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defon
2 years ago
9

How many moles of al does 13.49 g of al represent?

Chemistry
1 answer:
Harlamova29_29 [7]2 years ago
7 0
Use equation
number of moles= sample mass/molar mass
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What gives the gem amethyst its purplish color?
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iron Presence of trace elements, irradiation and iron impurities give the gem amethyst its purplish color!
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2 years ago
Which of the changes gas to solid gas to liquid, liquid to solid liquid to gas,solid to liquid solid to gas will involve an outp
Verdich [7]

Answer:

Liquid to solid liquid to gas

Explanation:

Boiling a liquid result in a gas

7 0
2 years ago
Which of the following is the quantum number set for an electron in the 3rd energy level, dumb-bell shaped orbital, on the z-axi
Juli2301 [7.4K]

Answer:

Explanation:

Principal quantum no "n" = 3

Azimuthal quantum no "l"= 1

Magnetic quantum no "m"= +1/2

Over all is 3pz

5 0
2 years ago
Which of the following would have the largest pKa?
garri49 [273]

Answer:

CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

Explanation:

To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:

pKa of  CH3CH2NH3+ =  CH3CH2NH2;  C6H5NH3+ =  C6H5NH2

Also, Kw / Kb = Ka

Thus:

pKa of CH3CH2NH3+/CH3CH2NH2 is:

Kw / kb = Ka = 1.79x10⁻¹¹

-log Ka = pKa

pKa = 10.75

pKa of C6H5NH3+/ C6H5NH2 is:

Kw / kb = Ka = 2.5x10⁻⁵

-log Ka = pKa

pKa = 4.6

That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

5 0
3 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
2 years ago
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