An ionic compound is formed when there is a reaction between the elements

Answers

Answer 1
Final answer:

Ionic compounds are formed when a metal and a nonmetal react, leading to a transfer of electrons and the formation of ions. These ions are held together by ionic bonds, which demonstrate strong electrostatic attraction. These compounds display unique properties, including high melting and boiling points, and they can conduct electricity when dissolved or melted.

Explanation:

An ionic compound is formed when an element composed of atoms that readily lose electrons (a metal) reacts with an element composed of atoms that readily gain electrons (a nonmetal). This process usually leads to a transfer of electrons, resulting in the production of ions. For example, sodium atoms can lose an electron to form positively charged sodium ions (Na+), while chlorine atoms can gain an electron to form negatively charged chloride ions (Cl-). A compound such as NaCl consists of these sodium and chloride ions held together by ionic bonds — the electrostatic attractions between oppositely charged ions.

Ionic compounds have certain distinct properties. They exhibit a crystalline structure and are usually rigid and brittle. Their melting and boiling points tend to be high, implying the strength of the ionic bonds. These compounds are poor conductors of electricity in their solid state due to the immobility of ions, but once dissolved or melted, they become excellent conductors because the ions are free to move.

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Related Questions

The natural tendence of an object to resist change in its state of motion is:


friction

force

inertia

gravity

Answers

Answer:

inertia

Explanation:

The principal of inertia explains why objects do not voluntarily change their speed or direction without the influence of an external force. Inertia is not unaffected by other forces in planetary systems such as Earth and is, as such, not really observable on planetary surfaces. This is due to the resistance forces of gravity and friction from both the air and the ground. The drag slows objects down and makes it appear as if a constant force were necessary to keep them in motion.

Answer:

c

Explanation:

Which of these incorrectly pairs a hormone with its function? A. thyroxin—regulates metabolism B. insulin—enables the body to use glucose (blood sugar) for energy C. adrenaline (epinephrine)—prepares the body to "fight or flee" D. progesterone—controls muscle mass

Answers

Answer:

The correct answer is D progesterone control muscle mass.

Explanation:

Progesterone is a steroid hormone synthesized from cholesterol and secreted from corpus lutenum and placenta.progesterone regulate the luteal phase of the menstrual cycle, progesterone helps in the development of secondary sexual characteristics such as breast development,pubic hair development etc.

    On the other hand muscle mass is controlled by the growth hormone released from the anterior lobe of the pituitary gland.

A) Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0- L container. At 23.0 C, the total pressure in the container is 4.60atm . Calculate the partial pressure of each gas in the container.
Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane.
B) A gaseous mixture of O2 and N2 contains 39.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 585mmHg ?

Answers

Answer:

A. PpCH4 = 1.21 atm, Pp C2H6 = 1.46 atm,  Pp C3H6 = 1.93 atm

B. Pp O2 = 352.17 mmHg

Explanation:

A. First step: Get the total mols of mixture to know the mass of propane.

P . V = n . R . T

T° K = T°C + 273 → 23°C + 273 = 296K

4.60 atm . 10L = n . 0.082 L.atm/mol.K . 296K

(4.60 atm . 10L) / (0.082 mol.K/L.atm . 296K) = n

1.89 mol = n

Mass/Molar weight = mol

Mass CH4 = 16 g/m

Mol CH4: 8g / 16g/m = 0.5 mol

Mass C2H6 = 30 g/m

Mol C2H6 = 0.6 mol

Total mols - Mol CH4 - Mol C2H6 = Mol propane

1.89 mols - 0.5 mol - 0.6 mol = 0.79 mol propane

2nd step: We can use the molar fraction with the partial pressure.

Partial pressure gas / Total pressure = mols gas / total moles

Sum of molar fraction = 1

0.5 / 1.89 + 0.6 /1.89 + 0.79/1.89 = 1

Partial pressure CH4 → Pp CH4 /4.60 atm =  0.5 / 1.89

PpCH4 = (0.5 /1.89) . 4.60 atm = 1.21 atm

Partial pressure C2H6 → Pp C2H6 /4.60atm = 0.6 /1.89

Pp C2H6 = (0.6 /1.89) . 4.60atm = 1.46 atm

Partial pressure C3H6 → Pp C3H6 /4.60atm = 0.79/1.89

Pp C3H6 = (0.79/1.89) . 4.60atm = 1.93 atm

B. 39.8% means a percent of the molar fraction, so the fraction is 0.398

N2 mass /total mass = Partial pressure N2/ Total pressure

0.398 . 585 mmHg = Partial pressure N2

232.83 mmHg = Partial pressure N2

Total pressure - Pp N2 = Pp O2

585 mmHg - 232.82 mmHg = 352.17 mmHg

Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%, respectively) were determined in earlier experiments. Calculate the average atomic mass of bromine based on these experiments.

Answers

Answer:

Average atomic mass = 79.9034 amu

Explanation:

The formula for the calculation of the average atomic mass is:

[tex]Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})[/tex]

Given that:

For first isotope:

% = 50.69 %

Mass = 78.9183 amu

For second isotope:

% = 49.31 %

Mass = 80.9163 amu

Thus,  

[tex]Average\ atomic\ mass=\frac{50.69}{100}\times {78.9183}+\frac{49.31}{100}\times {80.9163}\ amu[/tex]

[tex]Average\ atomic\ mass=40.0036+39.8998\ amu[/tex]

Average atomic mass = 79.9034 amu

Final answer:

The average atomic mass of bromine is calculated by taking the sum of the products of each isotope's mass and its abundance in decimal form. The result is an average atomic mass of 79.898 amu for bromine.

