Jonathan is working in his basement on a science fair project when his little sister closes and locks the door. Jonathan wants to let his parents know that he is stuck down in the basement. He can either yell as loudly as he can, bang on the metal pipes, or bang on the concrete wall. Which should he do if he wants someone to hear him? Explain your answer, and explain why the other options would not be as effective.

Answer:

The volume for the ideal gas is: 4647.5 Liters

Explanation:

Formula for the Ideal Gases Law must be applied to solve this question:

P . V = n .  R . T

We convert the T° to K → 100°C + 273 = 373 K

We convert pressure value from kPa to atm.

2 kPa . 1atm/101.3 kPa = 0.0197 atm

We replace data in the formula.

V = ( n . R . T) / P → (3 mol . 0.082 . 373K) / 0.0197 atm =

The volume for the ideal gas is: 4647.5 Liters

The standard potential for a galvanic cell consisting of nickel and copper electrodes at 25 °C is calculated using the standard reduction potentials for each electrode, resulting in a cell potential of +0.77 V.

To calculate the standard potential for the galvanic cell at 25 °C, you need to know the standard reduction potentials for the nickel and copper electrodes. These potentials can be found in a table of standard reduction potentials. The cell potential (Ecell) is the difference between the reduction potential of the cathode and the anode.

In our case, for a cell composed of nickel and copper electrodes, the standard reduction potential for the nickel(II) ion to nickel reaction is approximately -0.25 V, and for the copper(I) ion to copper reaction, it is +0.52 V.

Considering that the copper electrode undergoes reduction (acts as cathode) and the nickel electrode undergoes oxidation (acts as anode), we use the following formula:

[tex]E_{cell} = E_{cathode} - E_{anode[/tex]

Therefore: Ecell = (+0.52 V) - (-0.25 V) = +0.52 V + 0.25 V = +0.77 V

The standard potential for this galvanic cell at 25 °C is +0.77 V.

Most nonmetals do not lose electrons when bonding with metal atoms; instead, they tend to gain electrons, forming anions. They share electrons in covalent bonds with other nonmetals.

The characteristic that is not typical of most nonmetals is their atoms lose electrons when bonding with metal atoms. When bonding occurs, nonmetals usually do the opposite; they tend to gain electrons to achieve more stable electron configurations. This is often seen in ionic bonds, where metals lose electrons to become positively charged cations, and nonmetals gain electrons to become negatively charged anions. However, in the case of bonding with other nonmetals, they typically share electrons, resulting in covalent bonds. Furthermore, nonmetals do indeed readily form compounds with other elements, either through ionic or covalent bonding.

Final answer:

Most nonmetals gain electrons when bonding with metal atoms to achieve stability, and they can also share electrons with other nonmetal atoms.

Explanation:

The characteristic that is not a characteristic of most nonmetals is that their atoms lose electrons when bonding with metal atoms. Nonmetals tend to gain electrons when bonding with metal atoms in order to achieve a stable electron configuration. They can also share electrons with other nonmetal atoms to form covalent bonds. Additionally, nonmetals readily form compounds with other elements.

Answer:

81mmHg

Explanation:

Since the total pressure is 2700mmHg and we are told that nitrogen occupies 3% of that pressure in the Venus atmosphere. We have to find the pressure of nitrogen using its percentage.

It should be noted that the total pressure of the atmosphere is the sum total of the partial pressures of all the respective gases in the atmosphere.

Hence partial pressure of nitrogen= 3.0%× 2700mmHg= 3.0/100× 2700

Partial pressure of nitrogen= 81mmHg

Answer: A. There are two hydrogen atoms and one oxygen atom per water molecule.

Explanation:

The chemical formular for water "H2O", means it contain 2 hydrogen atoms and one oxygen atom per molecule.

Final answer:

The molecular formula H2O indicates that there is a ratio of two hydrogen atoms to one oxygen atom in each water molecule, which makes option A the correct answer.

Explanation:

The formula of water, H2O, indicates several things about water molecules. The most direct interpretation is that for each water molecule, there are two hydrogen atoms and one oxygen atom. Therefore, the correct option in the provided question is A: there are two hydrogen atoms and one oxygen atom per water molecule. Although oxygen has a greater mass than hydrogen, the molecular formula does not reflect mass directly but rather the ratio of atoms within the molecule. Looking at the molecular mass, we know that the mass of an oxygen atom is approximately 16 times that of a hydrogen atom. Thus, while there are two atoms of hydrogen for every one of oxygen, oxygen contributes more to the overall mass of a molecule due to its greater atomic mass.

