Which pair of compounds are both products of photosynthesis?

Answer:

72 kJ of heat is removed.

Explanation:

First, the ethanol vapor will reduce its temperature until the temperature of the boiling point, then it will occur a phase change from vapor to liquid, and then the temperature of the liquid will decrease. The total heat will be:

Q = Q1 + Q2 + Q3

Q1 = n*cv*ΔT1, Q2 = m*Hl, and Q3 = n*cl*ΔT2

Where n is the number of moles, cv is the specific heat of the vapor (65.44 J/K.mol, cl is the specific heat of the liquid (111.46 J/K.mol), Hl is the heat of liquefaction (-836.8 J/g), m is the mass, and ΔT is the temperature variation (final - initial).

Q = n*cv*ΔT1 + m*Hl + n*cl*ΔT2

The molar mass of ethanol is 46 g/mol, and the number of moles is the mass divided by the molar mass:

n = 74.2/46 = 1.613 moles

Q = 1.613*65.44*(78.37 - 83) + 74.2*(-836.8) + 1.613*111.46*(26 - 78.37)

Q = -72000 J

Q = -72 kJ (because it is negative, it is removed)

Answer:

The percent yield of the reaction is 82%

Explanation:

First step: make the chemist equation.

2 Al (s) + Fe2O3 (s) → 2 Fe (s) + Al2O3 (s)

As the statement says that aluminun is in excess, the limiting reactant is the Fe2O3

Second step: Find out the moles in the reactant.

Molar weight Fe2O3: 159.7 g/m

Mass / Molar weight = moles

50 g /159.7 g/m = 0.313 moles

Third step: Analyse the reaction. 1 mol of Fe2O3 makes 2 moles of Fe.

1 mol Fe2O3 ____ 2Fe

0.313 mol Fe2O3 ____ 0.626 moles

Molar weight Fe = 55.85 g/m

Moles . molar weight = mass

55.85g/m . 0.626m = 34.9 grams

This will be the 100% yield of the reaction but we only made 28.65 g

34.9 g ____ 100%

28.65 g ____ 82.09 %

Answer:

83%

Explanation:

We first get the chemical reaction :

2 Al(s) + Fe2O3(s) ------> 2Fe(s) + Al2O3(s) 

From the reaction we can see that one mole of the oxide yielded 2 moles of iron.

Firstly, we need to calculate the theoretical yield of the iron. This is done as follows. The number of moles of the oxide equals the mass of the oxide divided by the molar mass of the oxide = 50g ÷ 159.7 = 0.313moles

From the first relation, one mole oxide yielded 2 moles iron, hence 0.313 mole oxide will yield 2 × 0.313 mole iron = 0.616 moles

The mass of iron thus generated = 0.616 × 56 = 34.496g

% yield = Actual yield/theoretical yield × 100%

%yield = 28.65/34.396 × 100% = 83%

Answer:

a) 1 litre of  10% solution and 7 litre of 20% solution

b) 1.67 litres of 50% solution and 8.33 litres of the 20% solution

c) 3.75 litres of 50% solution and 6.25 litres of 20% solution

Explanation:

Given:

chemist needs = 10 liters of a 25% acid solution

Concentration of three solutions that are to be mixed = 10%, 20% and 50%.

Solution:

A) Use 2 liters of the 50% solution

Let us mix this with 10% and 20% solution

They will have to equal 8 litres

Let x=20% solution

Then (8-x) =10%

So the equation becomes,

10%(8-x)+ 20%x+50%(2)=25(10)

(0.1)(8-x) +0.2x+0.50(2)= 0.25(10)

0.8-0.1x+0.2x+1.0=2.5

0.2x-0.1x=2.5-0.8-1.0

0.1x=0.7

[tex]x=\frac{0.7}{0.1}[/tex]

x= 7

so, 8-x = 8 -7= 1 litre of  10% solution and 7 litre of 20% solution

B)Use as little as possible of the 50% solution

Let x be the amount of 50% solution.

Then(10-x) be the 20% solution

Now the equation becomes,

50%(x)+20%(10-x)=25%(10)

0.50x+0.2(10-x)=0.25(10)

0.5x+2.0-0.2x=2.5

0.3x=2.5-2.0

0.3x=0.5

[tex]x=\frac{0.5}{0.3}[/tex]

x=1.67  

now (10-x)=(10-1.67)=8.33

so there will be 1.67 litres of 50% solution and 8.33 litres of the 20% solution

c) ) Use as much as possible of the 50% solution

Let x be the amount of 50% solution.

