 38

= 6.8172 x 10⁻²⁰ grams

f )

no of moles in .584 g of benzene = .584 / 78

= 7.487 x 10⁻³ moles

no of molecules = 6.02 x 10²³ x  7.487 x 10⁻³

= 45.07 x 10²⁰ molecules .

g )

moles of atoms = 5.09 x 10⁹ / 6.02 x 10²³

= .8455 x 10⁻¹⁴ moles

mass in gram = .8455 x 10⁻¹⁴ x 1

=  .8455 x 10⁻¹⁴ g

h )

.45 moles of Ca₃PO₄ = .45 x 6.02 x 10²³ molecules

= 2.709 x 10²³ molecules of Ca₃PO₄

no of atoms of Ca = 3 x 2.709 x 10²³

= 8.127 x 10²³ atoms of Ca .

Answer:

[tex]Sc_2O_3[/tex] reacts with an acidic solution

Explanation:

Scandium Oxide [tex]Sc_2O_3[/tex] is a basic metal oxide which therefore reacts with acidic solution. An oxide is  a compound that contains only two elements, one of which is oxygen .

The objective of this question is to Write a balanced chemical equation to support your answer.

The chemical equation to support the reaction of [tex]Sc_2O_3[/tex] with acidic solution is as follows:

Assuming the acidic solution to be HCl

[tex]\mathbf{Sc_2O_3_{(s)} + 6 HCl_{(aq)} ----> 2 ScCl_{3(aq)} + 3H_2O_{(l)}}[/tex]

The ionic equation :

[tex]\mathbf{Sc_2O_{3(s)} + 6H^+_{(aq)} ---> 2Sc^{3+}_{(aq)} + 3H_2O_{(l)}}[/tex]

Answer:

pH = 7.37

Explanation:

The buffer H₂PO₄⁻/HPO₄²⁻ has as pKa 7.21. To find the pH of a buffer you can use H-H equation:

pH = pKa + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]

Where [HPO₄²⁻] and [H₂PO₄⁻] are molar concentrations of each species. As volume is 1.00L, [HPO₄²⁻] and [H₂PO₄⁻] are MOLES.

Moles of 25.0g of KH₂PO₄ (Molar mass: 136.086g/mol):

25.0g KH₂PO₄ ₓ (1mol / 136.086g) = 0.1837 moles KH₂PO₄ = moles H₂PO₄⁻

Moles of 38.0g of Na₂HPO₄ (Molar mass: 141.96g/mol):

38.0g KH₂PO₄ ₓ (1mol / 141.96g) = 0.2677 moles Na₂HPO₄ = moles HPO₄²⁻

Replacing in H-H equation:

pH = pKa + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]

pH = 7.21 + log₁₀ [0.2677] / [0.1837]

Answer:

It will not achieve the desired separation

Explanation:

Chromatography is a separation method that involves the use of a stationary phase and a mobile phase. The stationary phase is immobile, in the particular instance of this question, the stationary phase is paper. The mobile phase is the appropriate solvent, in this case, a salt-water solution.

If the level of solvent is above the dye spots, it will introduce error into the separation. The solvent (if volatile) may evaporate without drawing up and separating the solute. Secondly, the solvent may simply dissolve the spots without achieving any meaningful separation of the components in the system. This second reason is particularly why the salt solution must be below the dye spots in this chromatographic separation.

Answer:

Your answer will either be 22.9897 or 22.990 !!

Explanation:

Answer:

0.900 V

Explanation:

Oxidation half cell;

2Al(s) -----> 2Al^3+(aq) + 6e

Reduction half equation;

3Zn^2+(aq) + 6e ----> 3Zn(s)

E°anode = -1.66V

E°cathode= -0.76 V

E°cell= E°cathode - E°anode

E°cell= -0.76-(-1.66)

E°cell= 0.900 V

Answer:

3.1 L

Explanation:

Step 1: Given data

Step 2: Calculate the moles of oxygen

The molar mass of oxygen is 32.00 g/mol.

[tex]3.1g \times \frac{1mol}{32.00g} = 0.097mol[/tex]

Step 3: Calculate the volume of the container

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 0.097 mol × (0.0821 atm.L/mol.K) × 390 K / 1.00 atm

V = 3.1 L

Answer:

Produces Wind Currents

Explanation:

Answer:

produces wind currents

Explanation:

i just took the test and got it right :}

Answer:

Poop Buttt.