Explanation:

To calculate the average atomic mass of bromine based on the given experimental results, we need to consider the masses and relative abundances of its isotopes. By multiplying the mass of each isotope by its abundance (converted to a decimal form) and adding these weighted masses together, we can determine the average atomic mass.

Calculation:

Adding the weighted masses of both isotopes gives us: 39.994 amu + 39.904 amu = 79.898 amu. Thus, the average atomic mass of bromine based on these experiments is 79.898 amu.

Checking for Accuracy:

It is important to make sure that this number makes sense and correlates with the given abundances, ensuring that no mathematical errors have been made.

B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 reacts with excess oxygen where the products are B2O3(s) and H2O(l). The standard enthalpy of formation of B5H9(l) is 73.2 kJ/mol, the standard enthalpy of formation of B2O3(s) is -1272 kJ/mol and that of H2O(l) is -285.4 kJ/mol. Express your answer in kJ.

Answers

Answer:

Heat realesed is - 2192.7 kJ

Explanation:

First lets have the chemical equation balanced, and then solve the question based on the fact that the enthalpy change for a reaction is the sum of the enthalpies of formation of products minus reactants.

    B₅H₉ (l) +     O₂ (g) ⇒       B₂O₃ (s) +      H₂O(l)

B atoms we have 5 reactants and 2 products, so the common mltiple is 10 and we have

  2  B₅H₉ (l) +     O₂ (g) ⇒      5 B₂O₃ (s) +      H₂O(l)

Now balance H  by multiplying by 9 the H₂O

  2  B₅H₉ (l) +     O₂ (g) ⇒      5 B₂O₃ (s) +    9 H₂O(l)

Finally, balance the O since we have 24 in products by multiplying by 12 the

O₂ ,

        2  B₅H₉ (l) +  12 O₂ (g) ⇒      5 B₂O₃ (s) +    9 H₂O

ΔHrxn = 5 x ΔHºf B₂O₃  + 9 x   ΔHºf H₂O  -  ( 2 x ΔHºf  B₅H₉ + 12 x  ΔHºf O₂  )

We have all the  ΔHºf s except oxygen but remember the enthalpy of formation of a pure element in its standard estate is cero.

ΔHrxn =  5 mol x ( -1272 kJ/mol )+ 9 mol  x ( -285.4 kJ/mol )   - (  2mol x 732 kJ/mol )

= -8928.6 kJ - 1464  kJ = -10,392 kJ

Now this enthalpy change was based in 2 mol  reacted according to the balanced equation, so for 0.211 mol of  B₅H₉  we will have:

-10,392 kJ/ 2 mol B₅H₉   x 0.211 mol B₅H₉ = - 2192.7 kJ

Answer:

ΔH=-957.41 kJ

Explanation:

The chemical equation is:

B₅H₉ (l) +     O₂ (g) ⇒       B₂O₃ (s) +      H₂O(l)

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

Then, you must balance the chemical equation. For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.  

So in first place you balance B

2 B₅H₉ (l) +     O₂ (g) ⇒   5 B₂O₃ (s) +      H₂O(l)

Then you balance H

2 B₅H₉ (l) +     O₂ (g) ⇒   5 B₂O₃ (s) +    9  H₂O(l)

Finally you balance O

2 B₅H₉ (l) +    12 O₂ (g) ⇒   5 B₂O₃ (s) +    9  H₂O(l)

Then

Left side: 2*5=10 boron (B), 2*9=18 hydrogen. and 12*2=24 oxygen

Right side: 5*2=10 boron (B), 9*2=18 hydrogen. and 5*3 + 9*1=24 oxygen

Since you have the same amount of elements on each side of the equation, the equation is balanced.

You want to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction. For that you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (quantity of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants

Knowing that:

Heat of formation of B₅H₉ = 73.2 kJ/mol Heat of formation of water = -285.4 kJ/mol Heat of formation of B₂O₃ = -1,272 kJ/mol

For the formation of one mole of a pure element the heat of formation is 0, in this case we have as a pure compound the oxygen O₂.

Then:

ΔH= 9 mol*(-285.4 kJ/mol) + 5 mol* (-1,272 kJ/mol) - [2 mol* 73.2 kJ/mol + 12 mol* 0 kJ/mol]

ΔH= -9,075 kJ

If you observe the previous balanced reaction, you can see that 2 moles of B₅H₉ (l) are necessary. And the calculation of the heat of reaction previously carried out is based on this reaction. This ultimately means that the energy that would result in the reaction of 2 moles of B₅H₉ (l) is -9,075 kJ.

To determine the heat that is released when 0.211 mol of B₅H₉ (l) react with excess oxygen, a rule of three is applied as follows: if 2 moles of B₅H₉ (l) produces a heat ΔH of -9,075 kJ, when reacting 0.211 mol of B₅H l (l) how much heat ΔH is released?

ΔH=[tex]\frac{0.211 moles*(-9075)kJ}{2 moles}[/tex]

ΔH=-957.41 kJ

In the laboratory, a student uses a coffee cup calorimeter to determine the specific heat of a metal. She heats 18.1 grams of lead to 99.50°C and then drops it into a cup containing 81.7 grams of water at 23.00°C. She measures the final temperature to be 23.67°C. Assuming that all of the heat is transferred to the water, she calculates the specific heat of lead to be _____ J/g°C.

Answers

Answer:

[tex]specificheat=0.177J/g^{0}C[/tex]

Explanation:

The heat evolved by lead will be absorbed by water.

