How many Noble gases we have in Periodic Table???

Answer:

[tex]{ \bf{PV = nRT}} \\ { \tt{(4.12 \times 3700) = 3.05 \times 0.083 \times T }} \\ { \tt{15244 = 0.25315 \: T}} \\ { \tt{T = 6.02 \times {10}^{4} \: kelvin }}[/tex]

The temperature of the given gas is  60.95 K when it is occupying 3.70 L at 4.12 atm.

The ideal gas law can be described as a general equation of the state of an ideal gas. This equation gives the relationship between the volume and pressure of one-mole of gas equal to the multiplication of the universal gas constant and temperature.

The mathematical relationship can be shown for the ideal gas equation as:

PV = nRT

Where P is the pressure of the gas, n is the moles, V is the volume of the gas, and R is the gas constant.

Given, the volume of gas, V = 3.70 L

The  pressure of the given gas, P = 4.12 atm

The value of the gas constant, R =0.082 atmL/K mol

The number of moles of the given gas, n = 3.05 mol

Substitute the values V, R, P, and n in the ideal gas equation, and we get:

T = PV/nR

T = 4.12 × 3.70/(0.082 × 3.05)

T = 60.95 K

Therefore, the temperature of the given gas is  60.95 K.

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Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

[12 + (2×1) + 16]n = 90

[12 + 2 + 16]n = 90

30n = 90

Divide both side by 30

n = 90/30

n = 3

Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

Answer:

MoO2

Explanation:

The empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.

To solve this question we need to find the moles of Mo2O3. Twice these moles = Moles of Mo. With the moles of Mo we can find its mass.

The difference in masses between mass of new oxide and mass of Mo = Mass of oxygen. With the mass of oxygen we can find its moles and the empirical formula as follows:

Moles Mo2O3 -Molar mass: 239.9g/mol-

12.37g * (1mol / 239.9g) = 0.05156 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.1031 moles of Mo

Mass Mo -95.95g/mol-:

0.1031 moles of Mo * (95.95g/mol) = 9.895g of Mo

Mass oxygen in the oxide:

13.197 - 9.895g = 3.302g Oxygen

Moles oxygen -Molar mass: 16g/mol-:

3.302g Oxygen * (1mol / 16g) = 0.206 moles O

Now, the ratio of moles O / moles Mo is:

0.206 moles O / 0.1031 moles Mo = 2

That means there are 2 moles of O per mole of Mo and the empirical formula of the new oxide is:

Answer:

See explanation

Explanation:

Truly, the second law of thermodynamics requires that spontaneous processes generate entropy, either in the system or in the surroundings.

When a solid is dissolved in a liquid, the solid dissociates into ions. These ions increases the number of particles and hence the entropy of the system thereby making the process spontaneous.

Hence, the dissolution of a substance via an endothermic process is spontaneous because of increase in the number of particles which in turn increases the entropy of the system.

Answer:

1. Nitrate ions, NaNO3 - Sodium nitrate.

2. Sulphide ions, K2S - Potassium sulphide.

3. Sulphate ions, CaSO4 - Calcium sulphate.

4. Hydrogensulphite ions, NaHSO3 - Sodium hydrogensulphite.

5. Carbonate ions, CaCO3 - Calcium carbonate.

6. Hydrogencarbonate ions, KHCO3 - Potassium hydrogencarbonate.

7. Phosphite ions, PH3 - Hydrogen phosphite.

8. Nitride ions, NH3 - Hydrogen nitride ( ammonia ).

9. Ethanoate ions, CH3COONa - Sodium ethanoate.

10. Methanoate ions, HCOONa - Sodium methanoate.

11. Fluoride ions, HF - Hydrogen fluoride.

12. Chloride ions, KCl - Potassium chloride.

13. Bromide ions, HBr - Hydrogen bromide.

14. Iodide ions, NaI - Sodium iodide.

15. Phosphate ions, K3PO3 - potassium phosphate.

Answer:

this is because spreading it makes more sunlight hit the cloth which results in it drying faster

The difference is because SiO2 is solid while CO2 is gaseous

The difference is the melting point values despite belonging to the same group is their STATE OF MATTER.

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Answer:

29.2 g  of H₂SO₄  are produced

0.263 moles of water remain after the reaction goes complete.

Explanation:

We make the reaction in the first step:

Reactants are water and SO₃

H₂O  +  SO₃  →  H₂SO₄

Let's determine moles of reactants:

23.9 g . 1 mol / 80.06g = 0.298 moles

We apply density, to determine mass of water:

D = m/ V so m = D . V

1 g/mL . 10.1 mL = 10.1 g

moles of water are: 10.1 g . 1 mol/ 18g = 0.561 moles

As ratio is 1:1, for 0.298 moles of SO₃ we need the same amount of water, and we have 0.561 moles. Then, water is the excess reagent and sulfur trioxide is the limiting.

