A sample of gas in a balloon has an initial temperature of 16.5°C and a volume of 4.01×10 to the third leader if the temperature changes to 72°C and there is no change of pressure or amount of gas what is the new volume of gas

Answers

Answer 1

Answer:

45.5

Explanation:

your question is not clear at all

your question has no unit for volume


Related Questions

A 1.00 liter solution contains 0.31 M sodium acetate and 0.40 M acetic acid. If 0.100 moles of barium hydroxide are added to this system, indicate whether the following statements are TRUE or FALSE . (Assume that the volume does not change upon the addition of barium hydroxide.)

a. The number of moles of CH3COOH will remain the same.
b. The number of moles of CH3COO- will increase.
c. The equilibrium concentration of H3O+ will decrease.
d. The pH will decrease.
e. The ratio of [CH3COOH] / [CH3COO-] will remain the same.

Answers

Answer and Explanation:

The buffer solution is composed by sodium acetate (CH₃COONa) and acetic acid (CH₃COOH). Thus, CH₃COOH is the weak acid and CH₃COO⁻ is the conjugate base, derived from the salt CH₃COONa.

If we add a strong base, such as barium hydroxide, Ba(OH)₂, the base will dissociate completely to give OH⁻ ions, as follows:

Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻

The OH⁻ ions will react with the acid (CH₃COOH) to form the conjugate base CH₃COO⁻.

Initial number of moles of CH₃COOH = 0.40 mol/L x 1 L = 0.40 mol

Initial number of moles of CHCOO⁻= 0.31 mol/L x 1 L = 0.31 mol

moles of OH- added: 2 OH-/mol x 0.100 mol/L x 1 L = 0.200 OH-

According to this, the following are the answers to the sentences:

a. The number of moles of CH₃COOH will remain the same ⇒ FALSE

The number of moles of CH₃COOH will decrease, because they will react with OH⁻ ions

b. The number of moles of CH₃COO⁻ will increase ⇒ TRUE

Moles of CH₃COO⁻ will be formed from the reaction of the acid (CH₃COOH) with the base (OH⁻ ions)

c. The equilibrium concentration of H₃O⁺ will decrease ⇒ FALSE

The equilibrium concentration of OH⁻ is increased

d. The pH will decrease⇒ FALSE

pKa for acetic acid is 4.75. We add the moles of base to the acid concentration and we remove the same number of moles from the conjugate base in the Henderson-Hasselbach equation to calculate pH:

[tex]pH= pKa + log \frac{[conjugate base + base]}{[acid - base]}[/tex]

pH = 4.75 + log (0.31 mol + 0.20 mol)/(0.40 mol - 0.20 mol) = 5.15

Thus, the pH will increase.

Which is not a compound? A.They are substances B. They have compositions that vary C. They can be physically separated into their component Elements. D. They have properties similar to those of their component elements.

Answers

Answer:

D.They have properties similar to those of their component elements.

Explanation:

When elements are joined, the atoms lose their individual properties and have different properties from the elements they are made of .

Consider an endothermic reaction C(s) CO2(g)2CO(g) If such a system at equilibrium is heated and compressed at the same time, what happens to the equilibrium position?
A. It will shift to the left
B. It will shift to the right
C. It will not change
D. There is not enough information to tell

Answers

Answer:

A. It will shift to the left

Explanation:

In the equilibrium:

C(s) ⇄ CO2(g)2CO(g)

The system will shift to the right if any change stimulate the production of gas -LeChatelier's principle-; in the same way, if a change doesn't favors the production of gas the system will shift to the left producing less gas.

The changes that increasing the pressure of the system, doesn't favors the gas production doing the system shift to the left.

A gas that is heated expands itsellf doing the pressure increases.

In the same way, if you compress the gas, the gas increases its pressure.

Thus, both changes increase pressure of the gas doing the system shift to the left.

A. It will shift to the left

HELPPP.
Which of the following is a property of matter?
O A. It takes up space.
OB. It is everywhere.
O C. It is constantly changing.
O D. It cannot be divided.

