Find the standard enthalpy of formation of iodine atoms. (Round your answer to one decimal place.) Standard enthalpy of formation

Answers

Answer 1

Answer:

Enthalpy of formation is the energy change when one mole of a substance is formed from its constituent atoms under standard conditions


Related Questions

what volume in liters of carbon monoxide will be required to produce 18.9 L of nitrogen in the reaction below


2co(g) + 2no(g) -> n2(g) + 2co2(g)

Answers

Answer:

37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.

Explanation:

Equation for the reaction:

2 CO + 2 NO ------> N2 + 2 CO2

2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen

At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.

So therefore, we can say:

2 * 22.4 L of CO produces  22.4 L of N2

44.8 L of CO produces 22.4 L of N2

Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:

44.8 L of CO = 22.4 L of N

x L = 18.9 L

x L = 18.9 * 44.8 / 22.4

x L = 18.9 * 2

x = 37.8 L

The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L

Answer:

37.8

Explanation:

How many atoms are in 65.0g of zinc?​

Answers

from

1moles=iatom

Mole=mass÷avogardos

Where

Avogadro's= 6.02×10²³

So moles = 65.0÷6.02×10²³

Atoms of zinc = 391.6 ×10²³

The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].

Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.

Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately  [tex]6.022 \times10^{23}[/tex]atoms per mole.

First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.

Number of moles = Mass / Molar mass

Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol

Next, multiply the number of moles by Avogadro's number to find the number of atoms.

Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]

Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol

Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.

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Most of the costs associated with using renewable resources are due to
а. overuse of resources
b. atmospheric pollution
C.lack of availability
d.global warming

Answers

Answer:

The answer is a.

Explanation:

Most of the costs associated with using renewable resources are due to overuse of the resources.

Most of the costs associated with using renewable resources are due to overuse of resources.

What are renewable resources?

Renewable resources are those resources which will be generated naturally and continously from the nature and these are also inexhaustible means non ended.

As from the definition it is clear that we can reuse or will use again and again these types of resources, that's why cost associated with these renewable resources are high.

Atmospheric pollution and global warming causes hazardous effect on the environment, so it will not be the reason with the associated cost.Lack of availability makes its important, not costly and in our daily life we used many kinds of renewable resorces so it is not possible to use costly resources daily.

Hence, overuse of resouces is one of the reason.

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Determine the molar solubility of AgBr in a solution containing 0.150 M NaBr. Ksp (AgBr) = 7.7 × 10-13.

Answers

Answer:

Molar solubility of AgBr = 51.33 × 10⁻¹³

Explanation:

Given:

Amount of NaBr = 0.150 M

Ksp (AgBr) = 7.7 × 10⁻¹³

Find:

Molar solubility of AgBr

Computation:

Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr

Molar solubility of AgBr = 7.7 × 10⁻¹³ / 0.150

Molar solubility of AgBr = 51.33 × 10⁻¹³

When The Molar solubility of AgBr is = 51.33 × 10⁻¹³

Calculation of Solubility of AgBr

Given as per question:

The Amount of NaBr is = 0.150 M

Then Ksp (AgBr) is = 7.7 × 10⁻¹³

Now we Find:

The Molar solubility of AgBr

The we Computation is:

The Molar solubility of AgBr is = Ksp (AgBr) / Amount of NaBr

After that Molar solubility of AgBr is = 7.7 × 10⁻¹³ / 0.150

Therefore, Molar solubility of AgBr is = 51.33 × 10⁻¹³

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At 25 °C, only 0.0990 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C? AB3(s)↽−−⇀A3+(aq)+3B−(aq)

Answers

Answer:

[tex]Ksp=2.59x10^{-3}[/tex]

Explanation:

Hello,

In this case, given the 0.0990 moles of the salt are soluble in 1.00 L of water only, we can infer that the molar solubility is 0.099 M. Next, since the dissociation of the salt is:

[tex]AB_3\rightleftharpoons A^{3+}+3B^-[/tex]

The concentrations of the A and B ions in the solution are:

[tex][A]=0.099 \frac{molAB_3}{L}*\frac{1molA}{1molAB_3} =0.0099M[/tex]

[tex][B]=0.099 \frac{molAB_3}{L}*\frac{3molB}{1molAB_3} =0.000.297M[/tex]

Then, as the solubility product is defined as:

[tex]Ksp=[A][B]^3[/tex]

Due to the given dissociation, it turns out:

[tex]Ksp=[0.099M][0.297M]^3\\\\Ksp=2.59x10^{-3}[/tex]

Regards.

