Which functional group is used in other functional groups?
A. Ester
B. Carbonyl
c. Hydroxyl
D. Amino

q9​

Answer:

the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions is 253.748 J/K

Explanation:

Given the data in the question;

CaCO₃(s)       →     CaO(s)     +     CO₂(g)  

1.58 moles        1.58 moles     1.58 moles

Since 1 mole of CaCO₃ gives 1 mole of CaO and 1 mole of CO₂

Thus, 1.58 mole of CaCO₃ gives 1.58 moles of CaO and 1.58 moles of CO₂.

Now,

At 298 K, standard entropy values are;

ΔS° ( CaCO₃ ) = 92.9 J/mol.K

ΔS° ( CaO )     = 39.8 J/mol.K

ΔS° ( CO₂ )      = 213.7 J/mol.K

So,

ΔS°[tex]_{system[/tex] = ∑ΔS°( product ) - ∑ΔS°( reactant )

ΔS°[tex]_{system[/tex] = [ ΔS°(CaO) + ΔS°( CO₂ ) ] - ΔS°( CaCO₃ )

we substitute

ΔS°[tex]_{system[/tex] = [ 39.8 J/mol.K + 213.7 J/mol.K ] - 92.9 J/mol.K

ΔS°[tex]_{system[/tex] = 160.6 J/mol.K

i.e, for 1 mol CaCO₃, ΔS°[tex]_{system[/tex] = 160.6 J/mol.K

Now, for 1.58 mol CaCO₃,

ΔS°[tex]_{system[/tex] = 1.58 mol × 160.6 J/mol.K

ΔS°[tex]_{system[/tex] = 253.748 J/K

Therefore, the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions is 253.748 J/K

Answer:

C) Both will require the same amount because the concentrations are equal.

Explanation:

The pH of a solution is defined as:

pH = -log [H+]

This H+ is the ion that reacts with OH- (From NaOH) as follows:

H+ + OH- → H2O

When all H+ reacts, we can say the solution was neutralized.

Now, as both, the solution with the weak acid and the solution with strong acid have the same pH, we can say that their [H+] is the same. Assuming the volume of both solutions is the same:

Both will require the same amount because the concentrations are equal.

Answer:

Precipitation

Explanation:

Let's consider the balanced chemical equation between barium acetate and sodium carbonate to form barium carbonate and sodium acetate.

Ba(C₂H₃O₂)₂(aq) + Na₂CO₃(aq) → BaCO₃(s) + 2 NaC₂H₃O₂(aq)

Both products and reactants are salts. But, among the products, barium carbonate is solid. This allows us to classify it as a precipitation reaction.

Answer:

a. 7278 K

b. Kc = 4.542 × 10⁻³¹

Explanation:

a.

The reaction is spontaneous when ΔG° < 0. We can calculate ΔG° using the following expression.

ΔG° = ΔH° - T × ΔS°

Then, the reaction will be spontaneous when,

ΔH° - T × ΔS° < 0

T > ΔH°/ΔS

T > (180.5 × 10³ J/mol)/(24.80J/mol⋅K)

T > 7278 K

b.

First, we will calculate ΔG° at 25 °C (298 K)

ΔG° = ΔH° - T × ΔS°

ΔG° = (180.5 × 10³ J/mol) - 298 K × (24.80J/mol⋅K) = 1.731 × 10⁵ J/mol

Then, we will calculate the equilibrium constant (Kc) using the following expression.

ΔG° = - R × T × ln Kc

-ΔG°/R × T =  ln Kc

-(1.731 × 10⁵ J/mol)/(8.314 J/mol.K) × 298 K =  ln Kc

Kc = 4.542 × 10⁻³¹

Answer:

γ−Hydrogen is easily replacable during bromination reaction in presence of light , because Allylic substitution is being preferred.

Explanation:

that's all

γ−Hydrogen is easily replacable during bromination reaction in presence of light , because Allylic substitution is being preferred.

Answer:

1.59 M

Explanation:

What is the concentration (M) of CH₃OH a solution prepared by dissolving 11.7 g of CH₃OH sufficient water to give exactly 230 mL of solution?

Step 1: Given data

Step 2: Calculate the moles corresponding to 11.7 g of CH₃OH

The molar mass of CH₃OH is 32.04 g/mol.

11.7 g × 1 mol/32.04 g = 0.365 mol

Step 3: Calculate the molarity of the solution

M = moles of solute / liters of solution

M = 0.365 mol / 0.230 L = 1.59 M

Explanation:

I have no idea about this

Answer:

ATP hydrolase

Explanation:

Enzymes are biological catalysts which perform diverse functions in the body. Enzymes are specific in their mode of action because an enzyme fits into its substrate as a key fits into a lock.

The particular enzyme that catalyzes the breakdown of ATP to ADP is ATP hydrolase. The phosphate released by the action of this enzyme is used in the phosphorylation of other compounds thereby making them more reactive.

Answer:

If you contact water with a gas at a certain temperature and (partial) pressure, the concentration of the gas in the water will reach an equilibrium ('saturation') according to Henry's law.