Answers

Answer:

a

Explanation:

Matter can be volume or density. So, this concludes that it is when it takes up space.

Answer: A.

Explanation:

it takes up space

The element nitrogen forms a(n) _______ with the charge . The symbol for this ion is , and the name is . The number of electrons in this ion is .

Answers

Answer:

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10.

Explanation:

The element nitrogen is in the Group 15 in the Periodic Table, so it tends to gain 3 electrons (3 negative charges) to fill its valance shell with 8 electrons.

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10 (the original 7 plus the 3 gained). It is isoelectronic with the gas Neon, which accounts for its stability.

Suppose that the NaOH sitting on the shelf had absorbed 1 g of water for every 1 g of NaOH. How would this have affected your results

Answers

Answer:

The concentration of NaOH will be lower and the titration will be affected.

Explanation:

Hello,

In this case, sodium hydroxide is acknowledged as a highly hygroscopic substance, which means that is able to absorb water to its molecules. In such a way, in any measurement, if sodium hydroxide has absorbed water, the results will be wrong in terms of accuracy. More specifically, for concentration, if we have for example 30 grams of NaOH and we dissolve it a 100-mL solution, as it absorbed 30 grams of water, the total volume could be now approximated to 130 mL, thus, the concentration will change as follows:

[tex]M_1=\frac{30g/40g/mol}{0.1L}=7.5M\\ \\M_2=\frac{30g/40g/mol}{0.13L}=5.77M[/tex]

It causes the actual molarity to be decreased, it means that in a titration procedure, less acid would be used to neutralize it or more of it would be needed to neutralize a given acid.

Best regards.

We have that  the NaOH sitting on the shelf had absorbed 1 g of water for every 1 g of NaOH will affect the Molarity of NaOH and its effectiveness

With the situation of NaOH sitting on the shelf having to absorbed 1 g of water for every 1 g of NaOH.

Means that for every  g of NaOH collected 1/2g is water and 1/2g is actual NaOH.

Hence this will cause a change in the molarity of NaOH thereby causing Molarity to drop by half as well. Giveing the resultant Molarity to be somewhere around half the regular molarity.

This many cause a variation also in the results of titration

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Santalbic acid, a fatty acid isolated from the seeds of the sandalwood tree, is an unusual fatty acid that contains a carbon-carbon triple bond.What orbitals are used to form each of the three indicate single bongs in santalbic acid?

Answers

Answer and Explanation:

The explanation is described below:-

a. Bond that is a which is created by sp3 - sp2 orbitals

b. Bond that is b which is developed by sp2-sp orbitals

c. Bond that is c which is created by sp-sp3 orbitals

Sp2 is hybridised by a double bonded carbon, and a triple bonded carbon is hybridised. Both single bonded carbons are hybridised to sp3.

"Calculate the pH during the titration of 30.00 mL of 0.1000 M C3H7COOH(aq) with 0.1000 M LiOH(aq) after 29.9 mL of the base have been added. Ka of butanoic acid

Answers

Answer:

pH = 7.29

Explanation:

Ka of butanoic acid is 1.54x10⁻⁵

To obtain the pH of the solution you must use H-H equation for butanoic acid:

pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]

Where pKa is defined as -log Ka = 4.81

Now, you need to find [C₃H₇COO⁻] and [C₃H₇COOH] concentrations (Also, you can find moles of each substance and replace them in the equation.

Butanoic acid reacts with LiOH, producing C₃H₇COO⁻, thus:

C₃H₇COOH + LiOH → C₃H₇COO⁻ + H₂O + Li⁺

Moles of both reactants, C₃H₇COOH and LiOH are:

C₃H₇COOH = 0.0300L ₓ (0.1000mol / L) = 0.003000moles of C₃H₇COOH

LiOH = 0.0299L ₓ (0.1000mol / L) = 0.00299 moles of LiOH.