Then(10-x) be the 20% solution

Now the equation becomes,

50%(x)+10%(10-x)=25%(10)

0.50x+0.1(10-x)=0.25(10)

0.5x+1.0-0.1x=2.5

0.4x=2.5-1.0

0.4x=1.5

[tex]x=\frac{1.5}{0.3}[/tex]

x=3.75

Now, (10-x)=(10- 3.75)=6.26

So there will be 3.75 litres of 50% solution and 6.25 litres of 20% solution

The amount of liters of each solution to satisfy each given condition are;

A) 8.33 liters of 20% solution

B) 6.25 liters of 10% solution

C) 1 liter of the 10% solution

A) Use as little as possible the 50% solution.

Mix it with 20% solution only

Let x be the amount of 50% solution

Thus;

(10 - x) = 20% solution

equation:

0.50x + 0.20(10 - x) = 0.25(10)

0.5x + 2 - 0.2x = 2.5

0.3x = 2.5 - 2

0.3x  = 0.5

x = 0.5/0.3

x = 1.67 liters of 50% solution required

Thus; 10 - 1.67 = 8.33 liters of 20% solution

B) Use as little as possible of the 50% solution;

Mix it with the 10% solution only

Let x be amount of 50% solution

Thus;

(10 - x) = 10% solution

equation:

0.50x + 0.10(10 - x) = 0.25(10)

0.5x + 1 - 0.1x = 2.5

0.4x = 2.5 - 1

x = 1.5/0.4

x = 3.75 liters of 50% solution required. Thus;

10 - 3.75 = 6.25 liters of 10% solution

C) Use 2 liters of the 50% solution

Mix it with 10% and the 20% and they will have to equal 8 liters.

Let x be the amount of 20% solution

Thus;

8 - x = 10% solution

Equation:

0.20x + 0.10(8 - x) + 0.50(2) = 0.25(10)

0.20x + 0.8 - 0.10x + 1 = 2.5

0.2x - 0.1x + 1.8 = 2.5

0.1x = 2.5 - 1.8

0.1x = 0.7

x = 0.7/0..1

x = 7 liters of 20% solution

Thus; 8 - 7 = 1 liter of the 10% solution

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Answer:

Maximum amount of [tex]CO_{2}[/tex] can be produced is 37.5 g

Explanation:

Balanced equation: [tex]2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O[/tex]

Molar mass of butane ([tex]C_{4}H_{10}[/tex])  = 58.12 g/mol

Molar mass of [tex]O_{2}[/tex] = 32 g/mol

Molar mass of [tex]CO_{2}[/tex] = 44.01 g/mol

So, 24 g of butane  = [tex]\frac{58.12}{24}mol[/tex] of butane = 2.422 mol of butane

Also, 44.3 g of [tex]O_{2}[/tex]  = [tex]\frac{44.3}{32}mol[/tex] of [tex]O_{2}[/tex] = 1.384 mol of [tex]O_{2}[/tex]

According to balanced equation-

2 moles of butane produce 8 mol of [tex]CO_{2}[/tex]

So, 2.422 moles of butane produce [tex](\frac{8}{2}\times 2.422)moles[/tex] of [tex]CO_{2}[/tex] = 9.688 moles of [tex]CO_{2}[/tex]

13 moles of [tex]O_{2}[/tex] produce 8 mol of [tex]CO_{2}[/tex]

So, 1.384 moles of [tex]O_{2}[/tex] produce [tex](\frac{8}{13}\times 1.384)moles[/tex] of [tex]CO_{2}[/tex] = 0.8517 moles of [tex]CO_{2}[/tex]

As least number of moles of [tex]CO_{2}[/tex] are produced from [tex]O_{2}[/tex] therefore [tex]O_{2}[/tex] is the limiting reagent.

So, maximum amount of [tex]CO_{2}[/tex] can be produced = 0.8517 moles = [tex](44.01\times 0.8517)g=37.5 g[/tex]

To calculate the maximum mass of carbon dioxide produced, use the balanced equation and determine the limiting reactant. Then, calculate the moles of carbon dioxide produced using the mole ratio from the balanced equation and convert it to grams using the molar mass of carbon dioxide.

To calculate the maximum mass of carbon dioxide that could be produced by the reaction between gaseous butane (C4H10) and gaseous oxygen (O2), we need to use the balanced equation for the reaction:

C4H10 + 13/2 O2 → 4 CO2 + 5 H2O

First, we need to determine the limiting reactant by comparing the moles of butane and oxygen. The molar mass of butane is 58.1 g/mol, so 24 g of butane is equal to 24/58.1 moles. The molar mass of oxygen is 32 g/mol, so 44.3 g of oxygen is equal to 44.3/32 moles.

Next, we calculate the moles of carbon dioxide produced using the mole ratio from the balanced equation. Since the mole ratio between butane and carbon dioxide is 1:4, the moles of carbon dioxide produced is 4 times the moles of butane. Finally, we convert the moles of carbon dioxide to grams by multiplying by the molar mass of carbon dioxide (44 g/mol).

Using this information, we can calculate the maximum mass of carbon dioxide that could be produced. The answer should be rounded to 3 significant digits to match the rounding specified in the question.

