QUESTION 2. What is the molar mass of a gas that takes three times longer to effuse than helium?​

The mass (in grams) of magnesium sulfate, MgSO₄ that would be produced from the reaction is 87.76 g

The mass of magnesium sulfate, MgSO₄ produced can be obtain as illustrated below:

Al₂(SO₄)₃ + 3Mg -> 2Al + 3MgSO₄ ΔH = +722 KJ

From the balanced equation above,

When 722 KJ of heat were absorbed, 360 g of MgSO₄ were produced

Therefore,

When 176 KJ of heat is absorbed = (176 × 360) / 722 = 87.76 g of MgSO₄ will be produce

Thus, the mass of MgSO₄ produced is 87.76 g

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Answer:

Molarity =[ number of moles* (1000mL/L)] / volume in mL

Explanation:

Given: Molarity =1.5M, volume = 100mL

1.5 =[ (no. of moles) * (1000mL/L) / 100mL

no of moles = (1.5* 100) / 1000

no of moles = 0.15

We know that, 1 mole = 6.023 *10²³ atoms

Since the compound contains 0.15moles, it has

0.15 * 6.023* 110²³

9.0345*10²² atoms of carbon are present in the compound.

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29 is not the best answer depends on the context and the rules for significant figures.

What is best answer?

The best answer depends on the context and the rules for significant figures. If we assume that we need to round to three significant figures:

Therefore, 29 is not the best answer.

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Answer:

um, not sure how to answer this haha

Explanation:

The volume of the tank is approximately 130.75 m³.

To solve this problem, we need to use the concept of air and water vapor mixture. The given data includes the mass of dry air and water vapor, temperature, and total pressure. We can calculate the specific humidity, relative humidity, and volume of the tank using the following steps:

(a) Specific humidity:

The specific humidity (ω) is defined as the ratio of the mass of water vapor (m_w) to the total mass of the air-water vapor mixture (m_t):

ω = m_w / m_t

Given that the mass of water vapor is 0.17 kg and the total mass of the mixture is 15 kg + 0.17 kg = 15.17 kg, we can calculate the specific humidity:

ω = 0.17 kg / 15.17 kg ≈ 0.0112

So, the specific humidity is approximately 0.0112.

(b) Relative humidity:

Relative humidity (RH) is the ratio of the partial pressure of water vapor (P_w) to the saturation vapor pressure of water (P_ws) at the given temperature, multiplied by 100:

RH = (P_w / P_ws) * 100

To find the relative humidity, we need to determine the saturation vapor pressure at 30°C. Using a vapor pressure table or equation, we can find that the saturation vapor pressure at 30°C is approximately 4.246 kPa.

Given that the total pressure is 100 kPa, the partial pressure of water vapor is 0.17 kg / 15.17 kg * 100 kPa = 1.119 kPa.

Now we can calculate the relative humidity:

RH = (1.119 kPa / 4.246 kPa) * 100 ≈ 26.34%

So, the relative humidity is approximately 26.34%.

(c) Volume of the tank:

To find the volume of the tank, we can 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 gas constant, and T is the temperature in Kelvin.

First, we need to calculate the number of moles of dry air and water vapor in the tank. The number of moles (n) can be obtained using the equation:

n = m / M

Where m is the mass and M is the molar mass.

The molar mass of dry air is approximately 28.97 g/mol, and the molar mass of water vapor is approximately 18.015 g/mol.

For dry air:

n_air = 15 kg / 0.02897 kg/mol ≈ 517.82 mol

For water vapor:

n_water = 0.17 kg / 0.018015 kg/mol ≈ 9.43 mol

Now we can calculate the volume using the ideal gas law:

V = (n_air + n_water) * R * T / P

Given that R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (30°C + 273.15 = 303.15 K), and P is the total pressure (100 kPa), we can calculate the volume:

V = (517.82 mol + 9.43 Mol) * 8.314 J/(mol·K) * 303.15 K / (100,000 Pa) ≈ 130.75 m³

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The sample response is : how do mass and the type of material affect thermal energy transfer?

So one need to have include the below factors in your answer:

The mass of an object influences allure thermal strength transfer traits. Generally, objects accompanying best public have greater warm strength depository volume.

They can absorb more heat strength before experience a meaningful change in hotness. If an equal amount of heat is used to both objects, the object accompanying better mass will demand more heat strength to raise allure temperature distinguished to the object accompanying tinier mass .

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. In this experiment, you need to examine the idea of the thermal energy transfer. Using a controlled experiment, what might a good question about the variables that affect thermal energy transfer be? Thermal energy transfer depends on many properties, be limit your question to only two. What did you include in your question?

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.