That means moles of C₃H₇COO⁻ produced are 0.00299 moles.

And moles of C₃H₇COOH that remains in solution are:

0.00300 - 0.00299 = 0.00001 moles of C₃H₇COOH

Replacing in H-H equation:

pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]

pH = 4.81 + log₁₀ [0.00299moles] / [0.00001moles]

pH = 7.29

Which of the following best explains why overfishing is considered a bad practice? a. Overfishing is not sustainable and ultimately depletes a fishery. b. Overfishing is a more expensive fishing practice. c. Overfishing results in species becoming overpopulated. d. Overfishing prevents the formation of protected habitats.

Answers

Answer:

The answer is A

Overfishing is not sustainable and ultimately depletes a fishery. Therefore, the correct option is option A among all the given options.

What is overfishing?

Overfishing is the activity of capturing more fish than their population can replenish. The population continues to diminish until it is completely wiped off. Bycatch is another phrase that is frequently used in conjunction with overfishing.

This refers to fish captured in nets designed for a certain breed. Dolphins, sea turtles, as well as other non-target fish, for example, are frequently caught in nets designed for anchoveta pollock. Trawling is another kind of overfishing. This is the process of scrapping the ocean floor and scooping up everything within reach within a certain region.  Overfishing is not sustainable and ultimately depletes a fishery.

Therefore, the correct option is option A.

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You use 10.0 mL of solution A, 10.0 mL of solution B, and 70.0 mL of water for your first mixture. What is the initial concentration of KIO3

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  initial concentration is  [tex]C_f = 0.0022 \ M[/tex]

Explanation:

From the question we are told that

    The volume of solution A is  [tex]V_i = 10.0 mL[/tex]

    The concentration of A is [tex]C_i = 0.0200 \ M[/tex]

    The volume of solution B  is  [tex]V_B = 10.0mL[/tex]

    The volume of water is  [tex]V_{w } = 70.0 mL[/tex]

Generally the law of dilution is mathematically represented as

             [tex]C_i * V_i = C_f * V_f[/tex]

Where  [tex]C_f[/tex] is the concentration of  the mixture

            [tex]V_f[/tex] is the volume of the mixture which is mathematically evaluated as

            [tex]V_f = 10 + 10 + 70[/tex]

           [tex]V_f = 90mL[/tex]

So  

      [tex]C_f = \frac{C_i * V_i}{ V_f}[/tex]

substituting values

       [tex]C_f = \frac{0.0200 * 10 }{90}[/tex]

       [tex]C_f = 0.0022 \ M[/tex]

Note the mixture obtained is  [tex]KIO_3[/tex]

Note the dynamic equilibrium in the opening photo which solution changes color when the pH of both solutions is increased explain?

Answers

Answer:

The colour of the orange solution becomes yellow.  

Explanation:

1. Before adding NaOH

Assume the picture showed a beaker of potassium chromate and one of potassium dichromate.

Both solutions are involved in the same equilibrium:

[tex]\rm\underbrace{\hbox{2CrO$_{4}^{2-}$(aq)}}_{\text{yellow}} +2H^{+}(aq) \rightleftharpoons \, \underbrace{\hbox{Cr$_{2}$O$_{7}^{2-}$}}_{\text{orange}} + H_{2}O[/tex]

The first beaker contains mostly chromate ions with a few dichromate ions.

The position of equilibrium lies to the left and the solution is yellow.

The second beaker contains mostly dichromate ions with a few chromate ions.

The position of equilibrium lies to the right and the solution is orange.

2. After adding NaOH

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Beaker 1

If you add OH⁻ to the equilibrium solution, it removes the H⁺ (by forming water).

The system responds by having the dichromate react with water to replace the H⁺.  

At the same time, the system forms more of the yellow chromate ion.

The position of equilibrium shifts to the left.

However, the solution is already yellow, so you see no change in colour.

Beaker 2

The reaction is the same as in Beaker 1.