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Answer:

0.272g

Explanation:

To calculate the mass of oxygen collected, we can calculate the number of moles of oxygen collected and multiply this by the molar mass of oxygen.

To calculate the number of moles of oxygen collected, we can use the ideal gas equation I.e PV = nRT

Rearranging the equation, n =PV/RT

We now identify each of the terms below before substituting and calculating.

n = number of moles, which we are calculating.

R = molar gas constant = 62.64 L.Torr. K^-1. mol^-1

V = volume = 500ml : 1000ml = 1L, hence , 500ml = 500/1000 = 0.5L

T = temperature = 25 degrees Celsius = 273 + 25 = 298K

P = pressure. But since the gas was collected over water, we subtract the vapour pressure of water from the total pressure = 642- 23.8= 618.2torr

We substitute these values into the equation to yield the following:

n = (618.2 × 0.5) ÷ ( 62.64 × 298)

n = apprx 0.017moles

To calculate the mass of oxygen collected, we need the atomic mass of oxygen. = 16 amu

Thus the mass of oxygen collected = 0.017mole × 16g= 0.272g

To find the mass of oxygen collected over water, we can use the ideal gas law equation and the given values of total pressure, volume, temperature, and vapor pressure of water. By rearranging the equation and calculating the number of moles, we can then determine the mass of oxygen using its molar mass.

To find the mass of oxygen collected, we need to use the ideal gas law equation, PV = nRT, where P is the total pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin by adding 273.15: 25°C + 273.15 = 298.15K.

To calculate the number of moles, we need to subtract the vapor pressure of water (23.8 torr) from the total pressure (642 torr) to obtain the partial pressure of oxygen: 642 torr - 23.8 torr = 618.2 torr.

Now we can rearrange the ideal gas law equation to solve for n:

n = (PV) / (RT) = (618.2 torr * 0.5 L) / (0.0821 L·atm/mol·K * 298.15K)

Using this value of n, we can calculate the mass of oxygen using the molar mass of oxygen (32 g/mol):

mass = n * molar mass = 0.0126 mol * 32 g/mol = 0.4032 g.

Answer: Glucose is an example of carbon-based macromolecule known as carbohydrates

Explanation:

carbon based macromolecule are important cellular components and they perform a variety of functions necessary for growth and development of living organisms. There are 4 major types of carbon based molecules and these includes;

Carbohydrate

Lipids

Proteins and

Nucleic acids.

Carbon is the primary components of these macromolecules. Carbohydrate macromolecules are made up of monosaccharide, disaccharide and polysaccharides. Glucose is an example of a monosaccharide and it has two important types of functional groups: a carbonyl group and a hydroxyl group. I hope this helps. Thanks

Glucose is a simple sugar and is an example of a carbohydrate, which is a type of carbon-based macromolecule. Glucose is a monosaccharide that can join together with other sugars to form complex carbohydrates or polysaccharides. It serves as a vital energy source for cells.

Glucose is a type of carbon-based macromolecule known as a carbohydrate. Carbohydrates are organic molecules composed of carbon, hydrogen, and oxygen atoms and they serve as a vital energy source for cells. The family of carbohydrates includes both simple and complex sugars. Glucose and fructose are examples of simple sugars, whereas starch, glycogen, and cellulose are examples of complex sugars, also called polysaccharides.

Monosaccharides like glucose are the simplest form of carbohydrates. These small molecules can combine to form larger, more complex structures. When many monosaccharides join together, they form polysaccharides such as starch and glycogen, which serve as energy storage in plants and animals respectively.

While glucose serves as the most common fuel for ATP production within cells, excess glucose is either stored for future energy needs in the liver and skeletal muscles as glycogen, or it's converted into fat in adipose cells.

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Answer:

The enthalpy of the reaction is -2855.622 kilo Joules.

Explanation:

[tex]2C_2H_6(g) + 7O_2(g)\rightarrow 4CO_2(g) + 6H_2O(g)[/tex]

We are given:

[tex]\Delta H^o_f_{(C_{2}H_6(g))}= -84.667 kJ/mol[/tex]

[tex]\Delta H^o_f_{O_2((g))}= 0 kJ/mol[/tex]

[tex]\Delta H^o_f_{CO_2((g))}= -393.5 kJ/mol[/tex]

[tex]\Delta H^o_f_{H_2O((g))}= -241.826 kJ/mol[/tex]

The equation used to calculate enthalpy of reaction :

[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=(4 mol\times \Delta H^o_f_{(CO_2(g))}+6 mol\times \Delta H^o_f_{(H_2O(g)))}-(2 mol\times \Delta G^o_f_{(C_{2}H_6(g))}+7 mol\times \Delta H^o_f_{(O_2(g)))[/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[4 mol\times (-393.5 kJ/mol)+6 mol\times (-241.826 kJ/mol)]-[2 mol\times (-84.667 kJ/mol)+7 mol\times 0 kJ/mol][/tex]

[tex]=-2855.622 kJ[/tex]

The enthalpy of the reaction is -2855.622 kilo Joules.