Answer:

Temperature of gas: 3 °C Height of the column of gas: 5.7 cm Temperature of gas: 276k Volume of gas: .72 cm^3

Explanation:

edge 2021

The value of  Temperature is 276K

and of Volume is 0.71592 [tex]cm^{3}[/tex]

As per the formula of conversion of degree Celsius to kelvin temperature

1°C=273K

∴23 °c +273= 276 K

V=π[tex]r^{2}[/tex]h

V= volume

r = radius of capillary tube =0.2 cm

h=height of column=5.7 cm

∴V=3.14×[tex](0.2)^{2}[/tex]×5.7

∴V=0.71592[tex]cm^{3}[/tex]

Hence, the value of temperature and volume are 276K and 0.71592[tex]cm^{3}[/tex], respectively.

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Answer:

1.5 litre of C5H12 produce 9 litre of water vapour;

Explanation:

First balance the combustion reaction;

[tex]C_5H_12 +8O_2[/tex] → [tex]5CO_2 +6H_2O[/tex]

from the balance reacion it is clearly that,

one mole of C5H12 reacts completely with 8 mole of O2 gas;

Hence

one litre of C5H12 reacts completely with 8 litre of O2 gas;

there fore 1.5 litre C5H12 needs 12 litre oxygen gas but we have 15 litre.

so,

C5H12 is a limiing reactant and O2 is a excess reactant.

so quantity of H2O depends on limiting reactant;

one litre of C5H12 produce 6 litre of water vapour;

therefore,

1.5 litre of C5H12 produce 9 litre of water vapour;

Answer:

18.5 grams

Explanation:

The first step is to balance the equation.

2LiOH+CO2⇒Li2CO3+H2O+ some extra oxygen

For CO2, carbon has a molar mass of about 12, and oxygen has a molar mass of about 16, so the total is 44. Now that the equation is balanced and there are two LiOH's, it's molar mass is now doubled and it effectively has a molar mass of 48. 12/48=0.25 moles of both original substances used, and 0.25 moles of the final products. Li2CO3 has a molar mass of 7*2+12+16*3=74, and multiplying this by the 0.25 moles yields 18.5 grams. Hope this helps!

Answer:

18.5 grams [tex]Li_{2}CO_{3}[/tex]

Explanation:

First, we'll need to balance the equation:

[tex]LiOH_{s} + CO_{2(g)}[/tex] → [tex]Li_{2}CO_{3(s)} + H_{2}O_{(l)}[/tex]

There is one (Li) on the left and two on the right, so let's add a 2 coefficient on the left.

[tex]2LiOH_{s} + CO_{2(g)}[/tex] → [tex]Li_{2}CO_{3(s)} + H_{2}O_{(l)}[/tex]

Now there are 2 (Li) on the left, and 2 on the right, 4 (O) on both sides, 1 (C) on both sides, and 2 (H) on both sides. The equation is balanced!

Step 1: Find the limiting reactant

To find the limiting reactant, we need to convert the given masses of each reactant into moles. Multiply the given masses by the molar masses.

12g 2LiOh × [tex]\frac{1 mol}{48 g}[/tex](double the molar mass because you have two molecules) = 0.25 moles

12g CO₂ × [tex]\frac{1 mol}{44.01 g}[/tex] = 0.272 moles

We'll test both reactants to see which one limits us.

Since we don't have enough CO₂ to use all of our LiOH, CO₂ is the limiting reactant. We will use all of the CO₂ to perform our reaction and ignore the excess LiOH.

Step 2: Calculating the mass of the product

You can find the mass of either product with a mole ratio. (Remember, we have too much CO₂, so we'll need to use the given LiOH to perform this calculation)

0.25 moles LiOH × [tex]\frac{1 mole CO_{2}}{1 mole Li_{2}CO_{3}}[/tex] = 0.25 moles [tex]Li_{2}CO_{3}[/tex]

Now we convert back to grams!