This time, however, as the dichromate ion disappears, do does its orange colour.

Also, the yellow chromate is being formed and its yellow colour appears .

The colour changes from orange to yellow.

When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose from 25.00 °C to 31.75 °C. Determine the ∆Hrxn in kJ/mol for the reaction as written below. Assume that the density of the solutions is 1.000 g/mL, and that the specific heat capacity of the solutions is that of pure water. Must show your work (scratch paper) to receive credit.

Answers

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density  * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

Argon gas has a boiling point of -197 °C. Which of the following diagrams best represents the
distribution of argon atoms in a steel sphere at -190 °C?

Answers

Answer:

(a)

Explanation:

Hello,

In this case, the temperature required to boil argon, it means, transform it from liquid to gas is -197 °C. In such a way, since the temperature inside the steel sphere is -190 °C, which is greater than the boiling point, we realize argon is gaseous, therefore, the molecules will be spread inside the sphere as they will be moving based on the kinetic theory of gases.

For that reason, answer is scheme (a).

Best regards.

what is the electron configuration of the iodide ion?
A. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²
B. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶
C. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶
D. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰

Answers

Answer:

the answer to this question is C

The electron configuration of the iodide ion is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶. The correct option is option C.

What is electron configuration ?

The arrangement of an atom's or molecule's electrons in their respective atomic or molecular orbitals is known as its electron configuration; for instance, the electron configuration of the neon atom is 1s2 2s2 2p6.

According to electronic configurations, electrons move individually within each orbital while interacting with the average field produced by all other orbitals. The corrosion potential and reactivity of an atom may be calculated from its electron configuration. The electron configuration of the iodide ion is  1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶.

Therefore, the correct option is option C.

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how many moles of ions are produced by ionization of 2 moles of MgCl2

Answers

Answer:

number of ions = 12.04 x 10^²³

Explanation:

n = number of ions/Avogadro's constant

2 = number of ions/6.02 x 10^²³

number of ions= 2 x 6.02 x 10^²³

number of ions = 12.04 x 10^²³

The location of chlorine below fluorine on the periodic table implies which of the following?

a. Chlorine atoms have a larger radius than fluorine atoms.
b. Chlorine atoms have a lower mass than fluorine atoms.
c. Chlorine atones have fewer valence electrons than fluorine atoms.
d. Chlorine atoms are more reactive than fluorine atoms.

Answers

the answer is A because as the element goes down the group, it has larger radius

A 23.0% by mass LiCl solution is prepared in water. What is the mole fraction of the water? 0.113 0.887 0.873 0.127 0.911

Answers

Answer: The Answer should be B. 0.887 (:

Explanation:

The mole fraction of the solution is 0.887.

Mass percent of LiCl present =  23.0%

Mass of the LiCl can be obtained from;

23 = x/100 × 100/1

x = 23 g

Number of moles of LiCl = 23g/42 g/mol = 0.55 moles

Number of moles of water = (100 - 23) g/18 g/mol = 4.3 moles

Total number of moles = 0.55 + 4.3 = 4.85 moles

Mole fraction of water = 4.3 moles/4.85 moles = 0.887

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List the following compounds in order from strongest acid to weakest acid. Rank the acids from strongest to weakest.
CH2CHCH2COOH CH2CH2CH2COOH CH3CHCH2COOH CH3CH2CH2COOH
Strongest Weakest

Answers

Answer:

CH3CH2CH2COOH<CH2(F)CH2CH2COOH<CH3CH(F)CH2COOH<CH2(F)CH(F)CH2COOH

Explanation:

We know that the presence of highly electronegative elements in carboxylic acid molecules lead to -I inductive effect. This implies that electrons are withdrawn along the chain towards the electronegative element. As electrons are withdrawn towards the electronegative element, the electron cloud of the carbonyl- hydrogen bond in the acid weakens and the hydrogen can now be easily lost as a proton, that is , the molecule becomes more acidic.