The enthalpy change for the reaction is   -2855.622  kJ/mol

Recall that enthalpy(ΔH) is a state function so;

ΔHreaction = ∑ΔHproducts - ΔHreactants

So;

The equation of the reaction is; (Recall that the question specified that we should use the smallest whole number coefficients)

2C2H6(g) + 7 O2(g) -----> 4CO2(g) + 6H2O(g)

The enthalpy of each of the reactants and products are given below;

[C2H6(g)] = −84.667 kJ/mol

O2 g = 0 KJ/mol ( O2 exists in its standard state)

[CO2(g)]  =  −393.5 kJ/mol

[H2O(g)] = −241.826 kJ/mol

Hence;

ΔHreaction = ∑[4 × (−393.5) + 6 × (−241.826)] - [2 × (−84.667) + (7 × 0)]

ΔHreaction = -2855.622  kJ/mol

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Answer:

Final temperature of calorimeter is 25.36^{0}\textrm{C}

Explanation:

Molar mass of anethole = 148.2 g/mol

So, 0.840 g of anethole = [tex]\frac{0.840}{148.2}moles[/tex] of anethole = 0.00567 moles of anethole

1 mol of anethole releases 5539 kJ of heat upon combustion

So, 0.00567 moles of anethole release [tex](5539\times 0.00567)kJ[/tex] of heat or 31.41 kJ of heat

6.60 kJ of heat increases [tex]1^{0}\textrm{C}[/tex] temperature of calorimeter.

So, 31.41 kJ of heat increases [tex](\frac{1}{6.60}\times 31.41)^{0}\textrm{C}[/tex] or [tex]4.76^{0}\textrm{C}[/tex] temperature of calorimeter

So, the final temperature of calorimeter = [tex](20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}[/tex]

To determine the final temperature of the calorimeter, we need to calculate the heat released by the combustion of anethole using its enthalpy change. Then, we can use the equation q = mcΔT to find the change in temperature of the calorimeter. The final temperature is 16.65 °C.

To determine the final temperature of the calorimeter, we can make use of the equation q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we need to consider the heat capacity of the calorimeter as well. We can start by calculating the heat released by the combustion of anethole using the given enthalpy change of -5539 kJ/mol. Then, we can use the equation to find the change in temperature of the calorimeter.

First, we need to calculate the moles of anethole. Using the molar mass of anethole, which is 178.26 g/mol, we can find:

Next, we can calculate the heat released by combustion:

Now, we can consider the calorimeter's heat capacity:

Since the initial temperature of the calorimeter is 20.6 °C, the final temperature will be:

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Answer:

I would interpret the statement by using a formula.

Explanation:

In order to be scientifical in a research, the scientific method must be used, this means that in this case, Lavoisier should´ve follow a Hypothesis, Objectives, a Methodlogy, Results and made a Discussion and Conclusion.

To prove that the mass gained by the pipe plus the mass of the collected gas "exactly" equaled the lost mass of water, Lavoisier should´ve used this statement as a formula where it proves with numbers that is correct.

For example:

mg= mass gained by the pipe

mc=mass of the collected gas

ml=lost mass of water

mg+mc=ml

Answer:

The answer to your question is 1.1 moles of water

Explanation:

                     2Al(OH)₃  +   3H₂SO₄   ⇒   Al₂(SO₄)₃  +   6H₂O

                       0.45 mol      0.55 mol                                ?

Process

1.- Calculate the limiting reactant

Theoretical proportion

       Al(OH)₃ / H₂SO₄ = 2/3 = 0.667

Experimental proportion

       Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81

From the proportions, we conclude that the limiting reactant is H₂SO₄

2.- Calculate the moles of H₂O

                        3 moles of H₂SO₄ ----------------  6 moles of water

                        0.55 moles of H₂SO₄ -----------    x

                        x = (0.55 x 6) / 3

                        x = 3.3 / 3

                       x = 1.1 moles of water

Answer:

B.

Explanation:

Elemental analysis involves the actual percentage compositions of each element in the sample of the compounds. We know that there are two different compounds here that both contain copper and oxygen. What we don't know is the actual number of copper or oxygen composed in the compounds.

An elemental composition of these compounds would thus tell us this and will give an idea of the chemical formula of each since we now know quite well the percentage composition of each of these elements

The question about the compounds which is most likely to be answered by the results of the analysis is "What is the formula unit of each compound?"

According to the law of multiple proportions, when two elements combine to form different compounds, the proportion of the elements in the different compounds are in simple ratio.

In accordance to this law, the proportion of copper and chlorine in the white and brown compounds vary.

