what is a compound?And what are some common examples of a compound?​

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:

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

Answer:

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

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:

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.

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:

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:

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.

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:

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:

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.

Explanation:

here's the answer to your question

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:

CO2 is a trigonal planar.

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:

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.

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

Answer:

Milli, m.

Explanation:

Hey there!

In this case, according to the given information, it turns out possible for us to answer to this question by bearing to mind the attached file whereas the most common prefixes and their factors are shown both in standard and scientific notation.

In such a way, we will be able to infer that the prefix related to the numerical value of 0.001 is milli, m, for example 1 mm which is 0.001 m.

Regards!

Answer:

The molar heat of fusion for a given substance basically tells you how much heat is required to melt one mole of that substance at its melting point from two angles.

Explanation:

:)

Answer:

4.858 g

Explanation:

Start with the formula

density = [tex]\frac{mass}{volume}[/tex]

density = 1.98 g/mL

volume = 2.45 mL

mass = ??

rearrange the formula to solve for mass

(density) x (volume) = mass

Add in the substitutes and solve for mass

1.98 g/mL x 2.45 mL = 4.858 g

Answer:

Increase

Decrease

Explanation:While in solution, ionic substances produce ions. The ions in solution determine the conductivity of the solution.

The ionic strength of a solution shows the concentration of ions in a given solution. The more the number of ions in the solution, the greater the ionic strength of the solution and vice versa.

When HNO3 is added to a solution of KOH, the number of ions in the solution increases and so does the ionic strength of the solution.

When KOH is added to a solution of CaCl2 then Ca(OH)2 is formed. The formation of a solid precipitate decreases the concentration of ions in solution as well as the ionic strength of the solution.

Answer:

[tex]K=1.7x10^{-3}[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly setting up the equilibrium expression for the given reaction, in agreement to the law of mass action:

[tex]K=\frac{[NO]^2}{[N_2][O_2]}[/tex]

Next, we plug in the given concentrations on the data table to obtain:

[tex]K=\frac{(0.034)^2}{(0.69)(0.98)}\\\\K=1.7x10^{-3}[/tex]

Regards!

Answer:

[tex]C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to figure out the required net ionic equation by firstly writing out the complete molecular equation between aspirin and sodium acetate:

[tex]C_9H_8O_4+NaC_2H_3O_2\rightarrow C_2H_4O_2+NaC_9H_7O_4[/tex]

Whereas acetic acid and sodium acetylsalicylate are formed. Now, we write the complete ionic equation whereby sodium acetate and sodium acetylsalicylate are ionized because they are salts yet neither aspirin nor acetic acid are ionized as they are weak acids:

[tex]C_9H_8O_4+Na^++C_2H_3O_2^-\rightarrow C_2H_4O_2+Na^++C_9H_7O_4^-[/tex]

Finally, for the net ionic equation we cancel out the sodium spectator ions to obtain:

[tex]C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-[/tex]

Regards!

The volume of 1.5 moles of gas at STP  is 33.6 L.

The volume of the gas at STP is calculated as follows;

I mole of gas at STP = 22.4 L

1.5 moles of the gas at STP = ?

= 1.5 moles  x 22.4 L/mole

= 33.6 L

Thus, the volume of 1.5 moles of gas at STP  is 33.6 L.

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The properties of solids, liquids and gases regarding their shapes and volumes are:

                       Shape             Volume

Solids              Fixed                Fixed

Liquids           Variable            Fixed

Gases             Variable           Variable

Solids have strong attraction forces between their molecules. Thus, the molecules are closely packed with little movement. As a consequence, both shape and volume are fixed.

In liquids, attraction and repulsion forces are similar. They have a little more movement than the solid state. Then, they do have a fixed volume but they adopt the shape of the container.

Gases have very weak attraction forces between their molecules. They move very freely and expand trying to occupy as much volume as possible. So, they have a variable volume and shape (adopt the shape of the container).

You can learn more about states of matter here:

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Answer:

77%

Explanation:

First we convert 3.0 g of cyclohexanol (C₆H₁₂O) to moles, using its molar mass:

Then we convert 1.9 g of cyclohexene (C₆H₁₀) to moles, using its molar mass:

Finally we calculate the theoretical yield:

Answer:

Frequency and wavelength are inversely proportional. c=f⋅λ (The speed of light is directly proportional to f and λ)

Answer:

B

Explanation:

Answer:

0.696 m

Explanation:

We'll begin by calculating the number of mole in 16.7 g of C₆H₁₂O₆. This can be obtained as follow:

Mass of C₆H₁₂O₆ = 16.7 g

Molar mass of C₆H₁₂O₆ = (6×12) + (12×1) + (6×16)

= 72 + 12 + 96

= 180 g/mol

Mole of C₆H₁₂O₆ =?

Mole = mass / molar mass

Mole of C₆H₁₂O₆ = 16.7 / 180

Mole of C₆H₁₂O₆ = 0.093 mole

Next, we shall convert 133.6 g of water to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

133.6 g = 133.6 g × 1 Kg / 1000 g

133.6 g = 0.1336 Kg

Thus, 133.6 g is equivalent to 0.1336 Kg.

Finally, we shall determine the molality of the solution. This can be obtained as illustrated below:

Mole of C₆H₁₂O₆ = 0.093 mole

Mass of water = 0.1336 Kg

Molality =?

Molality = mole / mass of water (in Kg)

Molality = 0.093 / 0.1336

Molality = 0.696 m

Therefore, the molality of the solution is 0.696 m

Answer:

The heat absorbed in Trial 2 is about 1,240 J greater than the heat absorbed in Trial 1.

The heat absorbed in Trial 2 is about 1,674 J greater than the heat absorbed in Trial 1.

The amount of heat absorbed or released by a substance can be calculated using the formula Q = mcΔT, where Q represents heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

In this case, the mass (m) is the same in both trials, but the initial and final temperatures (ΔT) differ. By comparing the values of ΔT in both trials, we can determine the difference in the amount of heat absorbed.

In Trial 1, the initial temperature is 15.0 °C and the final temperature is 65.0 °C, resulting in a ΔT of 65.0 - 15.0 = 50.0 °C.

In Trial 2, the initial temperature is 10.0 °C and the final temperature is 65.0 °C, resulting in a ΔT of 65.0 - 10.0 = 55.0 °C.

Since the specific heat capacity of water is approximately 4.18 J/g°C, we can calculate the difference in heat absorbed:

ΔQ = mcΔT = (80.0 g)(4.18 J/g°C)(55.0 °C - 50.0 °C) = 1,674 J

Therefore, the heat absorbed in Trial 2 is approximately 1,674 J greater than the heat absorbed in Trial 1.

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