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ahrayia [7]
3 years ago
13

You mix 5.25 moles of A with 7 moles of B in the chemical equation 3A + 4B 2C + 3D. Which statement is true about the reactants?

B has a greater molar mass than
a. A has a greater molar mass than
b. B is the limiting reactant.
c. A is the limiting reactant.
d. A and B are in ideal stoichiometric proportion.
Chemistry
2 answers:
MaRussiya [10]3 years ago
7 0
The answer is d. A and B are in ideal stoichiometric proportion. In a balanced reaction equation, the ratio of mole number of the reactants is equal to the ratio of the coefficients. 5.25:7=3:4.
eduard3 years ago
5 0

E for Plato users

a and b are in ideal stoiciometric proportion

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Write the following numbers in standard notation, maintaining the same number of significant figures.
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Answer:

6.104x10^2=610.4

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Explanation:

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3 years ago
Consider these elements: N, Mg, O, F, Al. a. Write the electron configuration for each element. b. Arrange the elements in order
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Answer:

N- 1s2 2s2 2p3

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Explanation:

Order of decreasing atomic radius

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Order of increasing ionization energy

Mg,Al, N,O,F

Reason:

Atomic radius decreases with increase in nonmetallic character. Looking at the electronic configurations, as effective nuclear charge increases, the atom becomes smaller and the attractive force between the nucleus and the outermost electrons increases. Hence, the radius of the atom decreases and ionization energy increases. Note that the addition of more orbital electrons implies addition of more nuclear charge since the both must exactly balance for the atom to remain electrically neutral. The more the electrons in the outermost shell, the higher the first ionization energy.

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Consider the conversion of 2-naphthol and 1-bromobutane into an ether via the williamson ether synthesis. list the procedural st
Lemur [1.5K]

Answer:

Following are the solution to this question:

Explanation:

Please find the complete question in the attachment.

Start of Laboratory

Dissolve 2-naphthol in the round bottom flask with ethanol.

Add pellets of sodium hydroxide and hot chips. Attach a condenser.

Heat for 20 minutes under reflux, until the put a burden dissolves.

After an additional hour, add 1-Bromobutane and reflux.

Pour the contents into a beaker with ice from a round bottom flask.

On a Bachner funnel, absorb the supernatant by vacuum filtration.

Utilizing cold water to rinse the material and dry that on the filter.

Ending of the Lab

6 0
3 years ago
The half-life of radium-226 is 1590 years. if a sample contains 100 mg, how many mg will remain after 1000 years?
Katyanochek1 [597]

Answer:

a=64.7mg

Explanation:

Hello,

In this case, we need to remember that for the required time for a radioactive nuclide as radium-226 to decrease to one half its initial amount we are talking about its half-life. Furthermore, the amount of remaining radioactive material as a function of the half-lives is computed as follows:

a=a_0(\frac{1}{2} )^{\frac{t}{t_{1/2}} }

Therefore, for an initial amount of 100 mg with a half-life of 1590 years, after 1000 years, we have:

a=100mg(\frac{1}{2} )^{\frac{1000years}{1590years} }\\\\a=64.7mg

Best regards.

4 0
3 years ago
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