0.25 moles [tex]Li_{2}CO_{3}[/tex] × [tex]\frac{74 g}{1 mole}[/tex] = 18.5 grams

Answer:

Ecological Relationships are of two types - Oppositional Relationships and Symbiotic Relationships. Oppositional Relationships are of two types - Predation and Competititon. Symbiotic Relationships are of four types - Mutualism, Commensalism, Amensalism & Parasitism.

Explanation:

Answer:

See below

Explanation:

When a strip of zinc metal is placed into a blue solution of copper(II) nitrate, a reaction immediately begins as the zinc strip begins to darken. If left in the solution for a longer period of time, the zinc will gradually decay due to oxidation to zinc ions.

3.41 is  the pH value of a sample that has ten times fewer hydronium ions than an equal volume of a vinegar sample with a pH value of 2.4.

Explanation:

pH of the first sample of acetic acid = 2.4

to know the [H+] concentration in the acetic acid solution, the equation used is:

pH = -log [H+]

[H+] = [tex]10^{-pH}[/tex]

[H+] = [tex]10^{-2.4}[/tex]

[H+]  = 3.98 X [tex]10^{-3}[/tex] M

The concentration of H+ ion in first case is 3.98 X [tex]10^{-3}[/tex] M, the second sample has ten times less hydronium ion so concentration of first case is divided by 10.

3.98 x [tex]10^{-4}[/tex] M

Now pH of the sample having 10 times fewer ions of acetic acid:

pH = -log [H+]

putting the values in the above equation:

pH = -log [3.98 x [tex]10^{-4}[/tex] M]

pH = 3.41

If solution has ten times less hydronium ion the pH will change to 3.41.

3.41 is the pH value of a sample that has ten times fewer hydronium ions. pH gives the concentration of Hydronium ion.

What information we have:

pH of the first sample of acetic acid = 2.4

To know the [H+] concentration in the acetic acid solution, the equation used is:

[tex]pH = -log [H^+]\\\\H^+ = 10^{-pH}\\\\H^+ = 10^{-2.4}\\\\H^+ = 3.98 *10^{-4} M[/tex]

The concentration of H+ ion in first case is [tex]3.98 *10^{-4} M[/tex], the second sample has ten times less hydronium ion so the concentration of first case is divided by 10.

[tex]3.98 *10^{-4} M[/tex],

Now pH of the sample having 10 times fewer ions of acetic acid:

pH = -log [H+]

pH = -log [ [tex]3.98 *10^{-4} M[/tex],]

pH = 3.41

If the solution has ten times less hydronium ion the pH will change to 3.41.

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Answer:

D)

Explanation:

PV/nRT=1, because particles are unable to overcome intermolecular attractions

Answer:

PV < nRT, because the volume is smaller due to significant intermolecular force effects.

Explanation:

FLVS exam states that this is correct because gasses change their "ideal" behaviors at lower temperatures.

Answer:

0.81M

Explanation:

Molarity, M = number of moles, n / Volume, V

number of moles of HCl, n=

mass of HCl / molar mass of HCl

Mass of HCl = 13.3g

Molar mass of HCl = 1 + 35.5 = 36.5g/mol

n = 13.3/36.5=0.364mol

V = 447mL=447/1000 L = 0.447L

Note that M is mol/L, hence, we convert 447mL to L. 1 mL = 1/1000 L

M=0.364mol/0.447L

M= 0.81M

Therefore, the molarity of the HCl solution is 0.81M

Answer:

A. 68°F

B. 293K

Explanation:

The temperature (celsius) = 20.0°C

A. Conversion of the temperature in celsius to fahrenheit.

This is illustrated below:

°F = 9C/5 + 32

C = 20°C

°F = 9C/5 + 32

°F = 9x20/5 + 32

°F = 36 + 32

°F = 68°F

B. Temperature (Kelvin) = temperature (celsius) + 273

Temperature (celsius) = 20°C

Temperature (Kelvin) = 20°C + 273

Temperature (Kelvin) = 293K

Final answer:

To convert the temperature from Celsius to Fahrenheit, use the formula F = (C * 1.8) + 32. To convert the temperature from Celsius to Kelvin, add 273.15 to the Celsius temperature.