Write a balanced nuclear equation for the following: The nuclide thorium-234 undergoes beta decay to form protactinium-234 .

Answers

Answer:

²³⁴₉₀Th --> ²³⁴₉₁Pa + ⁰₋₁e

Explanation:

thorium-234 = ²³⁴₉₀Th

beta decay = ⁰₋₁e

protactinium-234 = ²³⁴₉₁Pa

The balanced nuclear equation is given as;

²³⁴₉₀Th --> ²³⁴₉₁Pa + ⁰₋₁e

The second-order decomposition of NO2 has a rate constant of 0.255 M-1s-1. How much NO2 decomposes in 8.00 s if the initial concentration of NO2 (1.00 L volume) is 1.33 M

Answers

Answer:

0.9718M

Explanation:

Rate constant, k =  0.255 M-1s-1

time, t = 8.00 s

Initial concentration, [A]o = 1.33 M

Final concentration, [A] = ?

These quantities are represented by the equation;

1 / [A] = 1 / [A]o + kt

1 / [A] = 1 /1.33 + (0.255 * 8)

1 / [A] = 0.7519 + 2.04

[A] = 1 / 2.7919 = 0.3582 M

How much of NO2 decomposed is obtained from the change in concentration;

Change in concentration = Initial - Final

Change = 1.33 - 0.3582 = 0.9718M

What is the concentration of A after 50.7 minutes for the second order reaction A → Products when the initial concentration of A is 0.250 M? (k = 0.117 M⁻¹min⁻¹)

Answers

Answer:

0.101 M

Explanation:

Step 1: Given data

Initial concentration of A ([A]₀): 0.250 MFinal concentration of A ([A]): ?Time (t): 50.7 minRate constant (k): 0.117 M⁻¹.min⁻¹

Step 2: Calculate [A]

For a second-order reaction, we can calculate [A] using the following expression.

1/[A] = 1/[A]₀ + k × t

1/[A] = 1/0.250 M + 0.117 M⁻¹.min⁻¹ × 50.7 min

[A] = 0.101 M

An electrolysis cell has two electrodes. Which statement is correct? A. Reduction takes place at the anode, which is positively charged. B. Reduction takes place at the cathode, which is positively charged. C. Reduction takes place at the dynode, which is uncharged. D. Reduction takes place at the cathode, which is negatively charged. E. Reduction takes place at the anode, which is negatively charged.

Answers

Answer:

D. Reduction takes place at the cathode, which is negatively charged.

Explanation:

In an electrolytic cell there are two electrodes; the cathode and the anode. The anode is the positive electrode while the cathode is the negative electrode. Oxidation occurs at the anode while reduction occurs at the cathode.

At the anode, species give up electrons and become positively charged ions while at the cathode species accept electrons and become reduced.

1. Explain what the police siren sounds like to Jane:
2. Explain what the police siren sounds like to John:
3. Explain why the police siren sounds different between Jane and John:

Answers

Answer:

1. the siren has a lower pitch to Jane

2. the siren has a higher pitch to John

3. sound different due to moving away from Jane making the sound wave lengths longer and moving toward John making the wave lengths shorter

Explanation:

The Doppler effect expresses that sound is comparative with the spectator or observer. This is demonstrated valid by the model given with Jane and John. To one individual it could sound low and to someone else it could sound high, in light of where they are tuning in from. To John, the police alarm playing is a higher pitch. Be that as it may, to Jane this equivalent alarm is a totally extraordinary pitch and is heard lower than in comparison to the john.

This is a prime case of the Doppler Effect. They sound distinctive on the grounds that the sound is moving far from Jane making the sound frequencies longer and it is advancing toward John making the frequencies shorter. This impacts how the sound is heard by the human ear.

