energy difference between t2g and EG level in octahedral complex is denoted by ​

Answer:

Explanation:

ethyl 3-methylbenzoate

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:

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:

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:

-The particles of the two reactants will gain kinetic energy and collide with one another more frequently and forcefully, which makes the reaction take place more quickly

Answer:

open system

Explanation:

Answer:

Isolated system

Explanation:

An isolated system is one that cannot exchange either matter or energy with its surroundings.

Answer:

+5

Explanation:

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:

[tex]V_2=482.5mL[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law due to the fact that we are dealing with variable volume, temperature and pressure:

[tex]\frac{P_2V_2}{T_2}=\frac{P_1V_1}{T_1}[/tex]

In such a way, we solve for the final volume, V2, considering that the initial volume, V1, is 500 mL, the initial temperature, T1, is 273 K (STP), the initial pressure, P1, is 1 atm (STP) and the final temperature, T2, is 325 K and the final pressure, P2, is 125 kPa (1.23 atm):

[tex]V_2=\frac{P_1V_1T_2}{T_1P_2} \\\\V_2=\frac{(1atm)(500mL)(325K)}{(273K)(1.23atm)} \\\\V_2=482.5mL[/tex]

Regards!

Answer:

dont buy cheap and off we went

KAnswer:

See explanation

Explanation:

It is more common to use H2SO4 for dehydration reaction rather than HCl because HCl contains a good nucleophile,the chloride ion.

Owing to the presence of the chloride ion, a substitution reaction involving the chloride ion may also proceed also thereby affecting the elimination reaction.

Also, concentrated H2SO4 is a very good drying agent thus, as long as it is used, the alcohol substrate is completely dehydrated to yield the alkene.

Note that HCl is not a dehydrating agent.

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:

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:

the regulation of Soviet production of missiles that carried nuclear weapons

Explanation:

The aim of the SALT I agreement in 1972 was to stop or greatly reduce the arms race where world powers were stockpiling ballistic missiles and other nuclear arsenal.

Therefore, this agreement brought about the regulation of Soviet production of missiles that carried nuclear weapons

Answer:

D. the regulation of Soviet production of missiles that carried nuclear weapons.

Explanation:

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:

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.

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.

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:

C) EtOH 1% AgNO3

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:

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.

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.

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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:

It is a Biological Molecule