The heat evolved by lead will be:

[tex]Q1=massXspecificheatXchangeintemperature[/tex]

[tex]Q1=18.1X(Specificheat)X(99.50-23.67)[/tex]

Heat absorbed by water will be:

[tex]Q2=massXspecificheatX(changeintemperature)[/tex]

[tex]Q2=81.7X4.184(23.67-23)=229.03J[/tex]

We know that

Q1=Q2

[tex]18.1X(Specificheat)X(99.50-23.67)=229.03[/tex]

[tex]1372.523Xspecificheat=229.03[/tex]

[tex]specificheat=0.177J/g^{0}C[/tex]

The heat of vaporization of water at 100°c is 40.66 kj/mol. Calculate the quantity of heat that is absorbed/released when 9.00 g of steam condenses to liquid water at 100°c.

Answers

Final answer:

When 9.00g of steam condenses to liquid water at 100°C, 20.3 kJ of heat is released. This is calculated by converting the mass of water to moles, then multiplying by the heat of vaporization.

Explanation:

The heat absorbed or released during the condensation of steam to liquid water is calculated by using the heat of vaporization (40.66 kJ/mol) and the moles of water. The mass of water given is 9.00g, which is converted to moles by dividing by the molar mass of water (18.02g/mol), resulting in 0.499 mol of water. The heat absorbed/released (Q) is calculated by multiplying the number of moles and the heat of vaporization. So, Q = 0.499 mol * 40.66 kJ/mol = 20.3 kJ (3 sig figs). So when 9.00g of steam condenses to liquid water at 100°C, 20.3 kJ of heat is released.

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Cesium chloride has a lattice energy -657.0 kJ/mol and a heat of hydration of -639.0 kJ/mol. Determine how much heat is evolved or absorbed when 30.0 g of cesium chloride completely dissolves in water.

Answers

Answer:

Q = 3.21 kJ

Explanation:

First, with the heat of hydration and the lattice energy, we can actually calculate the difference between those two to get the heat evolved per mole. In this case it would be:

657 - 639 = 18 kJ/mol

Now, the expression to get the heat is:

Q = m*ΔE

We already have the energy, now we need to get the moles of the cesium chloride.

The molar mass of cesium chloride is 168.36 g/mol, so the moles:

mole = 30/168.36 = 0.1781 moles

Finally the heat:

Q = 0.1781 * 18

Q = 3.21 kJ

A 12-liter tank contains helium gas pressurized to 160 \rm atm.How many 3-liter balloons could the 12-L helium tank fill? Keep in mind that an "exhausted" helium tank is not empty. In other words, once the gas inside the tank reaches atmospheric pressure, it will no longer be able to fill balloons.

Answers

Answer:

636 balloons

Explanation:

If we assume that helium gas follows an ideal gas behaviour, we can use the ideal gas law to solve this problem as follows:

We consider two different states, the initial given by the conditions of the problem statement and the final, when the tank reaches atmospheric pressure and it's no longer able to fill balloons:[tex]P_{1}=160 atm\\V_{1}=12 L\\P_{2}=1 atm\\V_{2}= ?[/tex]To find out what would be this volume 2, we use the Boyle's Law: [tex]P_{1}V_{1}=P_{2}V_{2}\\V_{2}=\frac{P_{1}V_{1}}{P_{2}} \\V_{2}=\frac{160 atm \times 12L}{1 atm}\\V_{2}=1920 L[/tex]Now we find the available volume to fill the balloons by substracting both, volume 2 and volume 1: [tex]V_{b}=V_{2}-V_{1}=1920L-12L=1908 L[/tex]Finally, we determine the quantity of ballons by dividing that available volume between the volume of each ballon:[tex]B=\frac{1908L}{3L} =636 balloons[/tex]
Final answer:

By applying Boyle's Law that states the pressure of a gas is inversely proportional to its volume, we find that the 12-L tank pressurized to 160 atm can fill a total of 640 3-L balloons to atmospheric pressure.

Explanation:

This is a typical problem related to the ideal gas law and mainly involves the understanding of pressure-volume relationship in gases. Since the tank and the balloons are at different pressures, the volume the helium will occupy in the balloons at atmospheric pressure will be larger than the volume it occupies in the tank.

Based on this, if the helium tank is pressurized to 160 atm and the balloons are to be filled until the pressure in the tank is equivalent to atmospheric pressure (1 atm), the helium in the tank will expand 160 times its initial volume. Hence, the 12-L tank will fill 12 * 160 = 1920 L of balloon space. Given each balloon has a volume of 3 L, the number of balloons that can be filled is thus 1920 / 3 = 640 balloons.

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The Goodyear blimp we usually see hovering above our heads at a football game contains 5600 m3 of helium gas. If there are 1070 kg of gas at an absolute pressure of 1.10 105 Pa in the blimp, what is the temperature of the gas? The atomic mass of helium 4.00260 g/mol.

Answers

Final answer:

The questioned tackled determining the temperature of gas in a Goodyear blimp using the provided specifications and the ideal gas law: PV=nRT.

Explanation:

The question is about determining the temperature of the helium gas in a Goodyear blimp given certain parameters. Using the ideal gas law: PV=nRT we can calculate the temperature (T). In this scenario, the volume (V) is 5600 m3, the pressure (P) is 1.10x105 Pa and the number of moles (n) can be calculated by dividing the total mass of helium gas by its molar mass. By rearranging the gas law formula, the temperature could be calculated as T=(PV)/(nR), where R is the gas constant.