The following reactions all have K < 1. 1) a. C6H5COO- (aq) + C6H5OH (aq) → C6H5COOH (aq) + C6H5O- (aq) b. F- (aq) + C6H5OH (aq) → C6H5O- (aq) + HF (aq) c. C6H5COOH (aq) + F- (aq) → HF (aq) + C6H5COO- (aq) Arrange the substances based on their relative acid strength.

Answers

Answer:

the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]

Explanation:

Given that :

a . [tex]\mathsf{C_6H_5COO^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5COOH _{(aq)} + C_6H_5O^- _{(aq)}}[/tex]

b.  [tex]\mathsf{ F^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5O^- _{(aq)} + HF _{(aq)} }[/tex]

c.  [tex]\mathsf{C_6H_5COOH _{(aq)} + F^- _{(aq)} \to HF _{(aq)} + C_6H_5COO^- _{(aq)} }[/tex]

Acid strength is the ability of an acid  to dissociate into a proton and an anion. Take for instance.

HA  ↔ H⁺ + A⁻

The  acid strength of the following compounds above are:

[tex]\mathsf{C_6H_5OH _{(aq)} }[/tex] = 1.00 × 10⁻¹⁰

[tex]\mathsf{HF _{(aq)} }[/tex] = 6.6 × 10⁻⁴

[tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] = 6.3 × 10⁻⁵

As the acid dissociation constant increases the relative acid strength also increases.

From above, the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]

[tex]\mathsf{C_6H_5COO^- }[/tex], [tex]\mathsf{C_6H_5O^- _{(aq)}}[/tex] and F⁻ are Bronsted- Lowry acid

Bronsted- Lowry acid are molecule or ion that have the ability to donate a proton.

?
Which statement about energy transfer in a wave is ture ​

Answers

what are the options

Why do prices increase when demand for a product is high? Companies know they can make more money by selling fewer products at higher prices. Companies know that people will be willing to spend more to get an in-demand product. Companies take advantage of the demand to make people spend more money on excess products. Companies know they can stop production and still make money on sales.

Answers

Answer:

Companies know that people will be willing to spend more to get an in-demand product.

Explanation:

When a product is really in demand, many customers are willing to part with more money order to purchase the product, as a result of that, many companies may take advantage of the increasing demand for the product to hike it's price.

Hence, the increase in price may not really have a negative impact on the quantity demanded because the demand for the product is high and customers are willing to spend more money in order to purchase an in-demand product, hence the answer above.

Prices increase when demand is high because companies know that people will be willing to spend more to get in-demand products.

Prices generally increase with higher demand for goods because the higher demand creates pressure for the supply to meet up.

Manufacturing companies can either increase their production to meet up with demand at the same price or capitalize on the situation to make more money by increasing the price without increasing the supply.

Since there is a buying pressure on the product in the market already, people would still be open to buying even at higher prices.

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A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
Br(g)
Cl2(g)
I2(g)
F2(g)
B. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2S(g)
H2O(g)
H2O2(g)
C. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous)
C(s, diamond)
C(s, graphite)

Answers

Answer:

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Explanation:

Hello,

In this case, we can apply the following principles to explain the order:

- The greater the molar mass, the larger the standard molar entropy.

- The greater the molar mass and the structural complexity, the larger the standard molar entropy.

- The greater the structural complexity, the larger the standard molar entropy.

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

This is due to the fact that the greater the molar mass, the larger the standard molar entropy.

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).

Regards.

If 75.4 J of energy is absorbed by 0.25 mol of CCl4 at constant pressure, what is the change in temperature? The specific heat of CCl4 is 0.861 J/g·°C.

Answers

Answer:

ΔT = 2.28°C

Explanation:

Heat, H = 75.4J

Number of moles = 0.25 mol

Specific heat capacity, c = 0.861 J/g·°C

Change in temperature, ΔT = ?