Explanation:

This means: if you increase the pressure (e.g. by keeping the vial closed), the CO2 concentration will increase. So it simply depends what concentration you need for your assay: 'CO2-saturated' water at low pressure or 'CO2-saturated' water at high pressure.

Explanation:

Normality is one of the concentration terms.

It is expressed as:

[tex]N=\frac{mass of the substance}{equivalent mass}* \frac{1}{volume of solution in L.}[/tex]

The volume of the solution is 425 mL.

Mass of sulfuric acid given is:

[tex]mass=volume * purity* density\\ = 75 mL * 0.966 * 1.83 g/mL\\\\=132.5 grams\\[/tex]

The equivalent mass of sulfuric acid is 49.0g/equivalents

Hence, the normality of the given solution is:

[tex]N=\frac{132.5g}{49.0g/equi.} *\frac{1000}{425mL} \\Normality=6.36N[/tex]

Answer is: 6.36N.

````````````````````

Answer:

+5

Explanation:

Answer:

Hydrogen Bonding

Explanation:

Hydrogen Bonding occurs when a hydrogen atom is bonded to N, O, and F atoms.

The molecules H₂O, HF, and NH₃ all experience hydrogen bonding, which is a relatively strong IMF, causing the molecules to have stronger attraction to each other. Having a stronger attraction between molecules results in more energy required to separate them, thus these molecules will have a higher boiling point than the rest of the molecules in their group.

Answer:

It is a Biological Molecule

Answer:

water is 9/10

chemical would be 1/10

Explanation:

180/200 would be water concentration in solution

and 20/200 would be chemical solution concentration in solution (if the chemical were to be polar and mix)

Answer:

Metals tend to form positive oxidation states. Here, Iron (I) has an oxidation state of +1 while Iron (II) has an oxidation state of +2. Similarly, Lead (I) has an oxidation state of +1 while Lead(II) has an oxidation state of +2. A change in oxidation state can rather cause significant changes in the compound.

Answer:

B

Explanation:

E = hc/[tex]\lambda[/tex]. Remember, it is in meters not nanometers so you have to convert. You end up with B.

Answer:

The correct solution is "3.32 gm/L".

Explanation:

Given:

Rate constant,

[tex]K = 0.05 \ hr^{-1}[/tex]

Time,

[tex]t = 24 \ hours[/tex]

Concentration of ethanol,

[tex]C_o= 1 \ mg/L[/tex]

Now,

The concentration of ethanol after 24 hours will be:

⇒ [tex]C_o=C\times e^{-K\times t}[/tex]

By putting the values, we get

    [tex]1=C\times e^{-0.05\times 24}[/tex]

    [tex]1=C\times 0.30119[/tex]

    [tex]C= 3.32 \ gm/L[/tex]

Answer:

121 °C

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 5.97 atm

Initial temperature (T₁) = 374 °C

Final pressure (P₂) = 3.64 atm

Final temperature (T₂) =?

NOTE: Volume = constant

Next, we shall convert 374 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 374 °C

Initial temperature (T₁) = 374 °C + 273

Initial temperature (T₁) = 647 K

Next, we shall determine the final temperature. This can be obtained as follow:

Initial pressure (P₁) = 5.97 atm

Initial temperature (T₁) = 647 K

Final pressure (P₂) = 3.64 atm

Final temperature (T₂) =?

P₁ / T₁ = P₂ / T₂

5.97 / 647 = 3.64 / T₂

Cross multiply

5.97 × T₂ = 647 × 3.64

5.97 × T₂ = 2355.08

Divide both side by 5.97

T₂ = 2355.08 / 5.97

T₂ = 394 K

Finally, we shall convert 394 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

Final temperature (T₂) = 394 K

Final temperature (T₂) = 394 – 273

Final temperature (T₂) = 121 °C

Thus, the new temperature is 121 °C

Answer:

Rock A will have far more chemical weathering than Rock B due to the rise in area effect

Explanation:

Rock A undergoes both Physical and Chemical weathering. So, thanks to physical weathering there'll appear cracks within the rock, which can, in turn, increase the area of rock on which weathering is occurring. So, Chemical weathering will happen much faster now as there's a rise in the area. within the case of Rock B, there's only chemical weathering therefore the increase in the area won't be that very much like compared to Rock A.

Answer:

C) EtOH 1% AgNO3

Answer:

Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Explanation:

In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.

For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.

Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Answer: The molar mass of solute is 115 g/mol.

Explanation:

Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.

The expression for the calculation of elevation in boiling point is:

[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]

OR

[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)

where,

Boiling point of pure solvent (benzene) = [tex]80.10^oC[/tex]

Boiling point of solution = [tex]81.20^oC[/tex]

i = Vant Hoff factor = 1 (for non-electrolytes)

[tex]K_b[/tex] = Boiling point elevation constant = [tex]2.53^oC/m[/tex]

