What are the two formulas for the ions that's make up AgC2H3O2 bromide?

Answers

Answer 1

Answer:

Answer:

KCI+AgC2H3O2

Explanation:

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Isopropyl methyl ether is slightly soluble with water because the oxygen atom of ethers with three or fewer carbon atoms can form a few hydrogen bonds with water.(A) True(B) False
WORTH A LOT OF POINTS! just copy what on the picture for notes so i can copy and paste i do not feel like writing all of that down
How many electrons are in Fe3+ ?

At 25.0 ⁰C the henry's law constant for hydrogen sulfide(H2S) gas in water is 0.087 M/atm. Caculate the mass in grams of H2S gas that can be dissolved in 400.0 ml of water at 25.00 C and a H2S partial pressure of 2.42atm.

Answers

Answer: The mass of hydrogen sulfide that can be dissolved is 2.86 grams.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the gas.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_(H_2S)=K_H* p_(liquid)$

where,

$K_H$ = Henry's constant = $0.087M/atm$

$p_(H_2S)$ = partial pressure of hydrogen sulfide gas = 2.42 atm

Putting values in above equation, we get:

$C_(H_2S)=0.087M/atm* 2.42atm\n\nC_(H_2S)=0.2105M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.2105 M

Molar mass of hydrogen sulfide = 34 g/mol

Volume of solution = 400.0 mL

Putting values in above equation, we get:

$0.2105M=\frac{\text{Mass of hydrogen sulfide}* 1000}{34g/mol* 400.0mL}\n\n\text{Mass of }H_2S=(0.2105* 34* 400)/(1000)=2.86g$

Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.

a chemist encounters an unknown metal. They drop the metal into a graduated cylinder containing water, and find the volume change is 3.2 mL. If the metal weighs 1.5g, what is the density in g/mL of the metal?

Answers

The density of metal will be "0.469 g/mL".

The given values in the question are:

Mass of metal = 1.5 g
Volume of metal or Volume change = 3.2 mL

Now,

The density of metal will be:

→ $Density = (Mass)/(Volume)$

By substituting the given values, we get

→ $= (1.5)/(3.2)$

→ $=0.469 \ g/mL$

Thus the above is the appropriate answer.

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The density of the metal is 0.47 g/mL

The density of an object is defined as the mass of the object per unit volume of the object.

Density = mass / volume

With the above formula, we can obtain the density of the metal. This is illustrated below:

Volume of metal = change in volume of water = 3.2 mL

Mass of metal = 1.5 g

Density of metal =?

Density = mass / volume

Density = 1.5 / 3.2

Density of metal = 0.47 g/mL

Therefore, the density of the metal is 0.47 g/mL

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Adding charts and graphs helps a scientistO To state the problem
O To determine trends
O To simplify results
O Both B and C
O All of the above

Answers

Answer:

b and c

Explanation:

the problem was solved through the experiment and tested

Adding charts and graphs helps a scientist to determine trends and to simplify results

Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate gives, after protonolysis, a good yield of a mixture containing mostly a dextrorotatory ketone A and a trace of an optically inactive isomer B. Treatment of A with zinc amalgam and HCl affords an optically active, dextrorotatory hydrocarbon C. Identify compounds A, B, and C, including stereochemical configurations. Be sure to explicitly draw H and both wedge/dash bonds at any configuration center.

Answers

Answer:

use google and use the first link

Explanation:

Final answer:

Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate followed by protonolysis produces dextrorotatory ketone A and a trace of isomer B. Compound A can yield optically active, dextrorotatory hydrocarbon C when treated with zinc amalgam and HCl.

Explanation:

Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate followed by protonolysis yields a mixture containing primarily a dextrorotatory ketone A and a trace of an optically inactive isomer B. Compound A can be treated with zinc amalgam and HCl to produce an optically active, dextrorotatory hydrocarbon C.

To provide a specific identification and stereochemical configuration of compounds A, B, and C, I would need additional information or further context about the starting components and reaction conditions, as well as any other relevant data or observations.

Learn more about Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate here:

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The quantity of mass of an object contained within its volume is a measure of

Answers

the answer is density

the quantity of mass of an object contained within its volume is a measure of density

Which statements correctly describe the decay rates of radioactive isotopes? a} It takes two half-lives for a sample to fully decay.

b} The exact time when an individual atom will decay can be accurately predicted.

c} After each half-life, the amount of radioactive material is reduced by half.

d) All radioactive isotopes have the same half-life.

e} The decay of individual atoms in a sample of radioactive material is random.

Answers

Answer: b} The exact time when an individual atom will decay can be accurately predicted.

c} After each half-life, the amount of radioactive material is reduced by half.

Explanation:

All radioactive decay follows first order kinetics.

Rate law expression for first order kinetics is given by:

$t=(2.303)/(k)\log(a)/(a-x)$

where,

k = rate constant

t = time taken for decay process

a = initial amount of the reactant

a - x = amount left after decay process

Expression for calculating half life, which is the time taken by the half of the reactants to decompose is:

$t_(1/2)=(0.693)/(k)$

Answer:

For plato, the answer is C: after each half tile, the amount of radioactive material is reduced by half

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15. List the substances A-E in order from most dense to least dense based on the facts provided.Please write the letters in the spaces provided below.Substance A: 8.2 g/cm3Substance B: 3.5 cm and 30.0gSubstance C: 10.0g and 40mLSubstance D: 0.5 g/cm3Substance E: 2.0cm by 3.0cm by 1.0cm and 4.0gMost Dense_ _ _ _ _Least Dense

wo reactions and their equilibrium constants are given. A + 2 B − ⇀ ↽ − 2 C K 1 = 2.57 2 C − ⇀ ↽ − D K 2 = 0.226 A+2B↽−−⇀2CK1=2.572C↽−−⇀DK2=0.226 Calculate the value of the equilibrium constant for the reaction D − ⇀ ↽ − A + 2 B .