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lina2011 [118]
4 years ago
13

Give four ions that have the same electron configuration as neon.

Chemistry
1 answer:
wariber [46]4 years ago
6 0
N3-
O2-
Na+
Mg2+


I hope this helps text me if you want the full names of them:)
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Calculate the number of formula units in 1.87 mol of NH4Cl
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3 years ago
Given the balanced equation 2C+ 3H2
mario62 [17]

Answer:

Explanation:

The answer is (4) 4.0 mol. This is a stoichiometry problem. You start with 2.0 mol of C2H6 and obtain the moles of C by multiplying 2.0 by the mole ratio, in this case 2. 2.0*2=4.0mol.

4 0
3 years ago
Calculate the molar solubility of CuX (Ksp=1.27×10−36) in each of the following. pure water
Natasha2012 [34]
<span>Answer: CuX = Cu2+ + X2- Ksp = [Cu2+] * [X2-] for each mole of CuX that dissolves we get x mol of each of the anions and cations Ksp = x^2 = 1.27 x 10 ^-36 x= 1.13 x 10 ^-18 moles of CuX per liter of pure water if the solution has [Cu2+]= 0.27 M Ksp becomes x ( x + 0.27) as we can see above x is extremely small so can be ignored inside the brackets 0.27 x = 1.27 x 10^-36 x = 1.27 x 10^-36 / 0.27 = 4.70 x 10 ^-36 moles per liter In 0.19M X2- we have Ksp = 0.19x = 1.27 x 10^-36 x = 1.27 x 10^-36 / 0.19 = 6.68 x 10 ^-36 moles per liter</span>
8 0
3 years ago
Read 2 more answers
What percentage of incoming college students did not consume any alcohol?
Katyanochek1 [597]

Answer:

Approximately 38%

What percentage of incoming students report not drinking at all in the past year? Approximately 38% of incoming college students reported not drinking at all in the past year.

Explanation:

4 0
2 years ago
Suppose that 2.5 mmol n2 (g) occupies 42 cm3 at 300 k and expands isothermally to 600 cm3. calculate δgm in kj/mol for the proc
andriy [413]

<u>Given:</u>

MilliMoles of N2 = 2.5 mmol = 0.0025 moles

Initial volume V1 = 42 cm3

Final volume V2 = 600 cm3

Temperature T = 300 K

<u>To determine:</u>

The change in Gibbs free energy, ΔG

<u>Explanation:</u>

The change in Gibbs free energy is related to the ration of the volumes:

ΔG = -nRTln(V2)/(V1)

      = -0.0025*8.314*300*ln(600/42) = -16.582 J/mol

Ans: The Gibbs free energy for the process is 0.0166 kJ/mol

     


8 0
3 years ago
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