Explanation:

To convert the temperature from Celsius to Fahrenheit, we can use the formula:

F = (C * 1.8) + 32

where F is the temperature in Fahrenheit and C is the temperature in Celsius. Plugging in the given temperature of 20.0°C:

F = (20.0 * 1.8) + 32 = 68.0°F

To convert the temperature from Celsius to Kelvin, we simply add 273.15 to the Celsius temperature:

K = C + 273.15

Therefore, the temperature in Kelvin would be:

K = 20.0 + 273.15 = 293.15K

Answer:

The amount of water in a 95 mL sample of the solution is 33.25 mL

Explanation:

65% (v/v) rubbing alcohol is a solution containing 65 mL rubbing alcohol in 100 mL solution of the rubbing alcohol and water

Therefore, 95 mL sample of the solution will contain x mL of rubbing alcohol;

x = 65% × 95 =  0.65 × 95 = 61.75 mL rubbing alcohol and therefore, the amount of water contained = 95 - 61.75 = 33.25 mL

The amount of water in a 95 mL sample of the solution = 33.25 mL.

The volume of water present in 95 ml of 65 % rubbing alcohol solution has been 33.25 ml.

The v/v has been the concentration representation of the volume of solute present in the total volume of the solution.

The given aqueous solution has 65 % v/v concentration of the rubbing alcohol. This states that there has been a presence of 65 ml of rubbing alcohol in 100 ml of the solution.

Since, in 100 ml sample there has been a presence of 65ml alcohol. The volume of water in 100 ml sample has been the remaining portion of the solution.

[tex]\rm Volume \;of \;water \;= \;Total \;volume - volume \;of \;alcohol\\Volume \;of\; water = 100 \;ml - 65 \;ml\\Volume \;of \;water = 35 \;ml[/tex]

The aqueous solution of rubbing alcohol has been consisted of 35 ml water per 100 ml solution.

Since, 100 ml solution = 35 ml water

[tex]\rm 95\;ml\;solution\;=\;\dfrac{35}{100}\;\times\;95\;ml\;water\\95\;ml\;solution=33.25\;ml\;water[/tex]

The volume of water present in 95 ml of 65 % rubbing alcohol solution has been 33.25 ml.

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Answer:

The volume occupies of neon gas is 335, 7 L

Explanation:

We use the formula PV=nRT. We convert the unit temperature Celsius into Kelvin: 0°C=273 K---> -28°C= -28+273= 245K. We calculate the number of mols in 322 grams of neon:

20,18 g---1 mol neon

322g---x= (322g x 1 mol neon)/ 20, 18g=15, 96 mol neon

PV=nRT    ----> V= (nRT)/P

V= (15,96 mol x 0,082 l atm/K mol x 245 K)/ 0,955 atm= 335, 7 L

Answer:

T = 0.38 K

Explanation:

Given data:

Number of moles = 4.75 mol

Pressure = 1.2 atm

Volume = 0.125 L

R = 0.0821 atm.L /mol.K

Temperature = ?

Solution:

Formula:

PV = nRT

T = PV/nR

T = 1.2 atm × 0.125 L /  4.75 mol × 0.0821 atm.L /mol.K

T = 0.15  / 0.39 /K

T = 0.38 K

Final answer:

To produce 22 g of CO₂, we calculate the moles of CO₂ and use the 1:1 molar ratio with O₂ to find that 16 g of oxygen gas is required, according to the balanced chemical equation for the reaction.

Explanation:

Calculating Mass of Oxygen Required to Produce Carbon Dioxide

To determine the mass of oxygen gas required to produce 22 g of carbon dioxide in the reaction C(s) + O₂(g)
ightarrow CO₂(g), we will first consider the molar masses of the substances involved. The molar mass of carbon dioxide (CO₂) is 44.009 g/mol, which means 1 mole of CO₂ weighs 44.009 grams. Since carbon is a pure element in its solid form, its molar mass is equivalent to its atomic mass, which is 12.0 g/mol. The molar mass of oxygen (O₂) is 32.0 g/mol because one mole of O₂ consists of 2 atoms of oxygen and the atomic mass of oxygen is 16.0 g/mol per atom.

Using the stoichiometry from the balanced chemical equation, we know that 1 mole of C will react with 1 mole of O₂ to produce 1 mole of CO₂. Therefore, if we have 22 g of CO₂, we can calculate the moles of CO₂ and then the moles of O₂ required:

Calculate moles of CO₂ produced: (22 g CO₂) / (44.009 g/mol) = 0.5 mol CO₂.

