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3 years ago

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A nuclide of Superscript 64 subscript 29 upper C u. absorbs a positron. Which is the resulting atom? Superscript 65 subscript 29

upper C u. Superscript 63 subscript 29 upper C u. Superscript 64 subscript 28 upper N i. Superscript 64 subscript 30 upper Z n.

1 answer:

vichka [17]3 years ago

6 0

Answer:

option) D

Explanation:

i just took the test

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Which of the following statements are TRUE of buffer solutions? 1. A buffer solution can be made by mixing equal concentrations

Leokris [45]

Answer:

Option 1 and 3 are correct.

Explanation:

Buffer solution is the solution which resists the change in the magnitude of the pH when small additions of either acid or base is added. Buffer solutions consist of weak acid and its conjugate base usually which are mixed in relatively equal and large quantities.

Hence,

Option 1. is correct because acetic acid is a weak acid and sodium acetate is the conjugate base of it which are mixed in equal concentrations.

Option 2. is incorrect because HCl and NaCl are both strong acid and salt respectively and will not form a buffer.

Option 3. is correct which is the definition of a buffer solution.

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Besides drinking water, how is water helpful for life and nature?

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2 years ago

What is formed when 2 non metals reacts?

ryzh [129]

Answer:

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Explanation:

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3 years ago

Please help with #2 and #3

Vaselesa [24]

Answer:

2. $V_2=17L$

3. $V=82.9L$

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

$\frac{V_2}{T_2}=\frac{V_1}{T_1}$

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

$V_2=\frac{V_1T_2}{T_1}$

Therefore, we plug in the given data to obtain:

$V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L$

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

$V=\frac{nRT}{P}$

Therefore, we plug in the data to obtain:

$V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L$

Best regards!

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3 years ago