6. A 90.0 gram sample of an unknown solid is heated to 80 °C and placed into a calorimeter containing 120 grams of water at 24 °C. If the final temperature of the solid sample and the water is 32 °C, what is the specific heat of solid?
PLEASE SHOW WORK​

Answer:

Explanation:

a) The balanced equation for the double replacement reaction between lead (II) nitrate and potassium bromide is:

Pb(NO₃)₂(aq) + 2KBr(aq) → PbBr₂(s) + 2KNO₃(aq)

In this reaction, lead (II) bromide (PbBr₂) will precipitate, while potassium nitrate (KNO₃) will remain in solution.

b) To determine the amount of precipitate produced, we need to first determine the limiting reactant. We can do this by calculating the number of moles of each reactant and comparing it to the stoichiometry of the balanced equation.

The molar mass of lead (II) nitrate is 331.21 g/mol and the molar mass of potassium bromide is 119.00 g/mol.

The number of moles of lead (II) nitrate is 32.5 g / 331.21 g/mol = 0.0981 mol The number of moles of potassium bromide is 38.75 g / 119.00 g/mol = 0.3256 mol

According to the balanced equation, one mole of lead (II) nitrate reacts with two moles of potassium bromide to produce one mole of lead (II) bromide. This means that if all the lead (II) nitrate were to react, it would require 0.0981 mol * 2 = 0.1962 mol of potassium bromide.

Since we have more than enough potassium bromide (0.3256 mol > 0.1962 mol), lead (II) nitrate is the limiting reactant.

The number of moles of lead (II) bromide produced will be equal to the number of moles of lead (II) nitrate consumed, which is 0.0981 mol.

The molar mass of lead (II) bromide is 367.01 g/mol, so the mass of lead (II) bromide produced will be 0.0981 mol * 367.01 g/mol = 36.0 g.

c) To determine the amount of excess reactant remaining, we need to subtract the amount consumed from the initial amount.

The number of moles of potassium bromide consumed is half the number of moles of lead (II) nitrate consumed, which is 0.0981 mol / 2 = 0.04905 mol.

The mass of potassium bromide consumed is 0.04905 mol * 119.00 g/mol = 5.84 g.

The mass of potassium bromide remaining is 38.75 g - 5.84 g = 32.91 g.

The sample of the unknown metal has the mass of the 135 grams. The sample cools from the 100.5 °C to the 35.5 °C, and it releases the 7500 joules of the energy. The specific heat of the sample is 0.854 J/g °C.

Th mass of the metal = 135 g

The initial temperature = 100.5 °C

The final temperature = 35.5 °C

The heat energy releases = - 7500 J

The heat energy is expressed as :

Q = mc ΔT

Where,

The m is mass of the metal = 135 g

The c is the specific heat capacity = ?

The Q is heat energy releases = - 7500 J

The ΔT is the change in the temperature = final temperature - initial temperature.

The ΔT is the change in the temperature = 35.5 - 100.5

The ΔT is the change in the temperature =  - 65 °C

The specific heat capacity, c = Q / m ΔT

The specific heat capacity, c = - 7500 / 135 × - 65

The specific heat capacity, c = 0.854 J/g °C

The specific heat capacity of metal is 0.854 J/g °C.

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According to the ideal gas law, a gas's pressure is inversely related to its volume and directly proportionate to its temperature. So, if a gas sample's volume is reduced, the gas sample's pressure must also increase.

As a result, in order to determine the pressure of the gas sample under the specified circumstances, we must first determine the ratio of the two volumes before multiplying the starting pressure of the sample by that ratio.

We may get the ratio of the two volumes using the ideal gas law as follows: V2/V1 = (100.0 mL/0.75 L) x (273 K/25oC) = 8.02 As a result, the gas sample's pressure at 25 oC with a volume of 100.0 mL is 8.02 times higher than the sample's original pressure.