We need 0.5 mol of O2 since the mole ratio is 1:1.

Calculate mass of O₂ required: (0.5 mol O₂) times (32.0 g/mol) = 16 g O₂.

So, the mass of oxygen gas required to produce 22 g of carbon dioxide is 16 g.

Answer:

The answer to your question is below

Explanation:

Data

density = 13.53 g/cm³

volume = 62.5 cm³

mass = ?

moles = ?

atoms = ?

Process

1.- Calculate the mass of mercury

formula

density = mass/volume

-Solve for mass

mass = density x volume

-Substitution

mass = 13.53 x 62.5

-Result

mass = 845.63 grams

2.- Calculate the moles of mercury

Atomic mass = 200.6 g

              200.6 g of mercury --------------------- 1 mol

              845.63 g of mercury --------------------  x

                   x = (845.63 x 1) / 200.6

                   x = 4.22 moles of mercury

3.- Calculate the number of atoms

               1 mol --------------------  6 .023 x 10 ²³ atoms

             4.22 moles -------------   x

                 x = (4.22 x 6.023 x 10²³) / 1

                x = 2.54 x 10²³ atoms

The effect of warm tropical waters meeting with cool air is hurricane.

A hurricane is formed when warm tropical water meet with air that is cool and humid. This creates a zone of pressure over the warm water causing the air to rotate.

Hence, the effect of warm tropical waters meeting with cool air is hurricane.

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Answer:

0.33 moles of Oxygen at stp would occupy a volume of 7.392 dm³

Explanation:

1 mole of every gas at standard temperature and pressure (stp) occupies 22.4 dm³

0.33 moles of Oxygen at stp would occupy a volume of 0.33 × 22.4 dm³ = 7.392 dm³

Hope this Helps!!!

Answer:

[tex]V = 7.391\,L[/tex]

Explanation:

Let suppose that oxygen behaves ideally. The equation of state of the ideal gas is:

[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex]

The volume is now cleared:

[tex]V = \frac{n\cdot R_{u}\cdot T}{P}[/tex]

[tex]V = \frac{(0.33\,mol)\cdot \left(0.082\,\frac{atm\cdot L}{mol\cdot K} \right)\cdot (273.15\,K)}{1\,atm}[/tex]

[tex]V = 7.391\,L[/tex]

Answer:

B. organ

Explanation:

Smooth muscle is found in the walls of hollow organs like your intestines and stomach. They work automatically without you being aware of them. Smooth muscles are involved in many 'housekeeping' functions of the body.

Answer:

15.33

Explanation:

Given parameters:

Mass of KOH = 121g

Volume of solution  = 100mL   = 0.1dm³

Molar mass of KOH  = 56.11g/mol

Unknown:

pH  of the solution = ?

Solution:

To find the pH, we must first know the concentration of the solution obtained by mixing KOH up to this volume.

This can be done by finding the molarity of the solution.

  Molarity  = [tex]\frac{number of moles }{volume of solution}[/tex]

  Number of moles of KOH  = [tex]\frac{mass}{molar mass}[/tex]    = [tex]\frac{121}{56.11}[/tex]   = 2.16mole

Input parameters;

 Molarity of solution = [tex]\frac{2.16}{0.1}[/tex]   = 21.6moldm⁻³

       KOH    →    OH⁻   +          K⁺

       21.6           21.6              21.6

In the solution we have 21.6moldm⁻³ of OH⁻ which is need to find the pH;

          pOH  = -log₁₀(OH⁻)

         pOH  =  -log₁₀21.6

          pOH  = -1.33

Since pH + pOH   = 14

          pH  = 14 - (-1.33)  = 15.33

Answer:

The energy absorbed when 6.20 grams of aluminum is heated from 10.0°C to 65.0°C is 306.9 Joules.