0.561 - 0.298 = 0.263 moles of water that remain after the reaction goes complete.

As ratio is 1:1, again, 0.298 moles of SO₃ can produce 0.298 moles of acid.

We determine the mass: 0.298 mol . 98.06 g /mol = 29.2 g

Answer:

CO2 is a trigonal planar.

Answer:

*The molar volume of any gas at standard conditions of temperature and pressure is 22.4 L/mol

V= 101.3 L , n=? , Vm = 22.4 L/mol

101.3 = n × 22.4

n=101.3 / 22.4

n = 4.52 mol

I hope I helped you ^_^

Answer:

[tex]\boxed {\boxed {\sf 4.522 \ mol}}[/tex]

Explanation:

We are asked to find how many moles of gas occupy a volume of 101.3 liters.

1 mole of any gas at STP (standard temperature and pressure) has a volume of 22.4 liters. We can use this information to make a proportion.

[tex]\frac {1 \ mol}{22.4 \ L}[/tex]

We are converting 101.3 liters to moles, so we multiply the proportion by that value.

[tex]101.3 \ L *\frac {1 \ mol}{22.4 \ L}[/tex]

The units of liters (L) cancel.

[tex]101.3 *\frac {1 \ mol}{22.4}[/tex]

[tex]\frac {101.3}{22.4} \ mol[/tex]

Divide.

[tex]4.52232143 \ mol[/tex]

The original value of liters (101.3 L) has 4 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 3 in the ten-thousandths place to the right tells us to leave the 2 in the thousandths place.

[tex]4.522 \ mol[/tex]

101.3 liters of gas is equal to approximately 4.522 moles of gas.

Answer:

a) [tex]K_{2} S[/tex] and [tex]NH_{4} Cl[/tex] :

There are no insoluble precipitate forms.

b) [tex]Ca Cl_{2}[/tex] and [tex](NH_{4} )_{2} Co_{3}[/tex] :

There are the insoluble precipitates of [tex]CaCo_{3}[/tex]  forms.

c) [tex]Li_{2}S[/tex] and [tex]MnBr_{2}[/tex] :

There are the insoluble precipitates of [tex]MnS[/tex]  forms.

d) [tex]Ba(No_{3} )_{2}[/tex] and [tex]Ag_{2} So_{4}[/tex] :                        

As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.

e) [tex]Rb_{2}Co_{3}[/tex] and [tex]NaCl[/tex]:

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- [tex]K_{2} S[/tex] ⇒ soluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.

                                          KCl ⇒ soluble, [tex](NH_{4})_{2} S[/tex]  ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- [tex]Ca Cl_{2}[/tex] ⇒ soluble, [tex](NH_{4} )_{2} Co_{3}[/tex] ⇒ soluble.

                                        [tex]CaCo_{3}[/tex] ⇒ insoluble, [tex]NH_{4} Cl[/tex]  ⇒ soluble.

There are the insoluble precipitates of [tex]CaCo_{3}[/tex]  forms.

c)

Solubility rule suggests:- [tex]Li_{2}S[/tex] ⇒ soluble, [tex]MnBr_{2}[/tex] ⇒ soluble.

                                        [tex]LiBr[/tex] ⇒ soluble, [tex]MnS[/tex]  ⇒ insoluble.

There are the insoluble precipitates of [tex]MnS[/tex]  forms.

d)

Solubility rule suggests:- [tex]Ba(No_{3} )_{2}[/tex] ⇒ soluble, [tex]Ag_{2} So_{4}[/tex] ⇒insoluble.

As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- [tex]Rb_{2}Co_{3}[/tex] ⇒ soluble, [tex]NaCl[/tex] ⇒ soluble.

                                        [tex]RbCl[/tex] ⇒ soluble, [tex]Na_{2} Co_{3}[/tex]  ⇒ soluble.

There are no insoluble precipitates forms.

Answer:

A. 4+.

B. 4+

C. 2+.

Explanation:

Hey there!

In this case, according to the given substances, it turns out possible for us to find the oxidation number of each element by applying the concept of charge balance in all of them as shown below:

A. sulfur in S032- : overall charge is 2- and the oxidation number of oxygen is 2-, thus:

[tex]x-6=-2\\\\x=6-2\\\\x=4+[/tex]

B. nickel in NiO2 : overall charge is 0 and the oxidation number of oxygen is 2-, thus:

[tex]x-4=0\\\\x=4+[/tex]

C. iron in Fe(OH)2: overall charge is 0 and the oxidation state of the OH ion is 1-, thus:

[tex]x-2=0\\\\x=2+[/tex]

Regards!