Answer:If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid. The most energetic particles at the surface escape from the surface of the liquid as a vapour as it gets warmer. Liquids evaporate faster as they heat up and more particles have enough energy to break away.

Explanation:

Answer:

0.53g

Explanation:

We'll begin by converting 175mL to L. This is illustrated below:

1000mL = 1L

Therefore 175mL = 175/1000 = 0.175L

Next, we shall calculate the number of mole of O2 that occupy 0.175L. This is illustrated below:

1 mole of O2 occupy 22.4L at stp.

Therefore, Xmol of O2 will occupy 0.175L i.e

Xmol of O2 = 0.175/22.4

Xmol of O2 = 7.81×10¯³ mole

Therefore, 7.81×10¯³ mole of O2 occupy 175mL.

Next, we shall determine the number of mole of H2O2 that decomposed to produce 7.81×10¯³ mole of O2. This is illustrated below:

2H2O2 —> 2H2O + O2

From the balanced equation above,

2 moles of H2O2 decomposed to produce 1 mole of O2.

Therefore, Xmol of H2O2 will decompose to produce 7.81×10¯³ mole of O2 i.e

Xmol of H2O2 = 2 x 7.81×10¯³

Xmol of H2O2 = 1.562×10¯² mole

Therefore, 1.562×10¯² mole of H2O2 decomposed in the reaction.

Finally, we shall convert 1.562×10¯² mole of H2O2 to grams. This is illustrated below:

Molar mass of H2O2 = (2x1) + (16x2) = 34g/mol

Mole of H2O2 = 1.562×10¯² mole

Mass of H2O2 =..?

Mole = mass /Molar mass

1.562×10¯² = mass /34

Cross multiply

Mass of H2O2 = 1.562×10¯² x 34

Mass of H2O2 = 0.53g

Therefore, 0.53g of Hydrogen peroxide, H2O2 were decomposition in the reaction.

Answer:

The formula is :  C3H8 + 5O2 → 3CO2 + 4H2O.

Explanation:

good luck

The mass of chlorine gas product formed is 61.4 g.

What is the mass ?

First, we need to determine which reactant is limiting and which is in excess. To do this, we can use the given mass of HCl and convert it to moles:

63.2 g HCl × (1 mol HCl/36.46 g HCl) = 1.73 mol HCl

Next, we can use stoichiometry to determine how many moles of \(Cl_{2}\) will be produced from 1.73 mol of HCl:

1.73 mol HCl × (2 mol \(Cl_{2}\)/4 mol HCl) = 0.865 mol \(Cl_{2}\)

Finally, we can convert the moles of \(Cl_{2}\) to grams using its molar mass:

0.865 mol \(Cl_{2}\) × (70.91 g \(Cl_{2}\)/1 mol \(Cl_{2}\)) = 61.4 g \(Cl_{2}\)

Therefore, the mass of chlorine gas product formed is 61.4 g.

What is stoichiometry ?

Stoichiometry is a branch of chemistry that deals with the calculation of the quantities of reactants and products involved in a chemical reaction. It involves using balanced chemical equations to calculate the amounts of reactants required to produce a certain amount of product, or the amount of product that can be obtained from a given amount of reactant. Stoichiometry is an important tool for predicting the outcomes of chemical reactions and for designing chemical processes in industry.

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Previously cooked food must reheated to temperatures of at least 165°F.

When food is cooked, it should be eaten and any leftovers preserved.

Leftovers can be preserved by refrigeration.

However, when this refrigerated food is to be eaten again, it should be reheated before eating.

Previously cooked food must be heated to prevent spoilage and poisoning.

The addition of heat will prevent spoilage by:

Previously cooked food must reheated to temperatures of at least 165 °F.
The process of reheating the food may be done in microwave oven or over a stove.

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Answer:

B. the number of protons added to the number of neutrons in the nucleus.

Explanation:

Sana makatulong

i think it's a simple covalent structure

Heat is added to ice at 0 °C. Explain why the temperature of the ice does not change. What does change?When heat is added to ice at 0°C, the temperature of the ice does not change. This happens because all the heat energy is used up in overcoming the intermolecular forces of attraction (hydrogen bonds) that exist between the water molecules in ice.

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Answer:

C15H30 + 23(O2) -> 15(CO2) + 15(H2O)

Complicated one!

Answer:

The number of these organisms in the soil would decrease.

Explanation:

The actual amount of substance each required or produced by the reaction is obtained by stoichiometry.

From the reaction equation;

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

Number of moles in  1.56 grams of H2(g) = 1.56 grams/2.00g/mol = 0.78 moles

If 2 moles of water yields 1 mole of H2

x moles of water yields 0.78 moles of H2

x = 2 × 0.78/1 = 1.56 moles of water

Mass of water required = 1.56 moles of water × 18 g/mol = 28 grams of water

The statement that the reaction requires 27.9 grams of H2O is true.