Explanation:

[tex]Q=mc\Delta T=mc\times (T_2-T_1)[/tex]

Where:

Q = heat absorbed  or heat lost

c = specific heat of substance

m = Mass of the substance

ΔT = change in temperature of the substance

[tex]T_1,T_2[/tex] : Initial and final temperature of the substance

We have mass of aluminium = m = 6.20 g

Specific heat of aluminium= c = 0.900 J/g°C

Initial and final temperature of the aluminium [tex]T_1=10.0^oC [/tex]

Final temperature of the aluminium = [tex]T_2=65.0^oC[/tex]

Heat absorbed by the aluminium:

[tex]Q=6.20 g\times 0.900J/g^oC\times (65.0^oC-10.0^oC)=306.9 J[/tex]

The energy absorbed when 6.20 grams of aluminum is heated from 10.0°C to 65.0°C is 306.9 Joules.

Answer:

94.8% is the percentage yield of this reaction.

Explanation:

Mass of aluminium = 3.8 g

Moles of aluminium = [tex]\frac{3.8 g}{27 g/mol}=0.1407 mol[/tex]

[tex]2Al+3Cl_2\rightarrow 2AlCl_3[/tex]

According to reaction, 2 moles of aluminum gives 2 moles of aluminium chloride, then 0.1407 moles of aluminium will give:

[tex]\frac{2}{2}\times 0.1407 mol=0.01407 mol[/tex] aluminium chloride

Mass of 0.1407 moles of aluminum chloride:

= 0.1407 mol × 133.5 g/mol = 18.78 g

Theoretical yield of aluminum chloride = 18.78 g

Experimental yield of aluminum chloride = 17.8 g

The percentage yield of reaction:

[tex]=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

[tex]=\frac{17.8 g}{18.78 g}\times 100=94.8\%[/tex]

94.8% is the percentage yield of this reaction.

Answer:

All these statements are true.

Explanation:

Many describe it as an "educated guess," based on prior knowledge and observation. While this is true, the definition can be expanded. A hypothesis also includes an explanation of why the guess may be correct.

hope this helps :)

The true statement about hypothesis is: All these statements are true. The option (4) is correct.

A hypothesis is a testable statement that is formulated based on observations and background research.

It is typically written in an "if... then" form, which clearly states the relationship between two variables: the independent variable (the one at that the researcher manipulates) and the dependent variable (the one that the researcher measures).

A well-constructed hypothesis will also include a prediction regarding the dependent variable, which is an educated guess about the outcome of the experiment or study. Therefore, all the given statements about a hypothesis are true.

1. "It is written in an 'if...then' format." - This is true because a hypothesis often takes the form of ""If [independent variable], then [dependent variable]"" to clearly indicate the cause-and-effect relationship being tested.

2. "It is an educated guess based on background research." - This is true because a hypothesis is not just a random guess; it is formulated after conducting a thorough literature review and observing patterns or phenomena that suggest a certain relationship or outcome.

3. "It includes a prediction regarding the dependent variable." - This is true because a hypothesis predicts the effect that changing the independent variable will have on the dependent variable.

Since all three statements accurately describe the characteristics of a hypothesis, the correct answer is that all these statements are true.

The complete question is:

Which statement about a hypothesis is true?

1) It is written in an “if…then” format.

2) It is an educated guess based on background research.

3) It includes a prediction regarding the dependent variable.

4) All these statements are true.

Answer : The volume of gas occupy at [tex]1.00\times 10^2^oC[/tex] is, 1.25 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=1.00L\\T_1=25.0^oC=(25.0+273)K=298K\\V_2=?\\T_2=1.00\times 10^2^oC=((1.00\times 10^2)+273)K=373K[/tex]

Putting values in above equation, we get:

[tex]\frac{1.00L}{298K}=\frac{V_2}{373K}\\\\V_2=1.25L[/tex]

Therefore, the volume of gas occupy at [tex]1.00\times 10^2^oC[/tex] is, 1.25 L

Answer:

OK, I don't know if this is right but this is my thinking of it. I hope this helps you out.

Let's do problem "a"

Step 1.  Write a balanced equation.

                                       2 HCl  +  Ba(OH)2  -------------->  BaCl2  + 2 HOH  

Step 2.  Underneath the HCl and Ba(OH)2 in the equation write what you are given and what you need to find out.