The -I inductive effect increases with increase in the number of electronegative elements present in the molecule and the proximity of the electronegative element to the carbonyl group. The closer the electronegative element is to the carbonyl group, the greater the acidity of the molecule since the -I inductive effect dies out with increasing distance from the carbonyl group. Also, the more the number of electronegative elements in the molecule, the greater the - I inductive effect and the greater the acidity of the molecule, hence the answer.

If the charge on the chloride ion is -1, what is the charge on the magnesium ion in the compound MgCl2?

Answers

Answer:

+2

Explanation:

Because the charge of the chloride ion is negative, that means that the charge of the magnesium ion must be positive since cations and anions go together, not cation and cation nor anion and anion. Using the "reverse criss-cross method", since the subscript of Mg is 1, that means that this is the lowest whole number ratio so we don't need to worry about simplifying. Therefore, since the charge of Cl is 2, the answer is +2.

(4Ga + 3S2 → 2Ga2S3)



1. How many moles of Sulfur are needed to react with 100.0 grams of Gallium?

Answers

Answer:

[tex]n_S=1.076molS[/tex]

Explanation:

Hello,

In this case, given the undergoing chemical reaction, we can see a 4:3 mole ratio between the consumed moles of gallium and sulfur respectively, therefore, the consumed moles of sulfur, from the 100.0 g of gallium (use its atomic mass) turn out:

[tex]n_{S}=100.0gGa*\frac{1molGa}{69.72gGa}*\frac{3molS}{4molS} \\\\n_S=1.076molS[/tex]

Best regards.

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the surroundings

Answers

Answer:

[tex]V=27992L=28.00m^3[/tex]

Explanation:

Hello,

In this case, the combustion of methane is shown below:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]

Best regards.

4
Select the correct answer.
Which of the following describes a compound? (Hint: Carbon and oxygen both appear on the periodic table.)
ОА.
a piece of pure carbon, containing only carbon atoms
OB. oxygen gas surrounding a solid piece of carbon
a substance made of two oxygen atoms for each carbon atom
OD
carbon and oxygen atoms mixed without being bonded together

Answers

A substance made of two oxygen atoms for each carbon atom

Look at the reaction below. Upper H subscript 2 upper S upper O subscript 4 (a q) plus upper Upper M g (s) right arrow Uper M g upper S upper O subscript 4 (a q) plus upper H subscript 2 (g). Which substance is the acid in the reaction? Mg(s) H2(g) MgSO4(aq) H2SO4(aq)

Answers

Answer:

H2SO4(aq)

Explanation:

The balanced equation for the reaction is given below:

H2SO4(aq) + Mg(s) —> MgSO4(aq) + H2 (g)

An acid is a substance which dissolves in water to produce hydrogen ion, H+ as the only positive ion.

To know which of the substance is acid, let us dissolve them in water to see which will produce hydrogen ion, H+ as the only positive ion.

This is illustrated below:

H2SO4(aq) —> 2H+(aq) + SO4^2-(aq)

Mg(s) + 2H2O(l) —> Mg(OH)2(aq) + H2(g)

MgSO4(aq) —> Mg^2+(aq) + SO4^2-(aq)

H2 is insoluble in water.

From the above, only H2SO4 produces hydrogen ion H+ on dissolution in water. Therefore, H2SO4 is an acid

Answer:

D on edg 2021

Explanation:

Which of the following electron configurations is incorrect? A) S [Ne]3s³3p⁴ B) Sn [Kr]5s²4d¹⁰5p² C) Rb [Kr]5s¹ D) V [Ar]4s²3d³ E) I [Kr]5s²4d¹⁰5p⁵

Answers

Answer:

A S [Ne]3s³3p⁴ instead of [Ne]3s²3p⁴

Explanation:

Hello,

To find the incorrect electronics configuration, we have to refer back to our periodic table or simply writing the electronic configuration of each element down follow principles guiding it such as Aufbau principle and Hund's rule.

a) S = [Ne] 3s³ 3p⁴ instead of [Ne] 3s²3p⁴

Option A is wrong.