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The molecular formula of the strychnine, given that it is composed of 75.45% carbon, 6.59% hydrogen, 8.38% nitrogen, and the balance oxygen is C₂₁H₂₂N₂O₂

First, we shall obtain the empirical formula of compound. Details below:

Divide by their molar mass

C = 75.45 / 12 = 6.2875

H = 6.59 / 1 = 6.59

N = 8.38 / 14 = 0.5986

O = 9.58 / 16 = 0.59875

Divide by the smallest

C = 6.2875 / 0.5986 = 10.5

H = 6.59 / 0.5986 = 11

N = 0.5986 / 0.5986 = 1

O = 0.59875 / 0.5986 = 1

Multiply through by 2 to express in whole number

C = 10.5 × 2 = 21

H = 11 × 2 = 22

N = 1 × 2 = 2

O = 1 × 2 = 2

Thus, we can conclude that the empirical formula is C₂₁H₂₂N₂O₂

Now, we shall determine the molecular formula of strychnine. Details below

Molecular formula = empirical × n = mass number

[C₂₁H₂₂N₂O₂]n = 140.22

[(12×21) + (1×22) + (14×2) + (16×2)]n = 334

334n = 334

Divide both sides by 334

n = 334 / 334

n = 1

Molecular formula = [C₂₁H₂₂N₂O₂]n

Molecular formula = [C₂₁H₂₂N₂O₂]1

Molecular formula = C₂₁H₂₂N₂O₂

Thus, we can conclude that the molecular formula of strychnine is C₂₁H₂₂N₂O₂

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(NH4)3PO4 is insoluble in water. The correct option is D

According to their chemical formula and ionic charges, ionic compounds generally follow a set of solubility laws that define their solubility patterns in water. These guidelines aid in determining whether an ionic compound will dissolve in water or not as well as if it will precipitate when combined with other ionic compounds.

Therefore, (NH4)3PO4 is the compound that is expected to be insoluble in water based on the solubility rules.

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The centripetal acceleration experienced by the object can be calculated using the formula a = v^2/r, where v is the speed of the object and r is the radius of the circle. Substituting the given values, we get:

a = (50 cm/s)^2 / (250 cm)
a = 10 cm/s^2

Therefore, the centripetal acceleration experienced by the object is 10 cm/s^2.
To calculate the centripetal acceleration experienced by the object, you can use the formula:
Centripetal acceleration (a_c) = (velocity^2) / radius
Here, the velocity (v) is 50 cm/s and the radius (r) is 250 cm. Plugging in these values, we get:
a_c = (50^2) / 250 = 2500 / 250 = 10 cm/s²
So, the centripetal acceleration experienced by the object is 10 cm/s².

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Answer:

Unsaturated

Explanation:

A solution is unsaturated when it contains less than the maximum amount of solute that is capable of being dissolved.

Using the Ideal Gas Law, the pressure expected for 30 mol of  [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.

To calculate the pressure expected for 30 mol of [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C, we will use both the Ideal Gas Law and van der Waals equation.

Ideal Gas Law: PV = nRT
P = pressure
V = volume (6.00 L)
n = moles (30 mol)
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (20 °C + 273.15 = 293.15 K)

Solve for P (pressure):

P = nRT / V

P = (30 mol)(0.0821 L•atm/mol•K)(293.15 K) / 6.00 L

P = 1210.07 atm

Van der Waals equation:

(P + a(n/V)²)(V - nb) = nRT
a = 4.612 atm•L²/mol²
b = 0.0582 L/mol

Solve for P (pressure):
(P + (4.612)(30/6)²) (6 - 0.0582 * 30) = (30)(0.0821)(293.15)

P = 1179.71 atm

Using the Ideal Gas Law, the pressure is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.

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The ratio of the concentrations at equilibrium is as follows:

Chemical equilibrium is the point in a chemical reaction where both the forward and backward processes are occurring at the same rate.

The concentrations of the reactants and products are constant at equilibrium because the forward and reverse speeds are equal.

Considering the given statements based on the reaction equilibrium concentrations, the correct options are:

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The equilibrium constant is written as;

Keq = [tex][D]^2 [C]^5/[A] [B]^3[/tex]

The equilibrium constant's value is influenced by the reaction's chemical make-up and temperature.

The product of the product concentrations, each raised to the power of their stoichiometric coefficient, divided by the product of the reactant concentrations, each raised to the power of their stoichiometric coefficient, is known as the equilibrium constant.

The equilibrium constant is Keq = [tex][D]^2 [C]^5/[A] [B]^3.[/tex]

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The volume of the balloon once they have cooled to the outside temperature of -34.0 c is 1.53 L.

Charles' law predicts the relationship between the volumes and the temperatures of a sample of an ideal gas at different conditions. For this equation to hold true, the number of molecules and the pressure must remain constant despite changes in the environment.

Determine the volume of the balloon outside, V2. We do this by applying Charles' law, such that we relate the volume, V, and the temperature, T, of a sample of gas as

V₁ /T₁ = V₂/ T₂

at two conditions. We are given the following values for the variables:

• V₁ = 1.90 L

T₁ = 22.0+ 273.15 = 295.15 K

T₂= 34.0+273.15= 239.15 K

We proceed with the solution.

