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jeyben [28]
3 years ago
11

Which description of nitrogen-15 is correct?

Chemistry
1 answer:
VashaNatasha [74]3 years ago
8 0
The first one is correct cause first of all it's atomic number is 7 and second of all it's atomic mass is about 15
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Which of the following is a mixture?
KATRIN_1 [288]
C is a mixture!!!!!!!!!!!!!

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Which of the following interactions would be the strongest?
shepuryov [24]

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What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe
Deffense [45]

Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]

<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒  [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

7 0
3 years ago
If you gently shake a carbon dioxide fire extinguisher, you will feel the presence of liquid within the extinguisher What condit
scoundrel [369]

When we wish to convert a gas to liquid we have to either

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b) increase pressure

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the ideal pressure and temperature conditions when CO2 gas can be converted to CO2 gas

Pressure = 5 - 73 atm

Temperature = -57  to 31 degree Celsius


6 0
3 years ago
Radium-221 has a half-life of 30 sec. How long will it take for 96% of a sample to decay?
monitta
We can calculate how long the decay by using the half-life equation. It is expressed as:

A = Ao e^-kt

<span>where A is the amount left at t years, Ao is the initial concentration, and k is a constant.
</span><span>From the half-life data, we can calculate for k.
</span>
1/2(Ao) = Ao e^-k(30)
<span>k = 0.023
</span>
0.04Ao = Ao e^0.023(t)
<span>t = 140 sec</span>
3 0
3 years ago
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