14. Based on your previous observations, predict the impact of changing the number of moles of a gas sample on the volume of the gas sample (if pressure and temperature are held constant). What effect would changing the number of moles of a gas sample have on the temperature of a gas sample (if pressure and volume are held constant)? Explain

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:

Explanation:

[tex]2x^2+10x-48\\=2\left(x^2+5x-24\right)\\x^2+5x-24\\=\left(x^2-3x\right)+\left(8x-24\right)\\=x\left(x-3\right)+8\left(x-3\right)\\=\left(x-3\right)\left(x+8\right)\\=2\left(x-3\right)\left(x+8\right)\\2\left(x-3\right)\left(x+8\right)=0\\x-3=0\\x = 0+3\\x = 3\\x+8=0\\x+8-8=0-8\\x=-8\\x=3,\:x=-8[/tex]

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

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:

See explanation

Explanation:

The noble gas core electron configuration involves writing the inert gas core of an atom followed by the valence electrons. This is shown for the following atoms;

Bismuth;

[Xe]4f14 5d10 6s2 6p3

Chromium;

[Ar]4s1 3d5

Strontium;

[Kr]5s2

Phosphorus;

[Ne]3s2 3p3

2.

Bi

6p- n=6, l= 1, ml= 1, ms= 1/2

Cr

3d- n=3, l=2, ml=2,ms=1/2

Sr

5s- n=5, l=0, ml=0, ms=1/2

P

3p- n=3, l= 1, ml= 1, ms=1/2

3.

a) Tin (Sn) - [Kr] 5s2 4d10 5p2

b) Caesium (Cs)- [Xe] 6s1

c) Copper (Cu)- [Ar] 4s1 3d10

Answer:

7.37 mL of KOH

Explanation:

So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,

HNO3 + KOH → KNO3 + H2O

Step 1 : The moles of HNO3 here can be calculated through the given molar mass (  0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.

Mol of NHO3 = 0.140 M [tex]*[/tex] 30 / 1000 L = 0.140 M [tex]*[/tex] 0.03 L = .0042 mol

Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,

0.0042 mol HNO2 [tex]*[/tex] ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH

From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,

Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).

Volume of KOH = 0.0042 mol [tex]*[/tex] ( 1 L / 0.570 mol ) [tex]*[/tex] ( 1000 mL / 1 L ) = 7.37 mL of KOH

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.

Find  more information about Equilibrium constant here:

brainly.com/question/12858312

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.

Learn more:

https://brainly.com/question/21722989

Answer:

25.99mL is the volume internal volume of the flask

Explanation:

To complete the question:

The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask

The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.

To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:

Mass water = Mass filled flask - Mass of clean flask

Mass water = 60.167g - 34.232g

Mass water = 25.935g of water.

To convert this mass to volume:

25.935g × (1mL / 0.997992g) =

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:

2.81mL

Explanation:

Based on the reaction:

C₆H₃COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl

Benzoyl chloride + ammonia → Benzamide

1 mole of benzoyl chloride in excess of ammonia produce 1 mole of Benzamide.

Thus, assuming a theoretical yield, to produce 24.1mmoles of benzamide you require 24.1mmoles of benzoyl chloride.

As molar mass of benzoyl chloride is 141g/mol, mg you require are:

mg Benzoyl chloride: 24.1mmol × (141mg / 1mmol) = 3398.1mg = 3.3981g of benzoyl chloride.

to convert this mass to mL, you require density of Benzoyl chloride (1.21g/mL). Thus, mL you need are:

3.3981g × (1mL / 1.21g) =

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:

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:

–36 KJ.

Explanation:

The equation for the reaction is given below:

2B + C —› D + E. ΔH = – 24 KJ

From the equation above,

1 mole of D required – 24 KJ of energy.

Now, we shall determine the energy change associated with 1.5 moles of D.

This can be obtained as illustrated below:

From the equation above,

1 mole of D required – 24 KJ of energy

Therefore,

1.5 moles of D will require = 1.5 × – 24 = –36 KJ.

Therefore, –36 KJ of energy is associated with 1.5 moles of D.

Answer:

pH = 5.22

Explanation:

As you can see, your initial concentration of sodium acetate (NaCH₃COO) is 0.02M (0.02mol /L). When you add HCl, the reaction is:

NaCH₃COO + HCl → CH₃COOH + NaCl.