V₁/T₁ = V₂/T₂

V₁ /T₁ × T₂ = V₂

1.90 L/295.15 K x 239.15 K = V₂

1.53 L =V₂

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The complete question is

Balloons for a New Year's Eve party in Fargo, ND, are filled to a volume of 1.90 L at a temperature of 22.0 degrees Celsius and then hung outside where the temperature is -34.0 degrees Celsius. What is the volume of the balloons after they have cooled to the outside temperature? Assume that atmospheric pressure inside and outside the house is the same.

We need 21 grams of potassium chloride to make a 50 mL saturated solution at 40 degrees Celsius. It's important to note that if the temperature or volume of the solution were to change, the amount of solute needed to make a saturated solution would also change, as solubility is dependent on both temperature and volume.

According to the solubility table, the solubility of potassium chloride at 40 degrees Celsius is 42 grams per 100 mL of water. This means that we can dissolve 42 grams of potassium chloride in 100 mL of water at 40 degrees Celsius to make a saturated solution.

To make a 50 mL saturated solution, we can use the following formula:
mass of solute = (volume of solution x solubility)/100
mass of solute = (50 x 42)/100
mass of solute = 21 grams
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Statement that shows that the total energy of a system remains constant and is conserved would confirm the law of conservation of energy.

What is law of conservation?

The law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. Therefore, any statement that shows that the total energy of a system remains constant and is conserved would confirm the law of conservation of energy.

Here are some examples of statements that confirm the law of conservation of energy:

All of these statements confirm the law of conservation of energy by showing that the total energy of a system is conserved over time.

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Complete question is: "The total energy of a system remains constant and is conserved" statement would confirm the law of conservation of energy.

1. To solve for the grams of H2 needed, we need to use the given ΔH value to calculate the amount of moles of N2 that reacted. From the balanced chemical equation, we know that for every 3 moles of H2 that reacts, 1 mole of N2 reacts. Therefore, we can use the mole ratio to convert the moles of N2 to moles of H2 and then use the molar mass of H2 to convert to grams.

First, we need to calculate the moles of N2 that reacted to produce 150.9kJ of heat:
ΔH = -92.40 kJ/mol N2
150.9 kJ = (1 mol N2 / -92.40 kJ) x (-150.9 kJ)
mol N2 = 1.63 mol

Using the mole ratio from the balanced chemical equation:

1 mol N2 : 3 mol H2

We can calculate the moles of H2 needed:3 mol H2 = 1 mol N2

3 mol H2 = 1.63 mol N2
mol H2 = 0.543 mol

Finally, we can convert moles of H2 to grams:

mol H2 = 0.543 mol
molar mass of H2 = 2.02 g/mol
grams of H2 = (0.543 mol) x (2.02 g/mol)
grams of H2 = 1.10 g

Therefore, 1.10 grams of H2 are needed to involve 150.9kJ of heat.

2. To solve for the moles of NH3 produced, we can use the same mole ratio from the balanced chemical equation:

1 mol N2 : 2 mol NH3

From the moles of N2 that reacted calculated in part 1, we can calculate the moles of NH3 produced:
1 mol N2 = 2 mol NH3
1 mol N2 = 1.63 mol N2
mol NH3 = (2 mol NH3 / 1 mol N2) x (1.63 mol N2)
mol NH3 = 3.26 mol

Therefore, 3.26 moles of NH3 were produced in the process.

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Answer:

The main difference between practical work inside and outside a laboratory is that the practical work inside the lab includes good equipment and chemicals which are very advanced and the practical outside a laboratory is more about the safety of life.

Explanation:

Practicals are set up at stations with lab equipment and chemicals, where students can learn, and researchers can experiment and find different new things.

Thus, the practical work inside the lab includes lab equipment and chemicals, and the practical outside a laboratory is more about conserving nature.  

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A crucial trace mineral is chromium. Trivalent chromium, which is safe for people, and hexavalent chromium, which is toxic, are the two types.

Thus, Foods and dietary supplements both contain trivalent chromium. It might assist maintain normal blood sugar levels by enhancing the body's utilization of mineral.

Chromium is used by people to treat deficiencies. Additionally, it is used to treat bipolar disorder, diabetes, high cholesterol, and a variety of other conditions, but the majority of these uses are not well-supported by science.