One question that can be answered by an elemental analysis is the question; "What is the formula unit of each compound?"

The elemental analysis shows the amount of each element contained in each of the compounds from which its formula unit can be obtained.

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Final answer:

The aqueous solutions ranked from highest to lowest pH are: NaOH (most basic), Ba(OH)2, N2H2, HOCl, and HCl (most acidic).

Explanation:

To rank the following aqueous solutions in order of decreasing pH, from highest pH (most basic) to lowest pH (most acidic), we need to understand the nature of each compound in water. The ions or molecules that each solution releases in water determine their pH.

NaOH - Sodium hydroxide is a strong base, and will completely dissociate in water, releasing OH- ions, which leads to a high pH.

Ba(OH)2 - Barium hydroxide is another strong base, and it also fully dissociates in water, but being a dibasic base, it releases twice as many OH- ions per formula unit as NaOH, potentially leading to an even higher pH.

N2H2 - Hydrazine is a weak base, it does not completely dissociate in water, but will still increase the pH to some extent.

HOCl - Hypochlorous acid is a weak acid, so it only partially dissociates in water.

HCl - Hydrochloric acid is a strong acid and fully dissociates in water, releasing H+ ions and resulting in a low pH.

The solutions in order of decreasing pH (highest pH to lowest pH) are: NaOH, Ba(OH)2, N2H2, HOCl, HCl.

Answer:

see explanation below

Explanation:

First, you are not providing any data of the bromide solution to calculate the mass. So, in order to help you, I will take some random values from a similar exercise, so you can solve this later with your data.

Let's suppose you add 360 mL of a 1.45 mol/L of a calcium bromide solution into the flask. To calculate the mass it was added, you need to calculate first the moles added. This can be done with the following expression:

M = n/V

Where:

M: molarity of solution

n: moles of solution

V: volume (in liters) of solution

here, you have to solve for n, so:

n = M*V

replacing the above data you have:

n = 1.45 * (0.360) = 0.522 moles

Now that we have the moles, you can calculate the mass by the following expression:

m = n * MM

Where MM it's the molar mass of calcium bromide. The reported MM of calcium bromide is 199.89 g/mol, so replacing:

m = 199.89 * 0.522

m = 104.34 g

And this is the mass that was added of the solution. As I stated before, use your data in this procedure, and you should get an accurate result.

Answer:

a. K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)

b. Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)

c. Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)

d. NO REACTION

Explanation:

For the reactions, the cation and the anion of the compounds will be replaced. The reaction will occur if at least one of the products is insoluble and will form a precipitated.

a. Potassium carbonate = K₂CO₃

Lead(II) nitrate = Pb(NO₃)₂

Products = KNO₃ and PbCO₃.

According to the solubility rules, all K⁺ ions are soluble, with no exceptions, so KNO₃ is soluble. All CO₃⁻² ions are insoluble, and Pb⁺² is not an exception, so PbCO₃ will be insoluble and will form a precipitated, so the reaction happen:

K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)

b. Lithium sulfate = Li₂SO₄

Lead(II) acetate = Pb(C₂H₃O₂)₂

Products = Li(C₂H₃O₂) and PbSO₄

All Li⁺ are solubles, without exceptions, so Li(C₂H₃O₂) is soluble, and all SO₄⁻² are soluble, but Pb⁺² is an exception, so PbSO₄ is insoluble and will form a precipitated, then the reaction happens:

Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)

c. Copper(II) nitrate = Cu(NO₃)₂

Magnesium sulfide = MgS

Products = CuS and Mg(NO₃)₂

All NO₃⁻ are soluble, with no exceptions, so Mg(NO₃)₂ is soluble, and all S⁺² are insoluble, and Cu⁺² is not an exception, so CuS is insoluble, and will form a precipitated, then the reaction happens:

Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)

d. Strontium nitrate = Sr(NO₃)₂

Potassium iodi = KI

Products = K(NO₃)₂ and SrI₂

All K⁺ are soluble, with no exceptions, so K(NO₃)₂ is soluble, and all I⁻ are soluble, and Sr⁺² are not an exception, then SrI₂ is soluble. Therefore, no precipitated is formed and the reaction doesn't happen.

Precipitation reactions occur for some pairs of aqueous solutions, while others do not.

a. The combination of potassium carbonate and lead(II) nitrate will result in the formation of a precipitate. The balanced molecular equation for this reaction is:

Pb(NO3)2 + K2CO3 → PbCO3 + 2 KNO3

b. The combination of lithium sulfate and lead(II) acetate will not result in a precipitation reaction.

c. The combination of copper(II) nitrate and magnesium sulfide will result in the formation of a precipitate. The balanced molecular equation for this reaction is:

Cu(NO3)2 + MgS → CuS + Mg(NO3)2

d. The combination of strontium nitrate and potassium iodide will not result in a precipitation reaction.