[tex]m_{solute}[/tex] = Given mass of solute = 10 g

[tex]M_{solute}[/tex] = Molar mass of solute = ? g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 200 g

Putting values in equation 1, we get:

[tex]81.20-80.10=1\times 2.53\times \frac{10\times 1000}{M_{solute}\times 200}\\\\M_{solute}=\frac{1\times 2.53\times 10\times 1000}{1.1\times 200}\\\\M_{solute}=115g/mol[/tex]

Hence, the molar mass of solute is 115 g/mol.

Answer:

About 3500 grams of tin.

Explanation:

We want to determine amount of tin metal (in grams) that can be produced from smelting 4.5 kilograms of tin(IV) oxide.

First, write the chemical compound. Since our cation is tin(IV), it forms a 3+ charge. Oxygen has a 2- charge, so we will have two oxygen atoms. Hence, tin(IV) oxide is given by SnO₂.

By smelting it, we acquire elemental tin and oxygen gas. Hence:

[tex]\text{SnO$_2$}\rightarrow \text{Sn} + \text{O$_2$}[/tex]

(Note: oxygen is a diatomic element.)

The equation is balanced as well.

To convert from SnO₂ to only Sn, we can first convert from grams of SnO₂ to moles, use mole ratios to convert to moles of Sn, and then from there convert to grams.

Since Sn has a molar mass of 118.71 g/mol and oxygen has a molar mass of 15.999 g/mol, the molar mass of SnO₂ is:

[tex](118.71)+2(15.999) = 150.708\text{ g/mol}[/tex]

Therefore, given 4.5 kilograms of SnO₂, we can first convert this into grams using 1000 g / kg and then using the ratio:

[tex]\displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}[/tex]

We can convert this into moles.

Next, from the chemical equation, we can see that one mole of SnO₂ produces exactly one mole of Sn (and also one mole of O₂). So, our mole ratio is:

[tex]\displaystyle \frac{1\text{ mol Sn}}{1\text{ mol SnO$_2$}}[/tex]

With SnO₂ in the denominator to simplify units.

Finally, we can convert from moles Sn to grams Sn using its molar mass:

[tex]\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]

With the initial value and above ratios, we acquire:

[tex]\displaystyle 4.5\text{ kg SnO$_2$}\cdot \frac{1000 \text{ g SnO$_2$}}{1\text{ kg SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol Sn}}{1 \text{ mol SnO$_2$}} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]

Cancel like units:

[tex]=\displaystyle 4.5\cdot \frac{1000}{1}\cdot \displaystyle \frac{1}{150.708}\cdot \displaystyle \frac{1}{1} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1}[/tex]

Multiply. Hence:

[tex]\displaystyle = 3544.5696...\text{ g Sn}[/tex]

Since we should have two significant figures:

[tex]=3500 \text{ g Sn}[/tex]

So, about 3500 grams of tin is produced from smelting 4.5 kg of tin(IV) oxide.

Answer:

3546g

Explanation:

start w/ tin (IV) oxide n elemental tin and oxygen gas are the only products of this reaction

SnO2 -> Sn + O2

Sn molecular wt: 119

O2 molecular wt: 32

SnO2 molecular wt:  119+32 = 151

so Sn / SnO2 wt ratio = 119 / 151

4.5 kilograms of tin (IV) oxide will produce:

= 4.5 * 119 / 151

= 3.546 kg

or 3546 grams of tin metal

no need to involve moles ;)

Answer:

three electron shells

14 electrons

14 neutrons

Explanation:

Silicon has three electron shells arranged as follows; 2, 8, 4. This corresponds to the fact that silicon is a member of group 14 of the periodic table.

Note that, the number of protons in an atom is the same as the number of electrons in the neutral atom. Since Silicon has 14 protons, it also has 14 electrons likewise.

The mass number of silicon is 28 but number of neutrons= mass number - number of protons. Since mass number = 28, then there are 14 neutrons in silicon.

Answer:

It increases but less than double

Explanation:

As the temperature of a gas increase, the average kinetic energy of the gas increases. The kinetic energy of a gas is the thermal energy that the gas contains.

We know, the kinetic energy of an ideal gas is given by :

[tex]$V_{avg} = \sqrt{\frac{8R}{\pi M}}$[/tex]

where, R = gas constant

            T = absolute temperature

            M = molecular mass of the gas

From the above law, we get

[tex]$V_{avg} \propto \sqrt{T}$[/tex]

Thus, if we increase the temperature then the average kinetic energy of the ideal gas increases.

In the context, if the temperature of the ideal gas increases from 100°C to 200°C, then

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{T_2}{T_1}}$[/tex]

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{473.15}{373.15}}$[/tex]

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{1.26}$[/tex]

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =1.12$[/tex]

[tex]$(V_{avg})_2 = 1.12\ (V_{avg})_1$[/tex]

Therefore, [tex]$(V_{avg})_2 > (V_{avg})_1$[/tex]

Thus the average kinetic energy of the molecule increases but it increases 1.12 times which is less than the double.

Thus, the answer is " It increases but less that double".

Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.