These quantities are related by the following equation;

H = mc ΔT

Mass, m = Number of moles * Molar mass

m = 0.25mol * 153.82 g/mol

m = 38.455g

S back to the equation;

H = mc ΔT

Substituting the values;

75.4 = 38.455 * 0.861 * ΔT

ΔT = 75.4 /  33.11

ΔT = 2.28°C

The change in temperature is 2.28 °C

First, we will determine the mass of CCl₄ absorbed

From the given information,

Number of moles of CCl₄ absorbed = 0.25 mol

Using the formula

Mass = Number of moles × Molar mass

Molar mass of CCl₄ = 153.82 g/mol

∴ Mass of CCl₄ absorbed = 0.25 × 153.82

Mass of CCl₄ absorbed = 38.455 g

Now, using the formula

Q = mcΔT

Where Q is the quantity of heat

m is the mass

c is the specific heat of substance

and ΔT is the change in temperature

From the given information

Q = 75.4 J

c = 0.861 J/g.°C

Putting the parameters into the formula, we get  

75.4 = 38.455 × 0.861 ×ΔT

75.4 = 33.109755 × ΔT

∴ ΔT = 75.4 ÷ 33.109755

ΔT = 2.28 °C

Hence, the change in temperature is 2.28 °C

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A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup calorimeter fitted with a lid through which a thermometer passes. The acid-base reaction is as follows:
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
The temperature of each solution before mixing is 22.3 °C. After mixing, the temperature of the solution mixture reaches a maximum temperature of 31.4 °C. Assume the density of the solution mixture is 1.00 g/mL, its specific heat is 4.18 J/g.°C, and no heat is lost to the surroundings. Calculate the enthalpy change, in kj, per mole of H2SO4 in the reaction.
a. +85.6 kJ/mol.
b. -85.6 kJ/mol.
c. +5.71 kJ/mol.
d. -5.71 kJ/mol.
e. -114 kJ/mol.

Answers

Answer:

THE ENTHALPY CHANGE IN KJ/MOLE IS +114 KJ/MOLE.

Explanation:

Heat = mass * specific heat capacity * temperature rise

Total volume = 100 + 50 = 150 mL

Total mass = density * volume

Total mass = 1 * 150 mL = 150 g

So therefore, the heat evolved during the reaction is:

Heat = 150 * 4.18 * ( 31.4 - 22.3)

Heat = 150 * 4.18 * 9.1

Heat = 5705.7 J

Equation for the reaction:

2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)  

From the equation, 2 moles of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water

50 mL of 1 M of H2SO4 contains

50 * 1 / 1000 mole of acid

= 0.05 mole of acid

The production of 1 mole of water evolved 5705.7 J of heat and hence the enthalpy changein kJ per mole will be:

0.05 mole of H2SO4 produces 5705.7 J of heat

1 mole of H2SO4 will produce 5705.7 / 0.05 J

= 114,114 J / mole

In kj/mole = 114 kJ/mole.

Hence, the enthalpy change of the reaction in kJ /mole is +114 kJ/mole.

What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond

Answers

Answer:

A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable

Explanation:

Which of the following pairs of chemical reactions are inverses of each other? Answer options: a. Hydrogenation and alkylation b.Halogenation and hydrolysis c. Ammoniation and alkylation d. Oxidation and reduction

Answers

Answer:

d. Oxidation and reduction

Explanation:

For this question we have to remember the definition of each type of reaction:

-) Hydrogenation

In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.

-) Alkylation

In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.

-) Hydrolysis

In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.

-) Halogenation

In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.

-) Ammoniation

In this reaction, we have the addition of the ammonium ion ([tex]NH_4^+[/tex]). In the example, the ammonium ion is added to an acid.

-) Oxidation and reduction

In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:

[tex]Ag^+~+~e^-~->~Ag[/tex] Reduction

[tex]Al~->~Al^+^3~+~3e^-[/tex] Oxidation

16. The concentration of a solution of potassium hydroxide is determined by titration with nitric
acid. A 30.0 mL sample of KOH is neutralized by 42.7 mL of 0.498 M HNO3. What is the
concentration of the potassium hydroxide solution?

Answers

Answer:

[tex]M_{base}=0.709M[/tex]

Explanation:

Hello,

In this case, since the reaction between potassium hydroxide and nitric acid is:

[tex]KOH+HNO_3\rightarrow KNO_3+H_2O[/tex]

We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:

[tex]n_{acid}=n_{base}[/tex]

That in terms of molarities and volumes is:

[tex]M_{acid}V_{acid}=M_{base}V_{base}[/tex]

Thus, solving the molarity of the base (KOH), we obtain:

[tex]M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M[/tex]

Regards.