Following these steps, we can substitute the known values into the equation to solve for the temperature inside the Goodyear blimp. This problem involves concepts of physics, specifically gas laws, and would be covered typically in a college-level physics course.

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A buffer with a pH of contains M of sodium benzoate and M of benzoic acid. What is the concentration of in the solution after the addition of mol to a final volume of L? Assume that any contribution of to the volume is negligible.please Answer for H3O+. will rate

Answers

Answer:

benzoic acid(C7H6O2)--->

C7H5NaO2(sodium benzoate)+H30+

The answer is given per the balanced eqn

Hafnium, Hf, is also found in group 4B. Write the electron configuration for Hf. Express your answer in order of increasing orbital energy as a string without blank space between orbitals. For example, the electron configuration of Li could be entered as 1s^22s^1 or [He]2s^1.

Answers

Final answer:

The electron configuration for Hafnium (Hf) is [Xe]6s²4f¹⁴5d².

Explanation:

The electron configuration for Hafnium (Hf) can be determined by looking at its position in the periodic table. Hafnium is in group 4B, which means it has 4 valence electrons. To write the electron configuration, we start with the noble gas that comes before Hafnium, which is Xenon (Xe).

The electron configuration for Hafnium, therefore, is [Xe]6s24f145d2.

Hafnium (Hf) has an electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d², or [Xe] 4f¹⁴ 5d² 6s².

It's part of the transition metals in group 4B.

The element hafnium (Hf) is part of the group 4B elements in the periodic table, which includes transition metals like titanium (Ti) and zirconium (Zr). Hafnium has an atomic number of 72. To determine the electron configuration of hafnium in order of increasing orbital energy, we start by filling the orbitals based on the Aufbau principle:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d².

Alternatively, in shorthand notation, this can be written as [Xe] 4f¹⁴ 5d² 6s². This configuration follows the orderly filling of orbitals, ensuring that lower energy orbitals are filled before higher energy orbitals.

Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 170 g of glycerin to 337 mL of H2O at 40.0°C?
The vapor pressure of pure water at 40.0°C is 55.32 torr and its density is 0.992 g/cm3.

Answers

Answer:

mass water = 347 x 0.992 = 344.2 g

moles water = 344.2 g / 18.02 g/mol= 19.1

moles C3H8O3 = 164 g / 92.097 g/mol= 1.78

mole fraction water = 19.1 / 19.1 + 1.78 = 0.915

p = 0.915 x 54.74 = 50.07 torr

Explanation:

What representative units define molecular compounds and ionic compounds?

Answers

Answer:

The molecular formula of the compounds indicates the amount of atoms that the substance contains. For example, water whose formula is H2O contains 2 atoms of hydrogen and 1 of oxygen. The representative unit of a molecular compound is a molecule while ions are represented by a formula unit.

Mole can be a unit 6.02 × 10²³. The units may be electrons, atoms, ions, or molecules, depending on the nature of the substance

Explanation:

The molecular formula of the compounds indicates the amount of atoms that the substance contains. For example, water whose formula is [tex]H2O[/tex] contains 2 atoms of hydrogen and 1 of oxygen. The representative unit of a molecular compound is a molecule while ions are represented by a formula unit.

Mole can be a unit 6.02 × 10²³. The units may be electrons, atoms, ions, or molecules, depending on the nature of the substance.

Therefore, water whose formula is [tex]H2O[/tex] contains 2 atoms of hydrogen and 1 of oxygen. The representative unit of a molecular compound is a molecule while ions are represented by a formula unit.

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Consider the reaction: H2(g) + I2(g)->2HI(g)
A reaction mixture in a 3.67-L flask at a certain temperature initially contains 0.763 g H2 and 96.9 g I2. At equilibrium, the flask contains 90.4 g HI. Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Answers

To calculate the equilibrium constant (Kc) for the reaction H2(g) + I2(g) \ightarrow 2HI(g), the moles of each reactant and product at equilibrium were calculated and then converted to concentrations. Using the concentrations, the Kc was determined to be approximately 610.97.

The reaction in question is H2(g) + I2(g) \ightarrow 2HI(g). To calculate the equilibrium constant (Kc), we need to determine the moles of HI, H2, and I2 at equilibrium. The molecular weights of H2, I2, and HI are approximately 2 g/mol, 254 g/mol, and 128 g/mol, respectively.

Initial moles of H2 = 0.763 g / 2 g/mol = 0.3815 mol

Initial moles of I2 = 96.9 g / 254 g/mol = 0.3815 mol

At equilibrium, there are 90.4 g of HI:

Equilibrium moles of HI = 90.4 g / 128 g/mol = 0.70625 mol

Since the reaction produces 2 moles of HI for every 1 mole of H2 and I2 that react, 0.70625 mol of HI would require half that amount of H2 and I2 to react:

Moles of H2 and I2 reacted = 0.70625 mol HI / 2 = 0.353125 mol

Moles of H2 and I2 at equilibrium = initial moles - moles reacted

Equilibrium moles of H2 = 0.3815 mol - 0.353125 mol = 0.028375 mol

Equilibrium moles of I2 = 0.3815 mol - 0.353125 mol = 0.028375 mol

We then convert these moles to concentrations by dividing by the volume of the flask:

[H2] = 0.028375 mol / 3.67 L \hickapprox 0.007732 M

[I2] = 0.028375 mol / 3.67 L \hickapprox 0.007732 M

[HI] = 0.70625 mol / 3.67 L \ hickapprox 0.192398 M

Now, we can calculate the equilibrium constant Kc

Kc = [HI]² / ([H2][I2]) = (0.192398 M)² / (0.007732 M × 0.007732 M) \hickapprox 610.97

Nylon 6,6 may be formed by means of a condensation polymerization reaction in which hexamethylene diamine [NH2-(CH2)6-NH2] and adipic acid react with one another with the formation of water as a byproduct. What masses of________(a) hexamethylene diamine and(b) adipic acid are necessary to yield 26 kg of completely linear nylon 6,6? This polymerization reaction occurs according to the following equation:

Answers

Answer:

Mass of hexamethylenediamine= 13,470.62 g

Mass of adipic acid =  16,940.13 g

Explanation:

Nylon -6,6 is synthesized by poly-condensation of hexamethylenediamine [NH₂-(CH₂)₆-NH₂] and adipic acid (C₆H₁₀O₄).