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Answer:

Pressure is inversely proportional to the volume of gas.

Explanation:

According to Boyle's law,

The volume of given amount of gas is inversely proportional to the pressure applied on gas at constant volume and number of moles of gas.

Mathematical expression:

P ∝ 1/ V

P = K/V

PV = K

when volume is changed from V1 to V2 and pressure from P1 to P2 then expression will be.

P1V1 = K         P2V2 = K

P1V1 = P2V2

Final answer:

In gases, pressure and volume are inversely related at constant temperature. Furthermore, pressure is directly proportional to temperature and volume is directly proportional to absolute temperature.

Explanation:

At constant temperature, the volume of a fixed number of moles of gas is inversely proportional to the pressure. This means that when the pressure doubles, the volume will halve.

Moreover, pressure is directly proportional to temperature when volume is constant, and the volume of a gas sample is directly proportional to its absolute temperature at constant pressure.

Answer: Option (c) is the correct answer.

Explanation:

It is known that relation between charge, current and time is as follows.

   Total charge passed = current (A) x time (s)

                      = [tex]3.86 \times 16.2 \times 60[/tex]

                    = 3751.92 C

Moles of electrons passed = [tex]\frac{\text{total charge}}{F} [/tex]

                                             = [tex]\frac{3751.92}{96485}[/tex]

                                            = 0.03889 mol

As, the given metal salt is MCl. Therefore, the reduction reaction is as follows.

         [tex]M^{+} + e^{-} \rightarrow M [/tex]

Thus, moles of M = moles of electrons = 0.03889 mol

As we known that molar mass is calculated using the formula:

     Molar mass of M = [tex]\frac{mass}{moles}[/tex]

                                  = [tex]\frac{1.52}{0.03889}[/tex]

                                 = 39.1 g/mol

We know that potassium is the metal which has molar mass as 39 g/mol.

Thus, we can conclude that the metal is identified as K (potassium).

The current passed for cell in the unit time gives the charge passes to the cell. The metal deposited in the electrolysis is Potassium.

The electrolysis is given as the breaking of the salt for the formation of the ions under the influence of the electric current.

The charge transferred ([tex]Q[/tex]) to the cell in the given time is calculated as:

[tex]Q=\rm current\;\times\;time\\\\\textit Q=3.86\;amp\;\times\;16.2\;\times\;60\;sec\\\\\textit Q=3751.92\;C[/tex]

The moles of sample is given as:

[tex]\rm Moles=\dfrac{Charge}{Faraday} \\\\Moles=\dfrac{3751.92}{96,485} \\\\Moles=0.03889\;mol[/tex]

The mass of sample deposited is 1.5 grams, the molar mass of the sample is calculated as:

[tex]\rm Molar\;mass=\dfrac{mass}{moles} \\\\Molar\;mass=\dfrac{1.52}{0.03889}\\\\ Molar\;mass=39.1\;g/mol[/tex]

The molar mass of the compound is 39.1 g/mol. The element with molar mass 39.1 g/mol is Potassium.

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Answer:

Number of protons and electrons stay constant

Number of neutrons Differs

Explanation:

Isotopes are the different kinds of same element. Now, as we know quite well that an element can only have one atomic number, this means that the proton number is irrespective of the type of atom

Of the element. The proton number is the identity of the element.

As we know that the atom is electrically neutral, it means the number of electrons will always stay the same too.

Since isotopes are not alike in every respect, the number of neutrons differ. This means they have same atomic numbers but different mass numbers.

Answer:

Number of protons and electrons stay constant

Explanation:

Answer:

Chemoselectivity

Explanation:

Chemoselectivity is a term that refers to the preferred result of a chemical reaction between two different functional groups.

In this case, the chemical reaction would be the bromination, which is preferred (or chemoselective) towards non-aromatic alkene groups, compared to aromatic alkene groups.

The question is incomplete. The complete question is attached below.

Answer : The correct option is, (A) 7-chloro-3-ethyl-4-methyl-3-heptene

Explanation :

The rules for naming of alkene are :

First select the longest possible carbon chain.

The longest possible carbon chain should include the carbons of double bonds.

The naming of alkene by adding the suffix -ene.

The numbering is done in such a way that first carbon of double bond gets the lowest number.

The carbon atoms of the double bond get the preference over the other substituents present in the parent chain.

If two or more similar alkyl groups are present in a compound, the words di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.

The given compound name will be, 7-chloro-3-ethyl-4-methyl-3-heptene.

The structure of given compound is shown below.

The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene.

The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene. The name is determined by identifying the longest carbon chain, the substituents attached to it, and their positions. In this case, the longest carbon chain has 7 carbons, with a chlorine atom attached at position 7. There is an ethyl group at position 3 and a methyl group at position 4. The presence of double bonds is indicated by the -ene ending.

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Answer:

The electrons stripped from glucose in cellular respiration end up in compound water by the reduction of oxygen.