If the lead concentration in water is 1 ppm, then we should be able to recover 1 mg of lead from _____ L of water.

Answers

Answer:

1 L

Explanation:

ppm means parts per million. Generally the relationship between mass and litre is given as;

1 ppm = 1 mg/L

This means that 1 ppm is equivalent to 1 mg of a substance dissolved in 1 L of water.

The pH of an acid solution is 5.82. Calculate the Ka for the monoprotic acid. The initial acid concentration is 0.010 M.

Answers

Answer:

The answer is

[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]

Explanation:

The Ka of an acid when given the pH and concentration can be found by

[tex]pH = - \frac{1}{2} log(Ka) - \frac{1}{2} log(c) [/tex]

where

c is the concentration of the acid

From the question

pH = 5.82

c = 0.010 M

Substitute the values into the above formula and solve for Ka

We have

[tex]5.82 = - \frac{1}{2} log(Ka) - \frac{1}{2} log(0.010) [/tex][tex] - \frac{1}{2} log(Ka) = 5.82 + 1[/tex][tex] - \frac{1}{2} log(Ka) = 6.82[/tex]

Multiply through by - 2

[tex] log(Ka) = - 13.64[/tex]

Find antilog of both sides

We have the final answer as

[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]

Hope this helps you

1. The manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0% acetic acid. What is the percent error between your result and the manufacturer statement

Answers

Answer:

72.8 % (But verify explanation).

Explanation:

Hello,

In this case, with the following obtained results, the percent error is computed as follows:

Volume of vinegar= 7.0 mL

Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL

Used concentration of NaOH= 1.5M

Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M

Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol

Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g

% of acetic acid in vinegar=8.64 %

% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %

Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.

Regards.

What are periodic trends if ionic radii

Answers

Answer:

Explan ionization energy, atomic radius, and electron affinityation:

This question most likely has answer choices. The possible answer choices are as followed:

Ionic radii tend to increase down a group.Ionic radii tend to decrease across a period.Anionic radii tend to increase across a period.Cationic radii tend to decrease across a period.Ionic radii increase when switching from cations to anions in a period.

The answers are Ionic radii tend to increase down a group, Cationic radii tend to decrease across a period, and Ionic radii increase when switching from cations to anions in a period (1st, 4th, and 5th options).

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

Answers

Molarity is in units of moles/L. So you must determine the moles of CH3OH, then divide by the total volume.

To determine the moles of CH3OH, divide the weight in grams of CH3OH by the molecular weight

11.7g / 32g/mol = 0.366 mol CH3OH

0.366 mol CH3OH / .230 L = your molarity

How many moles of oxygen are in 49.2 grams of carbon dioxide?

Answers

Answer:

Number of mole in Oxygen = 2.24 mol

Explanation:

Given:

Amount of carbon dioxide (CO₂) = 49.2 gram

Find:

Number of moles in Oxygen

Computation:

Molecular weight of CO₂ = 44 gram (Approx)

Number of mole in CO₂ = 49.2 / 44

Number of mole in CO₂ = 1.11818182 mol

CO₂ has 2 mole of Oxygen,

So,

Number of mole in Oxygen = 2 × 1.11818182

Number of mole in Oxygen = 2.23636364

Number of mole in Oxygen = 2.24 mol

How many moles of gold are equivalent to 1.204 × 1024 atoms? 0.2 0.5 2 5

Answers

Answer:

c

Explanation:

How many moles of gold are equivalent to 1.204 × 1024 atoms?

0.2

0.5

2

5

C) 2 Is the correct answer, I took the test and it was correct.

According to the concept of Avogadro's number, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.

What is Avogadro's number?

Avogadro's number is defined as a proportionality factor which relates number of constituent particles with the amount of substance which is present in the sample.

It has a SI unit of reciprocal mole whose numeric value is expressed in reciprocal mole which is a dimensionless number and is called as Avogadro's constant.It relates the volume of a substance with it's average volume occupied by one of it's particles .