S orbital can only accommodate a maximum of 2 electrons and in this case, 3s orbital is carrying 3 electrons. This has violated the rule.

b) Sn = [Kr] 5s² 4d¹⁰ 5p²

Option B is correct

c) Rb = [Kr] 5s¹

Option c is correct

d) V = 4s² 3d³

Option D is correct

e) I = [Kr] 5s² 4d¹⁰ 5p⁵

From the above, we can see that the answer is option A = S

The following electron configurations incorrect is -  A) S [Ne]3s³3p⁴.

The electron configuration of an element describes how electrons are distributed in their atomic orbitals.

It follows a standard notation in which all electron-containing atomic subshells are placed in a sequence.The atomic number of sulfur - 16, which means it has 16 electrons

The electronic configuration:

=> 1s² 2s² 2p⁶ 3s² 3p⁴

=> [Ne] 3s² 3p⁴ (Ne has 10 electrons with 1s² 2s² 2p⁶ electron configuration)

The given electronic configuration has 3 electrons in 3-s orbital instead of 2 electrons.

Thus, the following electron configurations incorrect is -  A) S [Ne]3s³3p⁴.

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What type of reaction is shown below? Check all
that apply.
AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s)
synthesis
decomposition
combustion
single replacement
double replacement

Answers

Answer:

double replacement

Explanation:

Double replacement reactions are where both cations and anions of reactants switch.

For eg: AB + CD -----> AD + BC (Here we see both things are replaced)

Synthesis reactions is where two simple things make one complex thing.

So A + B --- > AB

Combustion is usually reaction where Oxygen is reactant but here we don't have that.

Single replacement only replaces either cation or anion.

So AB + C .------> AC + B

Answer:

Replacement

Explanation:

neeeeed helpppppppppp

Answers

Answer:

Option C. Will always.

Explanation:

A spontaneous reaction is a reaction that occurs without an external supply of heat.

This implies that spontaneous reaction will always occur as no external supply of heat is needed.

A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of moles of an unknown gas requires 6.34 minutes to effuse through the same barrier. The molar mass of the unknown gas is:________.
g/mol.

Answers

Answer:

The molar mass of the unknown gas is [tex]\mathbf{ 51.865 \ g/mol}[/tex]

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

[tex]R \ \alpha \ \dfrac{1}{\sqrt{d}}[/tex]

[tex]R = \dfrac{k}{d}[/tex]  where K = constant

If we compare the rate o diffusion of two gases;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}[/tex]

Since the density of a gas d is proportional to its relative molecular mass M. Then;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

Rate is the reciprocal of time ; i.e

[tex]R = \dfrac{1}{t}[/tex]

Thus; replacing the value of R into the above previous equation;we have:

[tex]\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}[/tex]

We can equally say:

[tex]{\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

[tex]{\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}[/tex]

[tex]M_2 = 32 \times ( \dfrac{6.34}{4.98})^2[/tex]

[tex]M_2 = 32 \times ( 1.273092369)^2[/tex]

[tex]M_2 = 32 \times 1.62076418[/tex]

[tex]\mathbf{M_2 = 51.865 \ g/mol}[/tex]

How many minutes would be required to electroplate 25.0 grams of chromium by passing a constant current of 4.80 amperes through a solution containing CrCl3

Answers

Answer:

483.27 minutes

Explanation:

using second faradays law of electrolysis

Without doing any calculations (just compare the correct Ksp values) complete the following statements: 1. nickel(II) carbonate is MORE soluble than ___ . 2. nickel(II) carbonate is LESS soluble than ___ . A. CaSO3 B. BaSO4 C. Ni(OH)2 D. Ag2CrO4

Answers

Answer:

1. nickel(II) carbonate is MORE soluble than barium sulfate, nickel hydroxide and silver chromate because its Ksp is higher than those of such compounds, it means more ions will be dissolved.

