What is the molarity of a 27.0% (v/v) aqueous ethanol solution? the density of ethanol (c2h6o, molar mass 46.07 g/mol) is 0.790 g/ml?

Answers

Answer 1
Answer:

Answer:

M=4.63M

Explanation:

Hello,

In this case, with the given by-volume percentage and considering the molarity as:

M=(n_(Ethanol))/(V_(solution))

We assume the solution having 100 mL of volume in total, thus, the volume of ethanol is 27.0 mL, therefore, the moles:

n_(Ethanol)=27.0mL*(0.790g)/(1mL)*(1mol)/(46.07g)=0.463molEthanol

Moreover, the volume of the solution in liters is:

V_(solution)=100mL*(1L)/(1000mL) =0.1L

Finally, the molarity is:

M=(0.463mol)/(0.1L)\n M=4.63M

Best regards.

Answer 2
Answer:

If the density of ethanol (c2h6o, molar mass 46.07 g/mol) is 0.790 g/ml, the molarity of the 27.0% (v/v) aqueous ethanol solution is 17.14 M.

Calculate the moles of ethanol contained in the solution, then divide that number by the volume of the solution in litres to determine the molarity of the ethanol solution.

To start, we must ascertain how much ethanol is included in each 100 millilitres of the solution. Given that ethanol has a density of 0.79 g/ml, the amount of ethanol in 100 millilitres is as follows:

Mass of ethanol = density × volume

Mass of ethanol = 0.790 g/ml × 100 ml = 79 g

Now,

Moles of ethanol = mass / molar mass

Moles of ethanol = 79 g / 46.07 g/mol = 1.714 mol

So,

Volume of solution = 100 ml / 1000 ml/L = 0.1 L

We know that:

Molarity = moles of solute / volume of solution

Molarity = 1.714 mol / 0.1 L = 17.14 M

Thus, the molarity is 17.14 M.

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Weathering and erosion can turn metamorphic rock into sediment. Which of the following processes would turn that sediment into sedimentary rock? * 10 points weathering and erosion compaction and cementation heat and pressure melting

Answers

Answer:

compaction and cementation

Explanation:

sedimentary rock is layered rock, in order to become sedimentary it most be compacted to form layers and cemented to become hard. This is what i think. Hope i can help :)

A biochemist studying breakdown of the insecticide DDT finds that it decomposes by a first-order reaction with a half-life of 12.0 yr. How long does it take DDT in a soil sample to decompose from 809 ppbm to 10.0 ppbm (parts per billion by mass)

Answers

Answer:

76.0 years

Explanation:

Step 1: Given data

  • Half-life (t1/2): 12.0 y
  • Initial concentration of DDT ([A]₀): 809 ppbm
  • Final concentration of DDT ([A]): 10.0 ppbm
  • Time elapsed (t): ?

Step 2: Calculate the rate constant (k)

We will use the following expression.

k = ln 2/ t1/2 = ln 2 / 12.0 y = 0.0578 y⁻¹

Step 3: Calculate the time elapsed

For a first-reaction order, we will use the following equation.

ln [A]/[A]₀ = -k × t

t = ln [A]/[A]₀ / (-k)

t = (ln 10.0 ppbm/809 ppbm) / (-0.0578 y⁻¹)

t = 76.0 y

PromptDescribe the general process by which new elements are synthesized.


Please help!!!!

Answers

New elements are synthesized from the nuclear reaction of already existing elements. The change in nuclear composition of an atom leads to the conversion to new element.

What are elements ?

Elements are the basic things of anything in the world. Elements join together to form molecule and compounds. There are 118 known elements in periodic table.

Many more elements  discovered by scientists are accidental. The properties of each elements are determined based on their atomic number. The nuclear reactions mostly for radioactive elements make new elements with different nuclear composition.

New elements are produced as sub products that are resulted from several synthesis reactions involving the nuclear change. Radioactivity involving bombardment of charged particles such as alpha, beta or their emission also leads to formation of new elements.

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Answer:

Describing Describe the general process by which new elements are synthesized.

The combustion of ethane (C2H6) produces carbon dioxide and steam. 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g) How many moles of CO2 are produced when 5.90 mol of ethane is burned in an excess of oxygen?

Answers

Final answer:

The combustion of ethane yields carbon dioxide, and with 5.90 moles of ethane being reacted, it results in the production of 11.8 moles of CO2.

Explanation:

The question pertains to the concept of stoichiometry in chemistry, and the chemical reaction in question is a combustion reaction involving ethane (C2H6). From the balanced reaction, it is evident that 2 moles of ethane (C2H6) produce 4 moles of carbon dioxide (CO2). Therefore, if we have 5.90 moles of ethane reacting, it's a straightforward calculation to determine that this would yield twice that many moles of CO2. We simply multiply the moles of ethane by the stoichiometric ratio (4/2) to get the moles of CO2.

Example Calculation: 5.90 moles of ethane x (4 moles CO2 / 2 moles C2H6) = 11.8 moles CO2

So, when 5.90 moles of ethane are burned in an excess of oxygen, 11.8 moles of CO2 are produced.

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Final answer:

In the combustion of ethane, for every mole of ethane burned, two moles of carbon dioxide are produced. Hence, when 5.90 moles of ethane are burned, 11.8 moles of carbon dioxide are produced.

Explanation:

The chemical reaction given, 2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g), states that 2 moles of ethane (C2H6) produce 4 moles of carbon dioxide (CO2). Thus, the mole-to-mole ratio of ethane to carbon dioxide is 2:4, or simplified, 1:2. So, for every mole of ethane burned, two moles of carbon dioxide are produced.

Given that 5.90 moles of ethane are burned, we can calculate the quantity of carbon dioxide produced by multiplying 5.90 moles by 2. Hence, when 5.90 moles of ethane are burned in an excess of oxygen, 11.8 moles of carbon dioxide are produced.

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Use the weather map to answer the question.Weather map with blue line with triangles pointing down and to the right. Red line with semi-circles pointing up and to the right. Red line is to the right of the blue line.
© Dorling Kindersley / Universal Images Group / Image Quest 2016

What does the red line indicate?

A. A cold front is moving to the north and east.
B. A cold front is moving to the south and west.
C. A warm front is moving to the north and east.
D. A warm front is moving to the south and west.

Answers

I think the answers to this is d

Answer: It's D. A warm front is moving to the south and west.

Explanation: I got it right on the test

A chemical equation is balanced when thesubscripts of the reactants equal the subscripts of the products,
coefficients of the reactants equal the coefficients of the products,
products and reactants are the same chemicals,
same number of each kind of atom appears in the reactants and in the products,

Answers

same number of each kind of atom appears in the reactants and in the products