Explanation:

During electron transport chain electrons are donated by various reducing equivalents such as NADH,FADH2.The donated electrons then moves through various electron carriers .

       During electron transport chain oxygen(O2) act as terminal electron acceptor which accept the electron from complex 4 and thereby get reduced to form water.(H2O).

In cellular respiration, electrons stripped from glucose are eventually incorporated into water. They are carried through the electron transport chain via NADH and FADH2 and combine with oxygen to form water in the process of oxidative phosphorylation.

In the process of cellular respiration, the electrons that are stripped from glucose ultimately end up in water. The process starts with glucose undergoing glycolysis and the Krebs cycle, forming NADH and FADH2 compounds. These compounds then donate their electrons to the electron transport chain in a series of redox reactions.

During this process, free oxygen acts as the final electron acceptor in the chain. The electrons combine with hydrogen ions and the accepted oxygen to form water. This is a part of the process called oxidative phosphorylation. The other product of this process is ATP, which is the main energy currency in cells.

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To find the total number of hours the battery can be used over its lifetime, we need to determine the number of times the battery can be recharged and calculate the cumulative usage hours. The battery loses 2% of its charge after each charging, retaining 98% of the previous charge. Using the formula for the sum of a geometric series, the total number of hours the battery can be used is calculated to be 1000 hours.

To find the total number of hours the battery can be used over its lifetime, we need to determine the number of times the battery can be recharged and calculate the cumulative usage hours. Since the battery loses 2% of its charge after each charging, it retains 98% of the previous charge. We can use this information to create a geometric sequence

First term (a) = 20 hours

Common ratio (r) = 98% or 0.98

Number of terms (n) = number of times the battery can be recharged

Using the formula for the sum of a geometric series, we can calculate the total number of hours the battery can be used:

Sum = a(1 - r^n) / (1 - r)

Substituting the given values:

Sum = 20(1 - 0.98^n) / (1 - 0.98)

This means that the battery can be used for a total of 1000 hours over its lifetime.

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Answer:

[tex]m_{Fe}=71.982 kg[/tex]

Explanation:

First of all, we need  to calculate the moles of H2. Assuming the H2 is an ideal gas:

[tex]n=\frac{P*V}{T*R}[/tex]

[tex]n=\frac{1atm*31100L}{294K*0.082 L*atm*mol^{-1}*K^{-1}}[/tex]

[tex]n=1290 mol[/tex]

Now, to produce 1 mol of H2 is required 1 mol of Fe:

[tex]m_{Fe}=n*M=1290mol*\{55.8 g}{mol}[/tex]

[tex]m_{Fe}=71982 g=71.982 kg[/tex]

The study of chemicals and bonds is called chemistry. There are two types of elements are there and these are metals and nonmetals.

The correct answer is 71.982kg.

The formula used as:-

[tex]n=\frac{P*V}{T*R} \\\\n=\frac{1*31100}{294*0.082} \\\\\\n=1290moles[/tex]

The mass will be:-

[tex]M= n*M\\1290*55.8\\\\=71.982[/tex]

Hence, the correct answer is 71.982.

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Answer:

16

Explanation:

The number of orbitals can be calculated when energy level is given.

For example:

n = 4

So in 4th energy level number of orbitals are,

n² = 4² = 16

There are 16 orbitals when n=4

One is 4s three are 4P five is 4d and seven are 4f.

while the number of electrons in energy level can be calculated as

2n²

n is energy level.

For n=4

number of electrons are,

2(4)² = 32

Answer:

9.82% of iron (II) will be sequestered by cyanide

Explanation:

We should first consider that Iron (II) and cyanide react to form the following structure:

[Fe(CN)₆]⁻⁴

Having considered this:

5.60 Lt Fe(II) 3.00x10⁻⁵ M ,this is, we have 5.60x3x10⁻⁵ =  1.68x10⁻⁴ moles of Fe⁺² (in 5.60 Lt)

Then , we have 9 ml NaCN 11.0 mM:

9 ml = 0.009 Lt

11.0 mM (milimolar) = 0.011 M (mol/lt)

So: 0.009x0.011 = 9.9x10⁻⁵ moles of CN⁻ ingested

As we now that the complex structure is formed by 1 Fe⁺² : 6 CN⁻ :

9.9x10⁻⁵ moles of CN⁻ will use 1.65x10⁻⁵ moles of Fe⁺² (this is, this amount of iron (II) will be sequestered

[(1.65x10⁻⁵ sequestred Fe⁺²)/(1.68x10⁻⁴ total available Fe⁺²)x100

% sequestered iron (II) = 9.82%

Answer : The correct option is, (d) [tex]5.0\times 10^{-12}mole[/tex]

Explanation :

First we have to calculate the [tex]H^+[/tex] concentration.

