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spin [16.1K]
3 years ago
6

Rank the ions in order of increasing basicity: CH3NH-, CH3CH2-, CH30-

Chemistry
1 answer:
Alexxx [7]3 years ago
3 0

In order to find out the ranking of ions basicity, check the pKa values of each ions. The principle that you need to remember is that the stronger the acid the weaker the corresponding conjugate base. The pKa dictates acid value of the compound. The answer would be CH3NH, CH3O-, and CH3CH2-. 

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A solution is made using 200.0 mL of methanol (density 0.792 g/mL) and 985.1 mL of water (density 1.000 g/mL). What is the mass
DaniilM [7]

<u>Answer:</u> The mass percent of methanol in solution is 13.85 %

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

  • <u>For water:</u>

Density of water = 1.000 g/mL

Volume of water = 985.1 mL

Putting values in above equation, we get:

1.000g/mL=\frac{\text{Mass of water}}{985.1mL}\\\\\text{Mass of water}=(1.000g/mL\times 985.1mL)=985.1g

  • <u>For methanol:</u>

Density of methanol = 0.792 g/mL

Volume of methanol = 200.0 mL

Putting values in above equation, we get:

0.792g/mL=\frac{\text{Mass of methanol}}{200.0mL}\\\\\text{Mass of methanol}=(0.792g/mL\times 200.0mL)=158.4g

To calculate the mass percent of methanol, we use the equation:

\text{Mass percent of methanol}=\frac{\text{Mass of methanol}}{\text{Mass of solution}}\times 100

Mass of solution = [158.4 + 985.1] g = 1143.5 g

Mass of methanol = 158.4 g

Putting values in above equation, we get:

\text{Mass percent of methanol}=\frac{158.4g}{1143.5g}\times 100=13.85\%

Hence, the mass percent of methanol in solution is 13.85 %

3 0
3 years ago
Calcium phosphate reacts with hydrochloride acid to produce calcium chloride and phosphoric acid. This is the balanced chemical
Mashcka [7]

Answer: 0.664 moles and 65.1 grams EDMENTUM USERS

Explanation: From the equation, we know that 1 mole of calcium phosphate produces 2 moles of phosphoric acid (H3PO4). So, if we know how many moles of calcium phosphate are present in 103 grams of Ca3(PO4)2, we can find the corresponding number of moles of H3PO4 that are produced during the reaction.

6 0
3 years ago
In a chemical formula, what is used to indicate the number of atoms of each element?
MrRa [10]
The coefficients next to the symbols of entities indicate the number of moles of a substance produced or used in the chemical reaction.
6 0
3 years ago
Question 6
kompoz [17]

Considering the combined law equation, the new temperature is -244.56 °C or 28.44 K.

<h3>Boyle's law</h3>

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant: if the pressure increases, the volume decreases while if the pressure decreases, the volume increases.

Mathematically, this law states that the multiplication of pressure by volume is constant:

P×V=k

<h3>Charles's law</h3>

Charles's law states that the volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles' law states that the ratio of volume to temperature is constant:

\frac{V}{T} =k

<h3>Gay-Lussac's law </h3>

Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature: increasing the temperature will increase the pressure, while decreasing the temperature will decrease the pressure.

Mathematically, Gay-Lussac's law states that the ratio of pressure to temperature is constant:

\frac{P}{T} =k

<h3>Combined law equation</h3>

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

\frac{PxV}{T} =k

Considering an initial state 1 and a final state 2, it is fulfilled:

\frac{P1xV1}{T1} =\frac{P2xV2}{T2}

<h3>New temperature</h3>

In this case, you know:

  • P1= 1 atm= 760 mmHg
  • V1= 15 L
  • T1= -30 °C= 243 K (being 0 °C= 273 K)
  • P2= 58 mmHg
  • V2= 23 L
  • T2= ?

Replacing in the combined law equation:

\frac{760 mmHgx15 L}{243 K} =\frac{58 mmHgx23 L}{T2}

Solving:

T2x\frac{760 mmHgx15 L}{243 K} =58 mmHgx23 L

T2 =\frac{58 mmHgx23 L}{\frac{760 mmHgx15 L}{243 K}}

<u><em>T2= 28.44 K= -244.56 °C</em></u>

Finally, the new temperature is -244.56 °C or 28.44 K.

Learn more about combined law equation:

brainly.com/question/4147359

#SPJ1

6 0
2 years ago
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Determine the init
Jlenok [28]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

                          H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

0.11 - x = 0.052

x = 0.11 - 0.052

x = 0.058 M

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0
3 years ago
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