Answer:

Mole = molecular weight / molecular mass

Mole = 16/16

Mole= 1

Answer:

11.9g remains after 48.2 days

Explanation:

All isotope decay follows the equation:

ln [A] = -kt + ln [A]₀

Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope

We can find k from half-life as follows:

k = ln 2 / Half-Life

k = ln2 / 27.7 days

k = 0.025 days⁻¹

t = 48.2 days

[A]  = ?

[A]₀ = 39.7mg

ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]

ln[A] = 2.476

[A] = 11.9g remains after 48.2 days

Explanation:

here's the answer to your question

Answer:

[tex]\boxed {\boxed {\sf 426 \ mL}}[/tex]

Explanation:

We are asked to find the volume of ammonia gas given a change in pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure of a gas. The formula is:

[tex]P_1V_1= P_2V_2[/tex]

The ammonia gas originally occupies a volume of 450 milliliters at a pressure of 720 millimeters of mercury. Substitute the values into the formula.

[tex]450 \ mL * 720 \ mm \ Hg = P_2V_2[/tex]

The pressure is changed to standard atmospheric pressure, which is 760 millimeters of mercury. The new volume is unknown.

[tex]450 \ mL * 720 \ mm \ Hg = 760 \ mm \ Hg * V_2[/tex]

We are solving for the volume at standard pressure. We will need to isolate the variable V₂. It is being multiplied by 760 millimeters of mercury. The inverse of multiplication is division. Divide both sides of the equation by 760 mm Hg.

[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= \frac{760 \ mm \ Hg * V_2}{760 \ mm \ Hg}[/tex]

[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= V_2[/tex]

The units of millimeters of mercury (mm Hg) cancel.

[tex]\frac {450 \ mL * 720 }{760} = V_2[/tex]

[tex]\frac {324,000}{760} \ mL = V_2[/tex]

[tex]426.3157895 \ mL =V_2[/tex]

The original values of volume and pressure have 3 significant figures. Our answer must have the same. For the number we calculated, that is the ones place. The 3 in the tenths place tells us to leave the 6 in the ones place.

[tex]426 \ mL \approx V_2[/tex]

The volume at standard atmospheric pressure is approximately 426 milliliters.

Answer:

Around 46° F (7.8° C)

Explanation:

"Scientists have predicted that the global average temperature during the ice age was around 46 degrees Farenheit (7.8 degrees Celsius.) However, the polar regions were far colder, around 25 degrees Fahrenheit (14 degrees Celsius) colder than the global average."

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Explanation:

Lanthanide series= E4

Boron=Si

Chalogen=O

Alkaline Earth metal =M9

Answer:

c. C3H9N2O2

Explanation:

The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:

Moles N -Molar mass: 14.01g/mol-

0.420g N * (1mol/14.01g) = 0.0300 moles N

Moles O -Molar mass: 16g/mol-

0.480g O * (1mol/16g) = 0.0300 moles O

Moles C -Molar mass: 12.01g/mol-

0.540g C * (1mol/12.01g) = 0.0450 moles C

Moles H -Molar mass: 1.0g/mol-

0.135g H * (1mol/1g) = 0.135moles H

Dividing in the moles of N (Lower number of moles) the ratio of atoms is:

N = 0.0300 moles N / 0.0300 moles N = 1

O = 0.0300 moles O / 0.0300 moles N = 1

C = 0.0450 moles C / 0.0300 moles N = 1.5

H = 0.135 moles H / 0.0300 moles N = 4.5

As the empirical formula requires whole numbers, multiplying each ratio twice:

N = 2, O = 2, C = 3 and H = 9

And the empirical formula is:

this is the difference between organic and inorganic if this doesn't help you can research more on it

Answer:

Explanation:

maleic acid ⇒ fumaric acid

ΔHreaction = ΔHproduct - ΔHreactant

ΔHproduct = -1336.0 kJ mol⁻¹

ΔHreactant = - 1359.2 kJ mol⁻¹.

ΔHreaction = -1336.0 kJ mol⁻¹ - ( - 1359.2 kJ mol⁻¹.)

=   1359.2 kJ mol⁻¹   -1336.0 kJ mol⁻¹

= 23.2 kJ mol⁻¹ .

Enthalpy of isomerization from maleic to fumaric acid is 23.2 kJ per mol.

Answer:

11.12 → pH

Explanation:

This is a titration of a weak base and a strong acid.

In the first step we did not add any acid, so our solution is totally ammonia.

Equation of neutralization is:

NH₃ + HCl → NH₄Cl

Equilibrium for ammonia is:

NH₃ + H₂O ⇄  NH₄⁺  +  OH⁻      Kb = 1.8×10⁻⁵

Initially we have 50 mL . 0.10M = 5 mmoles of ammonia

Our molar concentration is 0.1 M

X amount has reacted.

In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.

Expression for Kb is : x² / (0.1 - x)  = 1.8×10⁻⁵

As Kb is so small, we can avoid the x to solve a quadratic equation.