If you add HCl, and final concentration of NaCH₃COO is 0.015M, the concentration of CH₃COOH is 0.005M.

You can know the pH of this solution using H-H equation:

pH = pKa + log {NaCH₃COO} / {CH₃COOH}

pH = 4.74 +  log {0.015M} / {0.005M}

Answer:

Poop Buttt.

Explanation:

Answer:

26.87g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2Fe + O2 —> 2FeO

Next, we shall determine the masses of Fe and O2 that reacted and the mass of FeO produced from the balanced equation.

This is illustrated below:

Molar mass of Fe =56 g/mol

Mass of Fe from the balanced equation = 2 x 56 = 112 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32 g

Molar mass of FeO = 56 + 16 =72 g/mol

Mass of FeO from the balanced equation = 2 x 72 = 144 g

From the balanced equation above,

112 g of Fe reacted with 32 g of O2 to produce 144 g of FeO.

Next, we shall determine the limiting reactant.

This is illustrated below:

From the balanced equation above,

112 g of Fe reacted with 32 g of O2.

Therefore, 20.9 g of Fe will react with = (20.9 x 32)/112 = 5.97 g of O2.

From the calculations made above, we can see that only 5.97 g out of 9.19 g of O2 given were required to react completely with 20.9 g of Fe.

Therefore, Fe is the limiting reactant and O2 is the excess reactant.

Finally, we shall determine the mass of FeO produced from the reaction.

In this case, the limiting reactant will be used, as it will give the maximum yield of the reaction since all of it is used up in the reaction.

The limiting reactant is Fe and the maximum mass of FeO produced can be obtained as follow:

From the balanced equation above,

112 g of Fe reacted to produce 144 g of FeO.

Therefore, 20.9 g of Fe will react to produce = (20.9 x 144)/112 = 26.87g of FeO.

Therefore, the maximum mass of iron(II) oxide, FeO produced is 26.87g.

Answer:

The balanced equation is given below:

UO2(s) + 4HF(l) —› UF4(s) + 2H2O(l)

The coefficients are: 1, 4, 1, 2

Explanation:

UO2(s) + HF(l) —› UF4(s) + H2O(l)

The above equation can be balance as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balanced by writing 2 in front of H2O as shown below:

UO2(s) + HF(l) —› UF4(s) + 2H2O(l)

There are 4 atoms of F on the right side and 1 atom on the left side. It can be balanced by writing 4 in front of HF as shown below:

UO2(s) + 4HF(l) —› UF4(s) + 2H2O(l)

Now, the equation is balanced.

The coefficients are: 1, 4, 1, 2.

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

4 moles of Al produce 2 moles of Al₂O₃

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.0852 moles of Al₂O₃,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

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.

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:

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:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M

Answer:

The peaks are registered from tetramethyl silane (TMS)

Explanation:

Tetramethyl silane (TMS) is used as internal reference in proton nmr (H NMR) spectrometry.

Its peak is usually registered at about a 2.0 chemical shift means that the hydrogen atoms which caused that peak need a magnetic field two millionths less than the field needed by TMS to produce resonance. This is not affected by the chemical shift of the sample analysed.

I hope this helped.

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:

[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:

Explanation:

[tex]n = 5.01\\V = 70L\\P = ?\\T = 303K\\R = 0.08206\\\\PV =nRT\\Make \:P \:Subject\:of\:the\:formula\\P = \frac{nRT}{V} \\\\P = \frac{5.01\times0.08206\times303}{70} \\\\P =\frac{124.5695}{70}\\ P = 1.7795[/tex]

Answer:

B. CH3Br

Explanation:

Dipole -Dipole interactions take place in polar molecules.

CH3Br exhibits dipole -dipole forces as its strongest attraction between molecules because it is a polar molecule due to the slightly negative dipole present on the Br molecule.

While O2 is a nonpolar molecule due to its linear structure, CCl4 has zero resultant dipole moment, Helium is non-polar and BrCH2CH2OH is a non polar compound having net dipole moment is zero.

Hence, the correct option is B. CH3Br.

The compound that exhibits dipole -dipole forces is CH3Br

It should be taken place in polar molecules. Also, CH3Br should be the strongest attraction that lies between the molecules since it is treated as the polar molecule because of the slightly negative. While on the other hand, O2 should be non-polar molecule because of the linear structure.

Therefore, The compound that exhibits dipole -dipole forces is CH3Br

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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