Chromium by mouth doesn't help control blood sugar levels in people with prediabetes. Schizophrenia. Taking chromium by mouth doesn't affect weight or mental health in people with schizophrenia.

Thus, A crucial trace mineral is chromium. Trivalent chromium, which is safe for people, and hexavalent chromium, which is toxic, are the two types.

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When 28.2 grams of Mg is burned, 46.7 grams of MgO will be formed.

The balanced chemical equation for the combustion of magnesium is:

2 Mg + O2 --> 2 MgO

This equation shows that 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO.

To determine the amount of MgO formed when 28.2 grams of Mg is burned, we first need to convert the given mass of Mg to moles:

molar mass of Mg = 24.31 g/mol

moles of Mg = mass of Mg / molar mass of Mg

moles of Mg = 28.2 g / 24.31 g/mol

moles of Mg = 1.16 mol

According to the balanced chemical equation, 2 moles of Mg produce 2 moles of MgO. Therefore, we can use the mole ratio to calculate the moles of MgO formed:

moles of MgO = moles of Mg x (2 moles of MgO / 2 moles of Mg)

moles of MgO = 1.16 mol x 1

moles of MgO = 1.16 mol

Finally, we can convert the moles of MgO to grams using its molar mass:

molar mass of MgO = 40.31 g/mol

mass of MgO = moles of MgO x molar mass of MgO

mass of MgO = 1.16 mol x 40.31 g/mol

mass of MgO = 46.7 g

Therefore, when 28.2 grams of Mg is burned, 46.7 grams of MgO will be formed.

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the answer is 2.5 according to me

The rate constant (k) for a chemical reaction can be determined experimentally by measuring the reaction rate (v) at different concentrations of reactants and plotting the data using a suitable rate law equation.

There are different methods to determine the rate constant depending on the type of reaction, but a general approach is as follows:

Conduct the reaction under different initial concentrations of reactants while keeping other variables constant such as temperature, pressure, and pH.

Measure the reaction rate at each concentration by monitoring the change in concentration of reactants or products over time.

An overview was given based on incomplete information.

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If we label the compounds ABCD from left to right;

A - 12.10

B - 15.90

C - 12.66

D - 12.35

A molecule or compound's acidity is quantified by the pKa, which is the negative logarithm (base 10) of the acid dissociation constant (Ka). The lower the pKa value, the stronger the acid; it reflects a compound's propensity to give a proton (H+) in a solution.

The compound that has the highest number of attachment of the most electronegative elements would have the greatest pKa.

The justification of the answer above is that, seeing that the compound labelled B has three highly electronegative atoms hence it would have the most or the highest pKa of about  15.90 among the other compounds. The other compounds A, C and D have fewer electronegative atoms attached and thus a lower pKa as shown

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Answer:

The enthalpy of combustion of ethanol is -1430 kJ/mol, which means that for every mole of ethanol that is burned, 1430 kJ of heat is released.

To determine the amount of heat given off from the combustion of 1 dm³ of ethanol, we need to first calculate the number of moles of ethanol in 1 dm³.

1 dm³ is equivalent to 1000 cm³. Since the density of ethanol is 0.79 g/cm³, the mass of 1 dm³ of ethanol can be calculated as:

mass = density x volume

mass = 0.79 g/cm³ x 1000 cm³

mass = 790 g

To convert this mass to moles, we need to divide by the molar mass of ethanol:

moles = mass / molar mass

moles = 790 g / 46 g/mol

moles = 17.17 mol

Therefore, 1 dm³ of ethanol contains 17.17 moles of ethanol.

To calculate the heat given off from the combustion of 1 dm³ of ethanol, we can use the following equation:

heat = enthalpy of combustion x moles of ethanol

heat = -1430 kJ/mol x 17.17 mol

heat = -24,551 kJ

Therefore, the heat given off from the combustion of 1 dm³ of ethanol is -24,551 kJ, or approximately 24,551 kJ of heat is released.

To determine the change in enthalpy associated with the generation of 155.3 g of [tex]C_5H_1_0O[/tex] we must first calculate the moles of [tex]C_5H_1_0O[/tex] produced using its molar mass.

The molar mass of C₅H₁₀O is:

5(12.01 g/mol) + 10(1.01 g/mol) + 16.00 g/mol = 88.15 g/mol

Moles of C₅H₁₀O produced:

155.3 g / 88.15 g/mol = 1.763 mol C₅H₁₀O

The balanced chemical equation states that the formation of 1 mol of C₅H₁₀O results in an enthalpy change of -191.3 kJ.