Both of these are taken in equal equivalents and reacted to form nylon 6,6 and by-product water is also formed.

Molecular mass of hexamethylenediamine [NH₂-(CH₂)₆-NH₂] =116.21 g/mol

Molecular mass of adipic acid (C₆H₁₀O₄) = 146.1412 g/mol

Molecular mass of Nylon 6,6 (C₁₂H₂₀N₂O₂) = 224.3 g/mol

Using unitary method :

Mass of hexamethylenediamine required for 26,000 g of Nylon 6,6 if 116.21 g/mol is required for 224.3 g/mol, is:

= [tex]\frac{26,000 *  116.21}{224.3}[/tex]

= 13,470.62 g

Mass of adipic acid required for 26,000 g of Nylon 6,6 if 146.1412 g/mol is required for 224.3 g/mol, is:

= [tex]\frac{26,000 *  146.1412}{224.3}[/tex]

= 16,940.13 g

What is the molarity of a potassium triiodide solution, KI3(aq), if 30.00 mL of the solution is required to completely react with 25.00 mL of a 0.300 M thiosulfate solution, K2S2O3(aq)? The chemical equation for the reaction is 2 S2O32-(aq) + I3-(aq) → S4O62-(aq) + 3 I-(aq).

Answers

Answer:

Molarity of triiodide solution is 0.125M

Explanation:

According to balanced equation, 2 moles of [tex]S_{2}O_{3}^{2-}[/tex] completely react with 1 mol of [tex]I_{3}^{-}[/tex].

Moles of [tex]S_{2}O_{3}^{2-}[/tex] in 25.00 mL of 0.300 M of [tex]K_{2}S_{2}O_{3}[/tex] solution = [tex](0.300\times 0.02500)moles=0.0075moles[/tex]

So, 0.0075 moles of [tex]S_{2}O_{3}^{2-}[/tex] completely reacts with [tex](\frac{1}{2}\times 0.0075)moles[/tex] of [tex]I_{3}^{-}[/tex] or 0.00375 moles of [tex]I_{3}^{-}[/tex]

If molarity of [tex]KI_{3}[/tex] solution is S (M) then-

[tex]S\times 0.0300=0.00375[/tex]

or, [tex]S=0.125[/tex]

So, molarity of triiodide solution is 0.125M

Final answer:

To determine the molarity of the KI3 solution, calculate the moles of S2O32- that reacted and use the stoichiometry of the reaction to find the moles of I3-. Then, divide the moles of I3- by the volume of KI3 solution in liters to find the molarity.

Explanation:

The question asks for the molarity of a potassium triiodide (KI3) solution. To find the molarity of the KI3(aq) solution, we need to use the provided stoichiometry of the reaction between thiosulfate (S2O32-) and triiodide (I3-).

The reaction is given as: 2 S2O32-(aq) + I3-(aq) → S4O62-(aq) + 3 I-(aq). Since 25.00 mL of a 0.300 M thiosulfate solution is required to completely react with the KI3 solution, we first calculate the moles of S2O32- reacted.

Moles of S2O32- = 0.025 L * 0.300 mol/L = 0.0075 moles.

According to the balanced equation, 1 mole of I3- reacts with 2 moles of S2O32-. Therefore, we have:

Moles of I3- = 0.0075 moles S2O32- / 2 = 0.00375 moles.

Because 30.00 mL (or 0.03000 L) of KI3 solution contains 0.00375 moles I3-, the molarity of KI3 is calculated by:

Molarity of KI3 = 0.00375 moles / 0.03000 L = 0.125 M.

The enthalpy of combustion of ethane gas, C2H6(g), is about -1.5*103 kJ/mol. When ethane reacts with O2(g), the products are carbon dioxide CO2(g) and water H2O(l). How much heat is released during the combustion of ethane gas when 14 mols of O2(g) are consumed. Express your answer in kJ.

Answers

Answer:

[tex]6.0\times 10^3 kJ[/tex] heat is released during the combustion of ethane gas when 14 moles of oxygen are consumed.

Explanation:

[tex]C_2H_6(g)+\frac{7}{2}O_2(g)\rightarrow 2CO_2(g)+3H_2O(l),\Delta H_{rxn}=-1.5\times 10^3 kJ/mol[/tex]

According to reaction, when 7/2 moles of oxygen is consumed by 1 mol of ethane [tex]1.5\times 10^3 kJ[/tex] energy is released.

Energy released when 1 mole of oxygen are consumed: [tex]\frac{1.5\times 10^3 kJ\times 2}{7}[/tex]

Then energy released when 14 moles of oxygen are consumed:

[tex]\frac{1.5\times 10^3 kJ\times 2}{7}\times 14=6.0\times 10^3 kJ[/tex]

[tex]6.0\times 10^3 kJ[/tex] heat is released during the combustion of ethane gas when 14 moles of oxygen are consumed.

