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maksim [4K]
3 years ago
12

Calculate the volume of 5.00 mol of helium at 120C and 1520 mm Hg

Chemistry
1 answer:
levacccp [35]3 years ago
4 0

Answer:

80.7 L

Step-by-step explanation:

This looks like a case where we can use the Ideal Gas Law to calculate the volume.

pV = nRT       Divide both sides by p

V = (nRT)/p

=====

Data:

n = 5.00 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = (120 +273.15) K = 393.15K

p = 1520 mmHg × 1 atm/760 mmHg = 2.00 atm

=====

Calculation:

V = (5.00 × 0.082 06 × 393.15)/2.00

V = 161.3/2.00

V = 80.7 L  

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kirza4 [7]

Answer:

B. electron

Explanation:

B. electron

electron is a negatively charged particle

proton is a positively charged particle

7 0
3 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

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pKa of C6H5NH3+/ C6H5NH2 is:

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pKa = 4.6

That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

5 0
3 years ago
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Assuming all the Cr is contained in the BaCrO4 at the end. 
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8 0
3 years ago
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svetoff [14.1K]

Answer:

<em>For both cases the answer is C</em>

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We can see that the orbitals are not filled in the order of increasing energy and the Pauli exclusion principle is violated because it does not follow the correct order of the electron configuration; In the first exercise after the 2s2 orbital, the 2p2 orbital follows.

For the second exercise, you must start in order with level 1 and correctly filling each of the sublevels corresponding to each level until reaching level 7 and thus completing the desired number of electrons.

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2 years ago
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Bess [88]

Answer:

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