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Lisa studied the models of three carbon nuclei shown below.

valkas [14]

Answer:

The answer to your question is: letter B

Explanation:

The information given is showing us three kinds of carbons and their respective number of protons and neutrons. Now, let's analize the options given:

A.)Carbon Bond Types This option is about trhe kind of bonds carbon is able to form like single, double or triple and is different from the information descrive above. So ,this option is incorrect.

B.)Carbon Isotopes is option is about the carbon atoms that have the same number of protons and different amount of neutrons exactly as it is describe above, to this option is correct.

C.)The Carbon Cycle this option is about how carbon goes from one kind of matter to another and is completely different to the description given above, so this option is incorrect.

D.)Chemical Reactions this option is about reactants that after a process give products they could be from carbon -12, carbon-13 or carbon-14 and other elements, so this option is different from the description given above.

6 0

4 years ago

What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa? Ka = 1.8×10-5

Westkost [7]

Answer:

a) We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

b) We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

c) We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

Explanation:

a) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = pK = -log(1.8*10^-5) = 4.74

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4.74 = 4.74 + log(A-/HA)

0 = log(A-/HA)

A-/HA = 1

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1

X =0.9

We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4.74 = 4.74 + log(0.9/0.9) = 4.74

b) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 4

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4 = 4.74 + log(A-/HA)

-0.74 = log(A-/HA)

A-/HA = 0.182

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 0.182

X =0.277

We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4 = 4.74 + log(0.277/1.523)

c) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 5

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

5 = 4.74 + log(A-/HA)

0.26 = log(A-/HA)

A-/HA = 1.82

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1.82

X =1.16

We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

5 = 4.74 + log(1.16/0.64) = 5

3 0

4 years ago

What is the sum of the coefficients needed to balance the combustion reaction of methane?

pishuonlain [190]

Answer: The sum of the coefficients needed to balance the combustion reaction of methane is 6.

Explanation:

Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The chemical equation for combustion of methane will be :

$CH_4+2O_2\rightarrow CO_2+2H_2O$

The coefficient foer methane is 1 , 2 for oxygen , 1 for carbon dioxide and 2 for water. The sum is 6.

4 0

3 years ago

is the general formula of a certain hydrate. When 256.3 g of the compound is heated to drive off the water, 214.2 g of anhydrous

Marizza181 [45]

Answer:

The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.

Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams

The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,

Moles of water = weight of water/molecular weight

= 42.1 grams / 18 = 2.3

The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,

= 214.2 * 0.2190 = 46.91 grams

The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,

= 214.2 * 0.4314 = 92.40 grams

The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,

= 214.2 * 0.3497 = 74.91 grams

The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17

The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,

92.40/78.96 = 1.17

The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,

74.91/15.999 = 4.68.

Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O

6 0

4 years ago

How many moles of H2O are needed to produce 6.3 moles of H2

Goryan [66]

Answer:

H2 to O2

H2 to H2O

H20 to O2

H2O to H2

O2 to H20

O2 to H2

Explanation:

6 0

3 years ago