What volume of water must be added

Which of the following two conditions will favor a spontaneous reaction?

Alexxx [7]

Answer:

option A

an increase in entropy and a decrease in enthalpy

pls mark brainliest

8 0

2 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

2 years ago

Read 2 more answers

A propane stove burned 470 grams propane and produced 625 grams of water (this is the actual yield) C3H8 +5O2=3CO2+4H20. What wa

Liula [17]

Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

4 0

3 years ago

Urea, CO(NH2)2, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is th

Pachacha [2.7K]

Answer:

5.004kg

Explanation:

Combustion of carbon

C+O2=CO2

from the relationship of molar ratio

mass of carbon/molar mass of carbon=volume of CO2 produced\molar vol(22.4 dm3)

mass of carbon =1000kg

atomic mass of carbon =12

volume of CO2 produced=1000×22.4/12

volume of CO2 produced =1866.6dm3

from the combustion reaction equation provided

CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)

applying the same relationship of molar ratio

no of mole of CO2=no of mole of urea

therefore

vol of CO2\22.4=mass of urea/molar mass of urea

molar mass of urea=60.06g/mol

from the first calculation

vol of CO2=1866.6dm3

mass of urea=1866.6×60.06/22.4

mass of urea=5004.82kg

7 0

3 years ago

Read 2 more answers

. What is a carbonyl group? Draw the carbonyl groups that are characteristic of aldehydes, ketones, carboxylic acids, and esters

Elan Coil [88]

Answer :

Carbonyl group : It is a functional group composed of a carbon atom that double bonded to oxygen atom. It is represented as $C=O$

Carboxylic group : It is the class of organic compound in which the carboxylic (-COOH) group is attached to a hydrocarbon is known as carboxylic.

The general formula of carboxylic is, $C_nH_{2n+1}COOH$ . According to the IUPAC naming, the carboxylic are named as alkanoic acids.

Aldehyde group : It is the class of organic compound in which the (-CHO) group is attached to a hydrocarbon is known as aldehyde.

The general representation of aldehyde is, $R-CHO$ . According to the IUPAC naming, the aldehyde are named as alkanals.

Ketone group : It is the class of organic compound in which the (-CO) group is directly attached to the two alkyl group of carbon is known as ketone.

The general representation of ketone is, $R-CO-R^'$ . According to the IUPAC naming, the ketone are named as alkanone.

Ester group : It is the class of organic compound in which the (-COO) group is directly attached to the two alkyl group of carbon is known as ester.

The general representation of ester is, $R-COO-R^'$ . According to the IUPAC naming, the ester are named as alkyl alkanoate.

6 0

2 years ago