Answer:

6.0×10^3 KJmol-1

Explanation:

Equation of the reaction

C2H6(g) + 7/2 O2(g) ------> 2CO2(g) + 3H2O(g)

From the balanced reaction equation:

1 mole of ethane reacts with 3.5 moles of oxygen

x moles of ethane will react with 14 moles of oxygen

x= 14/3.5 = 4 moles of ethane

If heat of combustion of ethane= -1.5*103 kJ/mol

Then for 4 moles of ethane= 4× -1.5*103 kJ/mol= 6.0×10^3 KJmol-1

A chemistry student needs 65.0g of diethylamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of diethylamine is ·0.706gcm−3. Calculate the volume of diethylamine the student should pour out.

Answers

Final answer:

To find the volume of diethylamine needed for an experiment, dividing the required mass (65.0 g) by the density of diethylamine (0.706 g/cm3) calculates a volume of 92.1 mL that the student should measure out.

Explanation:

To calculate the volume of diethylamine that a student needs for an experiment, use the formula that relates mass, density, and volume: volume = mass / density. Here, the mass required is 65.0 g, and the density of diethylamine is given as 0.706 g/cm3.

Substitute the given values into the formula:

Volume = 65.0 g / 0.706 g/cm3 = 92.1 cm3 or 92.1 mL, since 1 cm3 = 1 mL.

Therefore, the student should pour out 92.1 mL of diethylamine for their experiment.

Calcium chloride, CaCl2, is commonly used as an electrolyte in sports drinks and other beverages, including bottled water. A solution is made by adding 6.50 g of CaCl2 to 60.0 mL of water at 25∘C. The density of the solvent at that temperature is 0.997 g/mL. Calculate the mole percent of CaCl2 in the solution.

Answers

Final answer:

The mole percent of CaCl2 in the solution would be 100% if we consider only the solute, but this does not account for the water's contribution. If water's contribution to total moles were considered, the mole percent for CaCl2 would be less than 100%.

Explanation:

To calculate the mole percent of CaCl2 in the solution, we must first determine the number of moles of CaCl2.

The molar mass of CaCl2 is approximately 110.98 g/mol (40.08 g/mol for calcium and 35.45 g/mol for each chlorine, with two chlorines). With 6.50 g of CaCl2, the number of moles of CaCl2 is:

6.50 g ÷ 110.98 g/mol = 0.05855 mol

To find the total mass of the solution, the mass of CaCl2 is added to the mass of water. With the water density given as 0.997 g/mL, the mass of 60.0 mL of water is:

60.0 mL × 0.997 g/mL = 59.82 g

The total mass of the solution is thus:

6.50 g (CaCl2) + 59.82 g (H2O) = 66.32 g

To find the mole fraction of CaCl2, we divide the moles of CaCl2 by the total moles (assuming the water in the solution contributes negligibly to the total moles):

Mole fraction of CaCl2 = 0.05855 mol ÷ 0.05855 mol = 1 (since there are no other solutes)

To express this as a mole percent, we multiply the mole fraction by 100%:

Mole percent of CaCl2 = 1 × 100% = 100%

However, this calculation does not account for the water's contribution to the mole fraction. To be precise, the mole fraction and mole percent should include the moles of water as well, but since the question seems to focus solely on CaCl2, the calculation reflects this perspective. If we were to include the moles of water, the mole percent would be significantly less than 100%.

Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO₃) into solid calcium oxide and gaseous carbon dioxide. Suppose 23.0 L of carbon dioxide gas are produced by this reaction, at a temperature of 380.0 °C and pressure of exactly 1 atm. Calculate the mass of calcium carbonate that must have reacted. Be sure your answer has the correct number of significant digits.

Answers

Answer:

CaCO3 (s)  → CaO (s) + CO2 (g)

The mass of carbonate that must have reacted was 43.03 grams

Explanation:

CaCO3  → CaO + CO2

Relation between reactant and product is 1:1

Let's apply the Ideal Gas Law to find out the moles of CO2 which were produced.

P . V = n . R . T

1 atm . 23 L = n . 0.082 L.atm/mol.K . 653K

(1atm . 23L) / (0.082 mol.K/L.atm . 653K) = n

0.43 moles = n

0.43 moles of CO2, were produced from 0.43 moles of CaCO3.

Molar weight of CaCO3 = 100.08 g/m

Mass = Molar weight . moles

Mass = 100.08 g/m 0.43 m = 43.03 g

The decomposition of calcium carbonate (CaCO₃) can be represented by the equation: CaCO₃(s) → CaO(s) + CO₂(g). Using the ideal gas law, we calculate that approximately 43.1 grams of CaCO₃ reacted to produce 23.0 liters of CO₂ at 380.0 °C and 1 atm pressure.

Decomposition of Calcium Carbonate:

The balanced chemical equation for the decomposition of solid calcium carbonate (CaCO₃) into solid calcium oxide (CaO) and gaseous carbon dioxide (CO₂) is:

CaCO₃(s) → CaO(s) + CO₂(g)

To calculate the mass of calcium carbonate that reacted to produce 23.0 L of CO₂ gas at 380.0 °C and 1 atm, we use the ideal gas law. The ideal gas law is expressed as:

PV = nRT

Where:

P = pressure (1 atm)V = volume (23.0 L)n = moles of gasR = ideal gas constant (0.0821 L·atm/mol·K)T = temperature in Kelvin (380.0 °C + 273.15 = 653.15 K)

Rearranging the ideal gas law to solve for n (moles of CO₂) gives:

n = PV / RT

Substituting the known values:

n = (1 atm * 23.0 L) / (0.0821 L·atm/mol·K * 653.15 K) ≈ 0.431 moles of CO₂

From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Hence, the moles of CaCO₃ that reacted is also 0.431 mol.

