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trasher [3.6K]
3 years ago
6

What is the electron configuration for a neutral atom in the ground state with a total of 6 valence electrons?

Chemistry
1 answer:
saveliy_v [14]3 years ago
8 0
Ns²np⁴

2+4=6
-------------------------
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andrezito [222]
Ndjsjdjdosixisoosxiosodjx xjsxhjsjxkaosmslwkskwksksowkdbxjsixsnnsjxosixis
5 0
3 years ago
Read 2 more answers
The osmotic pressure of a saturated solution of strontium sulfate at 25 ∘C∘C is 21 torrtorr. Part A What is the solubility produ
USPshnik [31]

Answer

solubility product = 3.18x 10^-7

Explanation:

We were given the pressure in torr then we need to convert to atm for consistency, ten we have

21torr/760= 0.0276315789 atm

21 Torr = .0276315789 atm

P = i M S T

M = P / iRT

Where p is osmotic pressure

T= temperature= 25C+ 273= 298K

for XY vanthoff factor i = 2

S = 0.0821 L-atm / mol K

M = .0276315789 atm / (2)(0.0821 L atm / K mole)(298 K)

M = 0.000564698046 mol/liters

solubility= 0.000564698046 mol/liters

Ksp = [X+][Y-]

Ksp = X^2

Ksp = [Sr^+2] * [SO4^-2]

Ksp = X^2

Ksp = (0.000564698046)^2

Ksp = 3.18883883 × 10-7

Ksp = 3.18x 10^-7

solubility product = 3.18x 10^-7

Therefore, the solubility product of this salt at 25 ∘C∘C is 3.18x 10^-7

7 0
3 years ago
Balance the following equation in acidic conditions. Phases are optional. Cu NO3,
Gwar [14]
If your equation is <span>Cu+NO^- 3-->Cu^2+NO, then the answer is 
</span><span>2 Cu + 1 NO3{-}  → 1 Cu^{2+} + 3 NO 
</span>
To check if it is balance, this is the solution:
2- Cu- 2
3- N -3
3- O -3 

5 0
3 years ago
Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.
zepelin [54]

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

3 0
3 years ago
107.854 how many significant
sattari [20]

Answer:

3

Explanation:

look after DP theres 3 digits

6 0
3 years ago
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