Also;

The number of moles in 1.55g of NaOH = 1.55g/40g/mol = 0.039 moles

If 2 moles of water produces 2 moles of NaOH

1.56 moles of water produces 1.56 × 2/2 = 1.56 moles of NaOH

Mass of NaOH = 1.56 moles of NaOH * 40 g/mol = 62.4 g of NaOH

The statement that the reaction also produces 1.55 grams of NaOH is false.

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To determine the original pressure of a 750 ml sample of helium (He) at 0 degrees Celsius, we can use the combined gas law, which relates the initial and final conditions of a gas sample. The combined gas law equation is:

(P1 × V1) / (T1) = (P2 × V2) / (T2)

Where:

P1 and P2 are the initial and final pressures, respectively.

V1 and V2 are the initial and final volumes, respectively.

T1 and T2 are the initial and final temperatures, respectively.

Let's assign the given values:

P1 = unknown (original pressure)

V1 = 750 ml (initial volume)

T1 = 0 degrees Celsius (initial temperature)

P2 = 2 atm (final pressure)

V2 = 500 ml (final volume)

T2 = 25 degrees Celsius (final temperature)

Before using the combined gas law equation, we need to convert the temperatures to Kelvin scale by adding 273.15 to both T1 and T2:

T1 = 0 + 273.15 = 273.15 K

T2 = 25 + 273.15 = 298.15 K

Now we can plug in the values into the combined gas law equation:

(P1 × 750 ml) / (273.15 K) = (2 atm × 500 ml) / (298.15 K)

To solve for P1, we can cross multiply and rearrange the equation:

P1 = (2 atm × 500 ml × 273.15 K) / (750 ml × 298.15 K)

P1 = 0.924 atm

Therefore, the original pressure of the 750 ml sample of helium at 0 degrees Celsius is approximately 0.924 atm.

Answer:

Compound 1 is a molecular compound

Compound 2 is an ionic compound

Compound 3 is a molecular compound

Explanation:

Let us review the properties of ionic and molecular compounds.

A molecular compound has a low melting and boiling point. This is as a result of weak intermolecular forces. Also, molecular compounds do not conduct electricity both in solid and liquid state because they are composed of molecules and not ions.

On the other hand, ionic substances have very high melting and boiling points. They are very strong solids that often have a dull appearance and do not conduct electricity in the solid state.

Compounds 1 and 3 have the properties of molecular substances hence they are classified as such. Compound 2 displayed the properties of an ionic substance hence it is classified as such.

Answer: I think its C I dont know  

Explanation: hApPy NeW yEaRsSsSs.

Answer:

Explanation:

the mass of an atom of a chemical element expressed in atomic mass units. It is approximately equivalent to the number of protons and neutrons in the atom (the mass number) or to the average number allowing for the relative abundances of different isotopes.

the total number of protons and neutrons in a nucleus.

to write the symbol for an isotope, place the atomic number as a subscript and the mass number (protons plus neutrons) as a superscript to the left of the atomic symbol

Answer:

B) Pressure and Temperature

Explanation:

Just did it

The mole fraction of sulfuric acid in the solution is 1.78 x 10^-4.

We need to know how many moles of sulfuric acid and how many moles there are in the solution as a whole in order to figure out the mole fraction of sulfuric acid.

To begin, we must convert the sulfuric acid's mass into moles. The molar mass of sulfuric corrosive is 98.08 g/mol. As a result, the number of moles of sulfuric acid is equal to 3.4 g divided by 98.08 g/mol, or 0.0347 mol.

Next, we need to figure out how many moles are all around the solution. We can expect that the volume of the arrangement is equivalent to the volume of water added (3500 ml). Notwithstanding, we want to switch the volume from milliliters over completely to liters since the unit of mole portion is moles per liter.

As a result, the volume of the solution is 3500 ml, or 3.5 L. Based on the assumption that water has a density of 1 g/mL, the mass of water in the solution can be calculated as follows:

The molar mass of water, which is 18.02 g/mol, can be used to determine the number of moles of water: mass of water = volume of water x density of water = 3500 ml x 1 g/mL = 3500 g

The mole fraction of sulfuric acid in the solution can be calculated as follows: 3500 g x 18.02 g/mol = 194.14 mol

Mole fraction of sulfuric acid is calculated by dividing the total number of moles in the solution by the number of moles of sulfuric acid: 0.0347 mol / (0.0347 mol + 194.14 mol) = 1.78 x 10-4.