                                    2 HCl        +       Ba(OH)2  -------------->  BaCl2  + 2 HOH

                                 C = 0.130 mol/L    v = 56 mL

                                  V = ?                   C = 0.109 mol/L

Step 3.  They ALWAYS provide a way to calculate the number of moles of one of the substances.  In this case it is Ba(OH)2 because you have C and V

                         n = C X V            n = 0.109 x 56 = 6.104 millimoles

Step 4.  Use the equation to figure out how many moles of the unknown (HCl) you will need.

               the equation says you need TWICE as many moles as you have of Ba(OH)2.  So you will need 6.104 millimoles x 2 or 12.208 milimoles.

Step 5.  Now that yo have moles and concentration of HCl you can now calculate volume

                  C = n / V, so V = n / C

                                 12.208 millimoles / 0.130 mol/L = 93.9 mL

Use this method for all these kind of problems.  

In B) you will have to convert he g of NaOH into moles by 0.240 / 40.00 g/mol and then following trhe rest of the procedure.

C) is done the same way.  convert the g of Na2SO4 into moles first.

d) is almost the same as a) except this time when you find the moles of Ca(OH)2 you just have to convert the moles into mass by multiplying moles x Molar mass of Ca(OH)2.

Best of luck.

Explanation:

To neutralize 15 mL of 2.0 M Ba(OH)₂, one would need 60 mL of 1.0 M HCl. The calculation is based on the mole ratio from the balanced chemical equation, with the final step being the conversion from liters to milliliters.

To determine how many milliliters of 1.0 M HCl are needed to neutralize 15 milliliters of 2.0 M Ba(OH)₂, you need to use the concept of mole ratios from the balanced chemical equation. The reaction between hydrochloric acid (HCl) and barium hydroxide (Ba(OH)₂) is:

2 HCl + Ba(OH)₂ ightarrow BaCl2 + 2 H₂O

The balanced equation shows that two moles of HCl react with one mole of Ba(OH)2. First, calculate the moles of Ba(OH)2:

moles Ba(OH)2 = volume (L) times molarity (M) = 0.015 L times 2.0 M = 0.03 moles

Since the ratio of HCl to Ba(OH)₂ is 2:1, you will need twice as many moles of HCl to neutralize Ba(OH)₂. Therefore, you'll need:

moles HCl needed = 0.03 moles Ba(OH)₂ times 2 = 0.06 moles HCl

Now, convert moles of HCl to volume:

volume HCl (L) = moles HCl / molarity HCl = 0.06 moles / 1.0 M = 0.06 L

Finally, convert liters to milliliters:

volume HCl (mL) = volume HCl (L) times 1000 mL/L = 0.06 L times 1000 mL/L = 60 mL

Therefore, 60 mL of 1.0 M HCl are needed to exactly neutralize 15 mL of 2.0 M Ba(OH)₂.

Answer:

50 mL

Explanation:

We can solve this problem by using Boyle's Law, which states that:

"For a fixed mass of an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume"

Mathematically:

[tex]p\propto \frac{1}{V}[/tex]

where

p is the pressure of the gas

V is the volume of the gas

The equation can be rewritten as

[tex]p_1 V_1 = p_2 V_2[/tex]

where in this problem:

[tex]p_1 = 100 kPa[/tex] is the initial pressure of the gas in the coke

[tex]V_1=25 mL[/tex] is the initial volume

[tex]p_2=50 kPa[/tex] is the final pressure

Solving for V2, we find the final volume:

[tex]V_2=\frac{p_1 V_1}{p_2}=\frac{(100)(25)}{50}=50 mL[/tex]

Answer:

ΔHr = -275 kj

Explanation:

It is possible to obtain the net change in enthalpy for the formation of one mole of lead(II) sulfate from lead, lead(IV) oxide, and sulfuric acid using the reactions:

(1) H₂SO₄(l) → SO₃(g) + H₂O (l) ΔH=+113kJ

(2) Pb(s) + PbO₂(s) + 2SO₃(g) → 2PbSO₄(s) ΔH=−775kJ

If you sum (1) + ¹/₂(2) you will obtain:

H₂SO₄(l) + ¹/₂Pb(s) + ¹/₂PbO₂(s) → PbSO₄(s) + H₂O(l)

Using Hess's law, the net change in enthalpy for this reaction could be obtained as:

ΔHr = ΔH(1) + ¹/₂ΔH(2)

ΔHr = +113kJ + ¹/₂ -775kJ

ΔHr = -275 kJ