According to the definitions, Avogadro's number depend on determined value of mass of one atom of those elements.It bridges the gap between macroscopic and microscopic world by relating amount of substance with number of particles.

Number of atoms can be calculated using Avogadro's number as follows: mass/molar mass×Avogadro's number.Number of moles=number of atoms/Avogadro's number=1.204×10²⁴ /6.023×10²³=1.999≅2

Thus, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.

Learn more about Avogadro's number,here:

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A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculate the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol. He next tried the same experiment with 100 ml of 1.0 M HCl and 100 ml of 1.0 M NaOH. The calculated molar enthalpy change of reaction for his second trial was:

Answers

Answer: The calculated molar enthalpy change of reaction for his second trial was -108 kJ.

Explanation:-

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]

Thus [tex]\text{no of moles}of HCl={1.0M}\times {0.05L}=0.05moles[/tex]

Thus [tex]\text{no of moles}of NaOH={1.0M}\times {0.05L}=0.05moles[/tex]

[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]

Given for second trial:

[tex]\text{no of moles}of HCl={1.0M}\times {0.1L}=0.1moles[/tex]

[tex]\text{no of moles}of NaOH={1.0M}\times {0.1L}=0.1moles[/tex]

0.05 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release  heat = 54 kJ

0.1 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release  heat  =[tex]\frac{54}{0.05}\times 0.1=108kJ[/tex]

Thus calculated molar enthalpy change of reaction for his second trial was -108 kJ.

For each bond, show the direction of polarity by selecting the correct partial charges. _________ Si-P _________ _________ Si-Cl _________ _________ Cl-P _________ The most polar bond is _______

Answers

Answer:

Siδ⁺ -- Pδ⁻⁻

Clδ⁻⁻ -- Pδ⁺

Siδ⁺ -- Clδ⁻⁻

Of the mentioned bonds the most polar bond is Si -- Cl

The polarity of the bond primarily relies upon the electronegativity difference between the two atoms that forms the bond. Therefore, if the electronegativity difference between the two atoms that forms the bond is more the bond will be more polar, and if it is less then the bond will be less polar. The electronegativity of the atoms mentioned is Si = 1.8 , P = 2.1 and Cl = 3.00.  

Therefore, the Si - Cl atoms exhibit more electronegativity difference, thus, the Si - Cl bond will be the most polar bond.  

Predict the order of acid strengths in the following series of cationic species: CH3CH2NH3 +, CH3CH=NH2

Answers

Answer:

CH3CH=NH2+>CH3CH2NH3 +

Explanation:

There are certain structural features that determine the stability of cationic species. These features that lead to the stability and higher acid strength of cations are those features that stabilize the cation.

CH3CH=NH2+ is more acidic than CH3CH2NH3 + owing to the fact that CH3CH=NH2+ contains a double bond in close proximity with the hydrogen that can be lost as a proton. Electron withdrawal by the double bond (greater s character) makes it easier for this hydrogen to be lost as a proton compared to CH3CH2NH3 +.

Describe how you would prepare 500ml of 40% (w/v) aqueous iodine solution.
[Atomic mass of iodine =127g/mol].​

Answers

Answer:

- Weight 333.3 grams of iodine.

- Measure 500 mL of water.

- Vigorously mix the resulting solution.

Explanation:

Hello,

In this case, since 500 mL of a 40% (w/v) aqueous solution iodine is required, we can compute the required mass of iodine by defining the given mass-volume percent:

[tex]\% w/v=\frac{m_{iodine}}{m_{solution}} *100%=\frac{m_{iodine}}{m_{water}+m_{iodine}} *100%[/tex]

In such a way, we need to find mass of iodine, which is computed as:

[tex]m_{iodine}=\frac{\%w/v*m_{water}}{100w/v-\%} \\\\m_{iodine}=\frac{40*500}{100-40}\\ \\m_{iodine}=333.3g\\[/tex]

Thereby, the procedure will be:

- Weight 333.3 grams of iodine.

- Measure 500 mL of water.

- Vigorously mix the resulting solution.

Best regards.

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