Therefore, the mole fraction of sulfuric acid in the solution is 1.78 x 10^-4.

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150 grams of NaOH is approximately equal to 2.256 x 10^24 particles of NaOH.

To convert grams of NaOH to particles of NaOH, we need to use the concept of molar mass and Avogadro's number. The molar mass of NaOH is calculated by adding the atomic masses of sodium (Na), oxygen (O), and hydrogen (H) together. It can be determined as follows:

Na: 22.99 g/mol

O: 16.00 g/mol

H: 1.01 g/mol

Molar mass of NaOH = (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 40.00 g/mol

Now, we can use the molar mass to convert grams of NaOH to moles. Since 1 mole contains Avogadro's number (approximately 6.022 x 10^23) particles, we can determine the number of particles as follows:

150 g NaOH * (1 mol NaOH / 40.00 g NaOH) * (6.022 x 10^23 particles / 1 mol NaOH) ≈ 2.256 x 10^24 particles

It's important to note that this calculation assumes the substance is pure NaOH and that the molar mass and Avogadro's number are accurate.

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The balanced molecular equation for the reaction of magnesium nitrate, Mg(NO₃)₂ and sodium phosphate, Na₃PO₄ is:

3Mg(NO₃)₂(aq) + 2Na₃PO₄(aq) → 6NaNO₃(aq) + Mg₃(PO₄)₂(s)

The balanced molecular equation for the reaction of magnesium nitrate, Mg(NO₃)₂ and sodium phosphate, Na₃PO₄ can be written as illustrated below:

Mg(NO₃)₂(aq) + Na₃PO₄(aq) → NaNO₃(aq) + Mg₃(PO₄)₂(s)

There is 1 atom of Mg on the left side and 3 atoms on the right. It can be balanced by writing 3 before Mg(NO₃)₂ as shown below:

3Mg(NO₃)₂(aq) + Na₃PO₄(aq) → NaNO₃(aq) + Mg₃(PO₄)₂(s)

There is 1 atom of P on the left side and 2 atoms on the right. It can be balanced by writing 2 before Na₃PO₄ as shown below:

3Mg(NO₃)₂(aq)  + 2Na₃PO₄(aq) → NaNO₃(aq) + Mg₃(PO₄)₂(s)

There are a total of 6 atom of Na on the left side and 1 atom on the right. It can be balanced by writing 6 before NaNO₃ as shown below:

3Mg(NO₃)₂(aq) + 2Na₃PO₄(aq) → 6NaNO₃(aq) + Mg₃(PO₄)₂(s)

Now, the equation is balanced.

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Answer:

activity coefficient \(\mathbf{\gamma =0.809}\)

activity coefficient \(\mathbf{\gamma = 0.791}\)

The change in pH in part A = 0.092

The change in pH in part B =  0.102

Explanation:

From the given information:

pH of HCl solution = 1.092

Activity of the pH solution [a] = \(10^{-1.092}\)

[a] = 0.0809 M

Recall that [a] = \(\gamma\) × C

where;

\(\gamma\)  = activity coefficient

C = concentration

Making the activity coefficient the subject of the formula, we have:

\(\gamma = \dfrac{[a]}{C}\)

\(\gamma = \dfrac{0.0809 \ M}{0.100 \ M}\)

\(\mathbf{\gamma =0.809}\)

B.

The pH of a solution of HCl and KCl = 2.102

[a] = \(10^{-2.102}\)

[a] = 0.00791 M

activity coefficient \(\gamma = \dfrac{0.00791 \ M}{0.01 \ M}\)

\(\mathbf{\gamma = 0.791}\)

C. The change in pH in part A = 1.091 - 1.0 = 0.092

The change in pH in part B = 2.102 -2.00 = 0.102

Answer:

3.25 moles of H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2HCl + Ca(OH)2 —> CaCl2 +2H2O

Now, we can obtain the number of mole H2O produced from the reaction as follow:

From the balanced equation above,

2 moles of HCl reacted to produce 2 moles of H2O.

Therefore, 3.25 moles of HCl will also react to produce 3.25 moles of H2O.

Therefore, 3.25 moles of H2O were produced from the reaction.

Answer: 71.43 M.

Explanation: To calculate the molarity of a solution, we divide the number of moles of solute by the volume of the solution in liters. The volume of the solution in this case is given in milliliters, so we need to convert it to liters before we can calculate the molarity.

35 mL = 0.035 L (since 1 mL = 0.001 L)

So, the molarity of the solution can be calculated as follows:

Molarity = moles of solute / volume of solution in liters

Molarity = 2.50 moles / 0.035 L

Molarity = 71.43 M

Therefore, the molarity of the solution is 71.43 M.