2. nickel(II) carbonate is LESS soluble than calcium sulfite because its Ksp is lower than it of such compound, it means, less ions will be dissolved.

Explanation:

Hello,

In this case, for calcium sulfite, barium sulfate, nickel hydroxide and silver chromate Kps is 6.13x10-5, 1.53x10-9, 1.63x10-16 and 9.03x10-12 respectively. Now, since Ksp for nickel (II) carbonate is 1.43 x10-7, we can notice that:

1. nickel(II) carbonate is MORE soluble than barium sulfate, nickel hydroxide and silver chromate because its Ksp is higher than those of such compounds, it means more ions will be dissolved.

2. nickel(II) carbonate is LESS soluble than calcium sulfite because its Ksp is lower than it of such compound, it means, less ions will be dissolved.

Best regards.

given the following quantities of reactants and the balance equations; assuming an 85.0% yield, determine how many grams of potassium aluminum sulphate dodecahydrate can be produced.

Answers

Answer:

[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=32.23gKAl(SO_4)_2\dot \ 12H_2O[/tex]

Explanation:

Hello,

In this case, we balance the given equations as shown below:

[tex]Al(s)+KOH(aq)+3H_2O(l)\rightarrow KAl(OH)_4(aq)+\frac{3}{2} H_2(g)\\\\KAl(OH)_4(aq)+2H_2SO_4(aq)\rightarrow KAl(SO_4)_2(aq)+4H_2O(l)\\\\KAl(SO_4)_2(aq)+12H_2O\rightarrow KAl(SO_4)_2\dot\ 12H_2O(aq)[/tex]

Now, with 3.00 grams of aluminium, 50.00 mL of water and 10.00 mL of 8.00M potassium hydroxide, the first step is to identify the limiting reactant by firstly computing the moles of all of them:

[tex]n_{Al}=3.00 gAl*\frac{1molAl}{27gAl}=0.111molAl\\ \\n_{KOH}=0.010L*8.00mol/L=0.08molKOH\\\\n_{H_2O}=50.00mL*\frac{1g}{1mL} *\frac{1mol}{18g}=2.78molH_2O[/tex]

Thus, we can notice that 0.111 mol of aluminium will consume 0.11. moles of potassium hydroxide and 2.78 moles of water will consume 0.927 moles of potassium hydroxide, for that reason, we can infer that since there are only 0.08 moles of potassium hydroxide, it is the limiting reactant, therefore, we compute the yielded moles of potassium aluminium hydroxide in the first reaction:

[tex]n_{KAl(OH)_4}=0.08molKOH*\frac{1molKAl(OH)_4}{1molKOH} =0.08molKAl(OH)_4[/tex]

Next, we compute the yielded moles of potassium aluminium sulfate in the second reaction assuming sulfuric acid is in excess:

[tex]n_{KAl(SO_4)_2}=0.08molKAl(OH)_4*\frac{1molKAl(SO_4)_2}{1molKAl(OH)_4}=0.08molKAl(SO_4)_2[/tex]

Finally, in the third reaction, we compute the yielded grams of potassium aluminum sulphate dodecahydrate by using its molar mass and its mole ratio with potassium aluminium sulfate:

[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}=0.08molKAl(SO_4)_2*\frac{1molKAl(SO_4)_2\dot \ 12H_2O}{1molKAl(SO_4)_2} *\frac{474.00gKAl(SO_4)_2\dot \ 12H_2O}{1molKAl(SO_4)_2\dot \ 12H_2O} \\\\m_{KAl(SO_4)_2\dot \ 12H_2O}=37.92gKAl(SO_4)_2\dot \ 12H_2O[/tex]

Which is the theoretical yield, thus, by using the percent yield the actual yielded mass turns out:

[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=0.85*37.92gKAl(SO_4)_2\dot \ 12H_2O\\\\m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=32.23gKAl(SO_4)_2\dot \ 12H_2O[/tex]

Best regards.

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