The molar mass of CaCO₃ is approximately 100.09 g/mol.

The mass of CaCO₃ that reacted is:

mass = moles * molar mass

mass = 0.431 mol * 100.09 g/mol ≈ 43.1 g

Thus, about 43.1 grams of calcium carbonate must have reacted.

Liquid hydrogen peroxide, an oxidizing agent in many rocket fuel mixtures, releases oxygen gas on decomposition.

2 H2O2(l) 2 H2O(l) + O2(g) Hrxn = -196.1 kJ
How much heat is released when 765 kg H2O2 decomposes?

Answers

Answer:

4.3 x 10⁶ kJ are released

Explanation:

From the balanced chemical equation we know that when 2 moles of hydrogen peroxide decompose, 191 kJ of heat are released. So what we need to calculate is the number of moles the 765 kg of H2O2 represent and calculate the heat released.

Molar Mass H2O2 = 34.01 g/mol

Mass H2O2 = 765 Kg x 1000 g/Kg = 765000 g

moles H2O2 = 1 Mol / 34.01 g x 765000  grams = 22.5 Mol

(-196.1 kJ/ 2 mol H2O2 )  x 22.5 mol H2O2 = -4.3 x 10⁶ kJ

The substance that dissolves to make a solution is called

Answers

Answer: Solute or solvent

Explanation: A solute is the substance that dissolves to make a solution. The solvent is the solution that does the dissolving.

Match the vocabulary terms to their definitions.
1. standard metric unit of length
2. a proposed explanation for a scientific problem
3. standard unit of volume
4. the curved top surface of a liquid column
5. a quantity in an experiment that remains unchanged or constant
6. amount of matter in an object
a.mass
b.hypothesis
c.liter
d.control
e.meter
f.meniscus

Answers

You may find bellow the association of therms with the definitions.

Explanation:

1. standard metric unit of length  - e. meter

2. a proposed explanation for a scientific problem  - b. hypothesis

3. standard unit of volume  - c. liter

4. the curved top surface of a liquid column  - f. meniscus

5. a quantity in an experiment that remains unchanged or constant  - d. control

6. amount of matter in an object  - a. mass

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Acidic solution In acidic solution, the iodate ion can be used to react with a number of metal ions. One such reaction is IO3−(aq)+Sn2+(aq)→I−(aq)+Sn4+(aq) Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: IO3−(aq)+Sn2+(aq)+ −−−→I−(aq)+Sn4+(aq)+ −−−

Answers

Answer:

[tex]\large \text{IO$_{3}$$^{-}$(aq) + 3Sn$^{2+}$(aq) + 6H$^{+}$(aq) $\longrightarrow \,$ I$^{-}$(aq) + 3Sn$^{4+}$(aq) + 3H$_{2}$O(l)}[/tex]

Explanation:

IO₃⁻ + Sn²⁺ ⟶ I⁻ + Sn⁴⁺

1: Separate the equation into two half-reactions.

IO₃⁻ ⟶ I⁻

Sn²⁺ ⟶ Sn⁴⁺

2: Balance all atoms other than H and O.

Done

3: Balance O.

IO₃⁻ ⟶ I⁻ + 3H₂O

Sn²⁺ ⟶ Sn⁴⁺

4: Balance H

IO₃⁻ + 6H ⟶ I⁻ + 3H₂O  

Sn²⁺ ⟶ Sn⁴⁺

5: Balance charge.

IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O  

Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻

6: Equalize electrons transferred.

1 × [IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O]

3 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]

7: Add the two half-reactions.

1 × [IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O]

3 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]                                  

      IO₃⁻ + 3Sn²⁺ + 6H⁺ ⟶ I⁻ + 3Sn⁴⁺ + 3H₂O

8: Check mass balance.

 On the left: 1 I, 3 O, 3 Sn, 6 H

On the right: 1 I, 3 O, 3 Sn, 6 H

Step 9: Check charge balance.

  On the left: 1- + 12+ = 11+

On the right: 1-  + 12+ = 11+

The equation is balanced.

[tex]\text{The balanced equation is }\\\large \textbf{IO$_{3}$$^{-}$(aq) + 3Sn$^{2+}$(aq) + 6H$^{+}$(aq) $\longrightarrow \,$ I$^{-}$(aq) + 3Sn$^{4+}$(aq) + 3H$_{2}$O(l)}[/tex]

Final answer:

The reaction in an acid solution involves IO3−(aq) reacting with Sn2+(aq) to form I−(aq) and Sn4+(aq). The complete balanced redox reaction is 2IO3−(aq) + 10H+(aq)+ 6Sn2+(aq) → I−(aq) + 4H2O(l) + 6Sn4+(aq). It is a redox reaction where iodate ions act as strong oxidizing agents.

Explanation:

The reaction under consideration involves the iodate ion (IO3−) reacting with a tin(II) ion (Sn2+) to create iodide (I−) and a tin(IV) ion (Sn4+) in an acidic solution. In an acid solution, water (H2O) and hydrogen ions (H+) play crucial roles, hence they have places in the reaction. The complete balanced redox reaction is:

2IO3−(aq) + 10H+(aq)+ 6Sn2+(aq) → I−(aq) + 4H2O(l) + 6Sn4+(aq)

. To balance this reaction, we consider the change in oxidation state, the law of conservation of charge, and the law of conservation of mass. This reaction is a prominent example of a redox reaction where iodate ions act as strong oxidizing agents, oxidizing Sn2+ to Sn4+.