[tex]pH=-\log [H^+][/tex]

[tex]11.5=-\log [H^+][/tex]

[tex][H^+]=3.16\times 10^{-12}M[/tex]

Now we have to calculate the [tex]OH^-[/tex] concentration.

[tex][H^+][OH^-]=K_w[/tex]

[tex]3.16\times 10^{-12}\times [OH^-]=1.0\times 10^{-14}[/tex]

[tex][OH^-]=3.16\times 10^{-3}M[/tex]

Now we have to calculate the molar solubility of [tex]Zn(OH)_2[/tex].

The balanced equilibrium reaction will be:

[tex]Zn(OH)_2\rightleftharpoons Zn^{2+}+2OH^-[/tex]

The expression for solubility constant for this reaction will be,

[tex]K_{sp}=[Zn^{2+}][OH^-]^2[/tex]

Now put all the given values in this expression, we get:

[tex]5.0\times 10^{-17}=[Zn^{2+}]\times (3.16\times 10^{-3})^2[/tex]

[tex][Zn^{2+}]=5.0\times 10^{-12}M[/tex]

Therefore, the molar solubility of [tex]Zn(OH)_2[/tex] is, [tex]5.0\times 10^{-12}mole[/tex]

Answer:

0.744 M

Explanation:

IO⁻⁴(aq) + 2H₂O(l) ⇌ H₄IO⁻⁶(aq)

Kc = 3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

First let's calculate the new concentration of IO⁻⁴ at equilibrium:

0.904 M * 26.0 mL / 500.0 mL = 0.047 M = [IO⁻⁴]

Now we can calculate [H₄IO⁻⁶] using the formula for Kc:

3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

3.5×10⁻²= [H₄IO⁻⁶] / 0.047 M

[H₄IO⁻⁶] = 0.744 M

Final answer:

The question asks for the equilibrium concentration of H4IO-6 after dilution and reaction has reached equilibrium, implying the use of equilibrium concepts and calculations surrounding concentration and reaction constants. However, without enough detail or context provided on changes in concentration or how the equilibrium constant is applied, an exact answer cannot be directly calculated from the given information.

Explanation:

The question involves calculating the equilibrium concentration of H4IO-6 after diluting NaIO4 and allowing the reaction IO-4(aq) + 2H2O(l) ⇌ H4IO-6(aq); Kc=3.5×10-2 to reach equilibrium. Firstly, calculate the initial concentration of IO-4 after dilution: C1V1 = C2V2, where C1 = 0.904 M and V1 = 26.0 mL, V2 = 500.0 mL. Solving gives C2, the concentration of NaIO4 after dilution. To find the concentration of H4IO-6 at equilibrium, you would typically use the equilibrium constant (Kc), but the question's information does not provide a direct route to calculate this without additional context regarding the relationship between concentrations of reactants and products at equilibrium. Normally, you would set up an ICE table and solve for the equilibrium concentrations using Kc, but without the concentration change (ΔC), this calculation cannot be directly completed.

To determine if the reactions are exothermic or endothermic, we look at the energy required or released when bonds are formed or broken. The formation of NH₃, SO₂, and F₂ is exothermic, whereas the decomposition of H₂O to H₂ and O₂ is endothermic.

To predict whether the following reactions will be exothermic or endothermic, we can consider the energy changes involved in the process of breaking and forming chemical bonds.

A. N₂(g) + 3H₂(g) ------> 2NH₃(g)

This reaction is known to be exothermic. When nitrogen gas reacts with hydrogen gas to form ammonia, energy is released in the process.

B. S(g) + O₂(g) ------> SO₂(g)

The formation of sulfur dioxide from sulfur and oxygen is typically an exothermic reaction because energy is released when the SO₂ molecules are formed.

C. 2H₂O(g) ------> 2H₂(g) + O₂(g)

This reaction involves the decomposition of water vapor into hydrogen and oxygen gas, which requires energy to break the bonds of H₂O molecules. Therefore, it is endothermic.

D. 2F(g) ------> F₂(g)

The formation of fluorine molecules from individual fluorine atoms is an exothermic process. Energy is released when the F₂ molecule is formed due to the formation of a strong bond between the two fluorine atoms.

Answer:

Decay series

Explanation:

A succession of radioactive decay is termed decay series. The radios decay of an unstable nuclei usually continues until a stable isotope is reached. This continuous decay of radioactive isotopes is also known as a radioactive cascade.

It is important to note that most radioisotopes do not decay directly to form a stable nuclei. Instead, they undergo a series of decay until a stable isotope is formed. An example of a decay series can be seen in the decay of uranium-238 to uranium-234.

U-238 is more radioactive than U-234. U-238 first undergoes an alpha particle decay to form thorium 234. This is known as the daughter nuclei. Afterwards, thorium 234 undergoes decay to give protactinium 234. This then undergoes a beta negative decay to form the uranium 234 nuclei.