1.8×10⁻⁵ = x² / 0.1

1.8×10⁻⁵  .  0.1 = x²

1.8×10⁻⁶ = x²

√1.8×10⁻⁶ = x → 1.34×10⁻³

That's the value for [OH⁻] so:

1×10⁻¹⁴ = [OH⁻] . [H⁺]

1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺] → 7.45×10⁻¹²

- log [H⁺] = pH

- log 7.45×10⁻¹² = 11.12 → pH

Answer:

A) Forms a racemic mixture of the two possible enantiomers

When carbonyl compounds are reduced with a reagent such as LiAlH₄ or NaBH₄ and  new stereogenic center is formed chemical change will lead to products that form a racemic mixture of the two possible enantiomers.

Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.

There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.

There are three types of chemical changes:

1) inorganic changes

2)organic changes

3) biochemical changes

During chemical changes atoms are rearranged and changes are accompanied by an energy change as new substances are formed.

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This question is incomplete, the complete question is;

In recrystallization from boiling water of benzoic acid contaminated with acetanilide, you begin with an impure sample of 5 grams. If the % composition of the acetanilide impurity in the sample is 6.3 %, what is the minimum amount in mL of solvent (water) required for the recrystallization

Compound      Solubility in water at 25°C      Solubility in water at 100°C

Benzoic Acid      0.34 g/100mL                           5.6 g/100mL

Acetanilide         0.53 g/100mL                           5.5 g/100mL

Answer:

The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL

Explanation:

Given the data in the question;

mass of sample = 5 g

percentage composition of the acetanilide impurity = 6.3%

mass of the acetanilide in impure sample will be;

⇒ 6.3% × 5 g = 0.315 g

Mass of benzoic acid in impure sample;

⇒ 5 g - 0.315 g = 4.685 g

now, solubility in water at 100°C for benzoic acid = 5.6 g/100mL

hence 4.685 g of benzoic acid is soluble in x mL

x = [ 100 mL × 4.685 g ] / 5.6 g

x = 83.66 ≈ 84 mL

Also, solubility in water at 100°C for acetanilide = 5.5 g/100mL

hence 0.315 g of benzoic acid is soluble in x mL

x = [ 100 mL × 0.315 g ] / 5.5 g

x = 5.727 ≈ 6 mL

So, the minimum amount in mL of solvent (water) required for the recrystallization will be;

⇒ 85 mL + 6 mL = 91 mL

The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL

Answer:

1.07 g of water.

Explanation:

A reaction between an acid and a base makes water and a salt as product.

Our reaction is:

H₂SO₄ +  2NaOH  →  Na₂SO₄  +  2H₂O

Reactants are the acid and the base. Which is the limiting?

2.9 g . 1mol /98 g =  0.0296 moles of acid

3.53 g . 1mol / 40 g = 0.088 moles of base

2 moles of base react to 1 mol of acid

0.088 moles may react to (0.088 . 1)/2 = 0.044 moles of acid

And we only have 0.0296, sulfuric acid is the limiting

Ratio is 1:2. 1 mol of acid can produce 2 moles of water.

Our 0.0296 moles may produce (0.0296 . 2) /1 = 0.0592 moles of water.

We convert moles to mass:

0.0592 mol . 18g /mol = 1.07 g

 9 is the element Florine

Florine has 9 electrons as well as the 9 protons that determine its atomic number.

The ground state configuration is the lowest energy configuration.

Answer:

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Explanation:

The chemical formula of propane is C₃H₈. When it burns in air, it reacts with gaseous oxygen (O₂), so carbon dioxide (CO₂) and water (H₂O) are formed. The combustion reaction is the following:

C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(g)

Finally, we have to balance the equation. For this, we write a coefficient 3 in CO₂, then a coefficient 4 in H₂O, and a coefficient 5 in O₂. We obtain the balanced equation:

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Answer:

640 g

Explanation:

Step 1: Given and required data

Step 2: Calculate the moles of oxygen gas

We will use the ideal gas equation

P × V = n × R × T

n = P × V / R × T

n = 25 atm × 20 L / (0.082 atm.L/mol.K) × 300 K = 20 mol

Step 3: Calculate the mass corresponding to 20 moles of oxygen

The molar mass of oxygen is 32.00 g/mol.

20 mol × 32.00 g/mol = 640 g

Answer:

Explanation:

¹⁹F₉ + ¹H₁ = ⁴He₂ + ¹⁶X₈

The atomic weight of fluorine is 19 and atomic weight of hydrogen is 1 . Total becomes 20 . Helium , the product has atomic weight of 4 . Therefore atomic weight of product X must be 16 . Similarly we can calculate atomic number of product X . It is found to be 8 .

Hence atomic weight and number of product will be 16 and 8 respectively.

It is similar to oxygen atom

Hence the other product will be oxygen.