Explanation:

Answer:

Hydroxyl

Explanation:

A hydroxyl group is a functional group that attaches to some molecules containing an oxygen and hydrogen atom, bonded together. Also spelled hydroxy, this functional group provides important functions to both alcohols and carboxylic acids.

The functional groups are the part of the organic chemistry that confers the characteristic feature of a molecule. The molecule has a hydroxyl group in its structure. Thus, option C is correct.

Hydroxyl functional groups are the atoms or molecules that provide a distinctive property to a compound. It has a chemical formula of -OH that has oxygen covalently bonded to the hydrogen atom.

The hydroxyl group is called the alcohol group that is seen in methanol, ethanol, propanol, etc. The presence of hydrogen allows the compound to form a water bond with other molecules and makes them soluble and polar.

Therefore, option C. the molecule has a hydroxyl or alcoholic functional group attached to its carbon atom.

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Answer:

The new volume will be 808 L

Explanation:

Charles's law is a law that says that the volume of gas at constant pressure is directly proportional to its absolute temperature (in degrees Kelvin), that is, when the amount of gas and pressure are kept constant, the quotient between volume and temperature will always have the same value:

[tex]\frac{V}{T}=k[/tex]

Having a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment, by varying the volume of gas to a new value V2, the temperature will change to T2 and the following will be fulfilled:

[tex]\frac{V1}{T1} =\frac{V2}{T2}[/tex]

In this case:

Replacing:

[tex]\frac{625 L}{273 K} =\frac{V2}{353 K}[/tex]

Solving:

[tex]V2=353 K*\frac{625 L}{273 K}[/tex]

V2= 808 L

The new volume will be 808 L

Answer:

This description shows a methyl group.

Explanation:

Answer:

ΔH =-47.3kJ

Explanation:

You must know standard enthalpy is defined as change of enthalpy during the formation of 1 mole of the substance from its constituent elements

You can find the standard enthalpy of any reaction from the sum of another similar reactions (Hess's law) as follows:

For methylamine, CH₃NH₂ is:

C(s) + 5/2 H₂(g) + 1/2 N₂(g) → CH₃NH₂ (l)

1. C(s) + O₂(g) → CO₂(g); ΔH=-393.5 kJ

2. 2H₂O(l) → 2H₂(g) + O₂(g); ΔH=571.6 kJ

3. 1/2N₂(g) + O₂(g) → NO₂(g); ΔH=33.10kJ

4. 4CH₃NH₂(l) + 13O₂(g) → 4CO₂(g) + 4NO₂(g) + 10H₂O(l); ΔH=-4110.4 kJ

The sum of (1)+(3) produce:

C(s) + 2O₂(g) + 1/2N₂(g) → CO₂(g) + NO₂(g) ΔH=-393.5kJ + 33.10kJ = -360.4kJ

-5/4 (2):

C(s) + 13/4O₂(g) + 1/2N₂(g) + 5/2H₂(g) → CO₂(g) + NO₂(g) + 5/2 H₂O(l)

ΔH= -360.4kJ -5/4 (571.6kJ) = -1074.9kJ

And this reaction -1/4 (4):

C(s) + 5/2 H₂(g) + 1/2 N₂(g) → CH₃NH₂(l)

ΔH= -1074.9kJ -1/4(-4110.4kJ)

Now, you can confirm the calculated answer!

Answer:

92.93 g

Explanation:

Number of half lives that have elapsed in eight days =8/14.3 = 0.559

Fraction of the radioactive nuclide that remains after 0.559 half lives is given by

N/No=(1/2)^0.559

Where N= mass of radioactive nuclides remaining after a time t

No= mass of radioactive nuclides originally present

N/No=(1/2)^0.559= 0.679

Mass of nuclides present eight days before= 63.1g/0.679

Mass of nuclides present eight days before=92.93 g

A heterogeneous mixture​

Answer:

The correct option is d

Explanation:

Nuclide is synonymous with groups of electrons or protons, that is, a nuclide is the grouping of nucleons.