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________ is an indication of how near the air is to saturation rather than the actual quantity of water vapor in the air.

Answers

Answer: Relative humidity

Explanation:

Relative humidity (RH) is the ratio of the partial pressure of water vapor to the equilibrium vapor pressure of water at a given temperature. Relative humidity depends on temperature and the pressure of the system of interest. The same amount of water vapor has higher relative humidity in cool air than in warm air. A related parameter to that regard is that of dew point.

Which one of the following statements is not true?
a. Basalt magmas, in general, have higher temperatures than rhyolite magmas.
b. When magma reaches the surface, its dissolved gas content increases.
c. Melting temperatures of silicate rocks are lowered by small amounts of water.
d. Melting temperatures of silicate rocks increase with increased pressure.

Answers

Answer:

When magma reaches the surface, its dissolved gas content increases is true

Explanation:

The volcanic eruptions happen because of magma that is expelled on the earth’s surface. At the earth’s depth, all magma have gas dissolved in liquid. When the pressure has decreased the magma rises towards the earth’s surface creating a separate vapour phase.  

As pressure reduces the volume of gas will  expands and giving magma its 'explosive character'. Thus, as magma reaches the surface the dissolved gas content decreases and magma comes out of earth’s surface.

Final answer:

The untrue statement is b, as the dissolved gas content in magma decreases as it reaches the surface due to reduced pressure. Basalt magma has higher temperatures than rhyolite magma. Water reduces the melting temperature of silicate rocks, and increased pressure raises it.

Explanation:

The subject here pertains to volcanic activity and magma characteristics. The statement that is not true is: b. When magma reaches the surface, its dissolved gas content increases. In fact, as magma rises to the surface, the pressure decreases which allows the dissolved gases to escape, thus its dissolved gas content actually decreases. Statements a, c and d are true - Basalt magmas do generally have higher temperatures than rhyolite magmas. Additionally, the presence of water does indeed lower the melting temperature of silicate rocks, and increased pressure results in a higher melting temperature for these rocks.

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An example of a synthesis is the Statue of Liberty turning green. Oxygen in the air combines with the copper statue to form copper oxide, which is green. Which of these is the correct chemical reaction for this?


A) Cu + O → CuO
B) Cu + O2 → CuO
C) CuO → Cu + O2
D) 2 Cu + O2 → 2CuO

Answers

Answer:

D. 2Cu + O2 = 2CuO

Explanation:

Then CuO, which is black, reacts with CO2 to form CuCO3 which is green.

Answer : The correct chemical reaction is, (D)

Explanation :

Synthesis reaction : It is defined as a chemical reaction where multiple substances or reactants combine to form a single product.

It is represented as,

[tex]X+Y\rightarrow XY[/tex]

When copper react with an oxygen then it react to give copper oxide as a product.

The balanced chemical reaction will be:

[tex]2Cu+O_2\rightarrow 2CuO[/tex]

Hence, the correct chemical reaction is, (D)

A moving company is called into a building and asked to relocate 520 bricks to a location 15 meters above the ground. If each brick is 4 kilograms, what amount of work will be needed to accomplish the task?A. 305,760 JB. 229,320 JC. 203,840 JD. 588,000 J

Answers

Answer: A. 305,760 J

Explanation:

Work is defined as a force causing the movement or displacement of an object.

work=[tex]mass\times acceleration\times height[/tex]

Given: m=  mass = 4 kg

g= acceleration due to gravity = [tex]9.80m/s^2[/tex]

h = height = 15m

Putting in the values we get,

work=[tex]4kg\times 9.80m/s^2\times 15m=588J[/tex]

[tex]1kgm^2s^{-2}=1Joule(J)[/tex]

Now 520 bricks are to be relocated, thus work done = [tex]520\times 588J=305760J[/tex]

Thus amount of work will be needed to accomplish the task is 305760 J.

A gas mixture used for anesthesia contains 2.83 mol oxygen (O2), and 8.41 mol of nitrous oxide (N2O). The total pressure of the mixture is 192 kPa. What is the partial pressure of N2O? Report your answer with three significant figures.

Answers

Answer:

The partial pressure of N2O is 143.6 kPa

Explanation:

Step 1: Data given

Moles O2 = 2.83 mol

Moles N2O = 8.41 mol

Total pressure = 192 kPa

Step 2: Calculate total number of moles

Total numberof moles = moles O2  + molesN2O = 2.83 + 8.41 = 11.24 mol

Step 3: Calculate mole fraction of N2O

Mole fraction N2O = mole N2O / Total moles

Mole fraction N2O = 8.41 / 11.24 = 0.748

Step 4: Calculate partial pressure of N2O

pN2O = 0.748 * 192 kPa

pN2O = 143.6 kPa

The partial pressure of N2O is 143.6 kPa

Answer: 144 kPa

Explanation:

To determine the partial pressure of N2O, first determine the mole fraction of N2O in the mixture, then multiply the mole fraction of N2O by the total pressure.

XN2O = nN2O/ ntotal = 8.41 mol / 2.83 mol + 8.41 mol = 0.7482

PN2O=XN2O×Ptotal = 0.7482 × 192 kPa = 143.7 kPa

Rounding the answer to three significant figures, the partial pressure of N2O is 144kPa.

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