When calcium carbonate reacts with hydrogen chloride, if the reaction occurs with 78.5% yield, what mass of carbon dioxide will be collected if 19.88g of CaCO3 is added to sufficient hydrogen chloride?

Answers

Answer 1

Answer: 97.95 is the answer to the question

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Draw 1,2,3,4,5,6-hexachlorocyclohexane with :a. all the chloro groups in axial positions.
b. all the chloro groups in equatorial positions.

Answers

Answer:

This is required answer.

Explanation:

Given that,

1,2,3,4,5,6-hexachlorocyclohexane

(a). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions

Using given data

We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions.

When we say that all the chloro groups in axial position that means axial bonds are vertical.

(b). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions

Using given data

We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions.

When we say that all the chloro groups in equatorial position that means axial bonds are horizontal.

Hence, This is required answer.

What reaction conditions most effectively conver a cabocxylic acid to a methly ester?

Answers

Answer:

Esterification reaction

Explanation:

When we have to go from an acid to an ester we can use the esterification reaction. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).

In this case, we need the methyl ester, therefore we have to choose the appropriate alcohol, so we have to use the methanol as reactive if we have to produce the methyl ester.

Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equations: C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = -1,123 kJ C(s) + O2(g) ---> CO2(g) ΔH° = -340 kJ H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -211 kJ

Answers

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

$C_2H_2(g) + (5)/(2)O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_(1) = -1,123 kJ$ ...[1]

$C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_(2) = -340 kJ$ ..[2]

$H_2(g) + (1)/(2)O_2(g)\rightarrow H_2O(l) ,\Delta H^o_(3) = -211 kJ$ ..[3]

$2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_(4) =?$ ..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

$\Delta H^o_(4)=2* \Delta H^o_(2)+\Delta H^o_(3) - \Delta H^o_(1)$

$\Delta H^o_(4)=2* (-340 kJ) + (-211 kJ) - (-1,123 kJ)$

$\Delta H^o_(4)=232 kJ$

The enthalpy for given reaction is 232 kilo Joules.

What is the molarity of a solution prepared from 25.0 grams of methanol (CH3OH, density = 0.792 g/mL) with 100.0 milliliters of ethanol (CH3CH2OH)? Assume the volumes are additive.

Answers

Final answer:

The molarity of a solution prepared from 25.0 grams of methanol and 100.0 milliliters of ethanol is approximately 7.80 M.

Explanation:

This is a question about calculating molarity, which is a measure of concentration using moles per liter. To calculate the molarity of a methanol in ethanol, we first have to convert the mass of methanol into moles. The molar mass of methanol (CH3OH) is about 32.04 g/mol. Therefore, 25.0 g of methanol equals about 0.780 moles (25.0 g ÷ 32.04 g/mol).

Next, the volume of ethanol needs to be converted from milliliters to liters. Thus, 100.0 mL becomes 0.100 L. Finally, the molarity is calculated by dividing the moles of methanol by the volume of the ethanol in liters, resulting in a molarity of approximately 7.80 M (0.780 moles ÷ 0.100 L).

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Ne ( g ) effuses at a rate that is ______ times that of Cl 2 ( g ) under the same conditions.

Answers

Answer: 1.88times as that of Cl2

Explanation:

According to Graham law of effusion , the rate of effusion is inversely proportional to the square root of the molar mass

Rate= 1/√M

R1/R2 =√M2/M1

Let the rate of diffusion of Ne= R1

And rate of diffusion of Cl2 = R2

M1 ,molar mass of Ne= 20g/mol

M2,molar mass of Cl2 =71g/mol

R1/R2 = √ (71/20)

R1/R2 = 1.88

R1= 1.88R2

Therefore the Ne effuses at rate that is 1.88times than that of Cl2 at the same condition.

A student dissolves 12.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.01/gmL . The student notices that the volume of the solvent does not change when the sucrose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits. molarity = molality =

Answers

Answer:

Molarity → 0.17 M

Molality → 0.11 m

Explanation:

The student notices that the volume of the solvent does not change when the sucrose dissolves in it; therefore we assume the volume of solvent as solution.

Molarity = Mol of solute/L

Let's calculate the mol of solute (mass / molar mass)

12 g / 342 g/mol = 0.0351 moles

Let's conver the volume (mL) to L

300 mL / 1000 = 0.3 L

Molarity (mol/L) = 0.0351 mol / 0.3L → 0.17 M

Molality = mol of solute / 1kg of solvent.

Let's find out the mass of solvent with the density

Solvent density = Solvent mass / Solvent volume

1.01 g/mL = Solvent mass / 300 mL

1.01 g/mL . 300 mL = Solvent mass →303 g

Let's convert the mass to kg