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Anestetic [448]
3 years ago
6

[Rn] 7s25f6d4 Which element is denoted

Chemistry
1 answer:
Colt1911 [192]3 years ago
4 0

Answer:

Explanation:

[Rn] 7s25f6d4

*-*

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2. What physical force causes suspensions to settle
MakcuM [25]

Hello!

Answer:

Gravity

Explanation:

It can settle down and separate over time due to gravity.

Hope this helps! Have a great day!

8 0
2 years ago
Does beryllium (Be) or Sodium (Na) have the same electron arrangement as Magnesium (Mg)? Why?
Finger [1]

Answer:

Na has the most similar configuration.

Explanation:

Na electron configuration: 1s²2s²2p⁶3s¹ or [Ne] 3s₁

Mg electron configuration: 1s²2s²2p⁶3s² or [Ne] 3s²

Be electron configuration: 1s²2s² or [He] 2s²

This is because Na and Mg are right next to each other in the same period (horizontal).

3 0
3 years ago
Do atoms lend and borrow electrons from inner shells
zaharov [31]
False. They don't borrow electrons at all. They already have their respective electron affinities. This is called as electronegativity, and it's an occurence where it already has its own from its actual structure. It never borrows any electrons at all.
4 0
3 years ago
What will happen if water is deionized​
aleksley [76]
If water is deionized and it is consumed, it may cause people to urinate more and eliminate more electrolytes from the body.
4 0
3 years ago
You placed 6.35 g of a mixture containing unknown amounts of BaO(s) and MgO(s) in a 3.50-L flask containing CO₂(g) at 30.0°C and
adelina 88 [10]

Mass of BaO in  initial mixture = 3.50g

Explanation:

Let mass of BaO in mixture be x g

mass of MgO in mixture be (6.35 - x) g

Initially CO_2

Volume = 3.50 L

Temp = 303 K

Pressure = 750 torr = 750 / 760 atm

Applying ideal gas equation

PV = nRT

n = PV / RT

(n)_CO_2 = ((750/760)* 3.50) / 0.0821 * 303

(n)_CO_2 = 0.139 mole

Finally; mole of CO_2

n= PV /RT

((245/760) *3.5) / 303* 0.0821

(n)_CO_2 = 0.045 mole

Mole of CO_2 reacted = 0.139 - 0.045

=0.044 mole

BaO + CO_2  BaCO_3

Mgo + CO_2  MgCO_3

moles of CO_2 reacted = ( moles of BaO + moles of MgO)

moles of BaO in mixture = x / 153 mole

moles of MgO in mixture = 6.35 - x mole / 40

Equating,

x/ 153 +6.35/40 = 0.094

= x/153 + 6.35 / 40 - x/40 =0.094

= x (1/40 - 1153) = (6.35/40 - 0.094)

= x * 10.018464

= 0.06475

mass of BaO in mixture = 3.50g

5 0
3 years ago
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