As a result, the enthalpy change during the formation of 1.763 mol of C₅H₁₀O is: -191.3 kJ/mol x 1.763 mol = -337.8 kJ

The enthalpy change for the production of 155.3 g of C5H10O is -337.8 kJ.

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Answer:

All is Correct

Explanation:

Fossil fuels have the following properties:

Therefore, the answer is to select all

Answer:

It's A, B, and D

Explanation:

Maybe not D, because that is burning wood like charcoal. Not sure about that. Hope this helps!

The net ionic equation of the reaction is as follows:

4 Al3+(aq) + 12 NO3-(aq) + 3 Ag(s) = 4 Al(s) + 12 Ag+(aq) + 12 NO3-(aq)

Ions which remain in their ground state and do not take part in the reaction are called spectator ions. The net ionic equation cancels out these ions, which are present on both the reactant and product sides of the equation.

Spectator ions, which can be found on both the reactant and product sides, but are not included in the finished reaction from the net ionic equation. The [tex]NO^3^-[/tex] ions are spectator ions in this example, thus taking them out of the equation. The net ionic equation makes up the rest of the species.

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The mass of hydrogen produced in the reaction is 2g.

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.

A mole is defined as the amount of substance containing the same number of atoms, molecules, ions, etc. as the number of atoms in a sample of pure 12C weighing exactly 12 g.

Given,

Mass of Zn = 65g

Mass of HCl = 72g

Moles of Zn = mass / molar mass

= 65 / 65 = 1 mole

Moles of HCl = 72 / 36.5

= 1.97 moles

Since moles of Zn is lesser, therefore it is the limiting reagent.

From the reaction, 1 mole of Zn gives 1 mole of hydrogen

Moles of hydrogen = 1 mole

mass of hydrogen = moles × molar mass

= 1 × 2 = 2g

Therefore, the mass of hydrogen produced in the reaction is 2g.

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A pair of molecules which exist in two forms that are mirror images of each other but cannot be superimposed one upon the other are called the enantiomers. They are present in pairs and have similar molecular shape.

The compounds with the same molecular formula but are non-superimposable non-mirror images are called diastereomers. They have distinct physical properties and molecular shape.

The constitutional isomers have the same molecular formula but have different bonding atomic organization and bonding patterns.

So here:

1st structure is constitutional isomers (c), 2nd structures are enantiomers (b) and the 3rd are completely different not isomers (d).

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The molarity of the product is  0.00368 M. Option B

When the rates of the forward and reverse reactions in a chemical reaction are equal and the concentrations of reactants and products are stable over time, this condition is referred to as reaction equilibrium.

The equilibrium constant is a measure of the relative concentrations of reactants and products at equilibrium.

[tex]Keq = [H_{2} O] [CO]/[H_{2}] [CO_{2} ]\\0.106 = x^2/(0.0113)^2\\x = \sqrt{} 0.106 (0.0113)^2\\x = 0.00368 M[/tex]

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20 g of calcium would release the same number of valence electrons as 23 g of sodium.

The atomic number of calcium (Ca) is 20, which means it has 20 electrons in its neutral state. When calcium loses two electrons, it becomes a Ca2+ ion with 18 electrons.

On the other hand, the atomic number of sodium (Na) is 11, which means it has 11 electrons in its neutral state. When sodium loses one electron, it becomes a Na+ ion with 10 electrons.

To release the same number of valence electrons as 23 g of Na, we need to calculate how many moles of Na there are in 23 g:

Molar mass of Na = 23 g/mol

Number of moles of Na = 23 g / 23 g/mol = 1 mol

Since each Na+ ion has lost one electron, 1 mol of Na+ ions has lost 1 mol of valence electrons.

To release the same number of valence electrons as 1 mol of Na+ ions, we need to calculate how many moles of Ca2+ ions are required:

1 mol of Na+ ions = 1 mol of valence electrons

1 mol of Ca2+ ions = 2 mol of valence electrons

Therefore, we need 0.5 mol of Ca2+ ions to release the same number of valence electrons as 1 mol of Na+ ions.

Finally, we can calculate the mass of calcium that would release the same number of valence electrons as 23 g of Na:

Molar mass of Ca = 40 g/mol

Mass of Ca required = 0.5 mol x 40 g/mol = 20 g

Therefore, 20 g of calcium would release the same number of valence electrons as 23 g of sodium.

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