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777dan777 [17]
3 years ago
11

Which Period is this element in?

Chemistry
1 answer:
Vesna [10]3 years ago
5 0
2nd if periods are the ones that are vertical
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What is the hybridization of the central atom in becl2?
Anuta_ua [19.1K]
Its hybridization would be sp because Be only has 2 covalent bonds with Cl
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3 years ago
Dry air is 78.08% nitrogen, 20.95% oxygen and 0.93% argon with the 0.04% being other gases. if the atmospheric pressure is 760.0
sveticcg [70]
Partial pressure (N2) = mole fraction * total pressure
 
{    1 mole of any ideal gas occupy same volume of 1 mole of any other ideal gas under same condition of temperature and pressure so mole fraction in the sample is simply 78.08%   =   0.7808   this is because equal volume of each gas has equal moles

partial pressure N2 = 0.7808 * 760 .0
partial pressure =  593.4 mmhg    (   1 torr = 1mmhg  )
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4 years ago
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4. How many weeks are equal to 2.94 x 10 to the power of 6 minutes?
OverLord2011 [107]

Answer:

292 weeks

Explanation:

5 0
3 years ago
For an atom’s electrons, how many energy sublevels are present in the principal energy level n = 4?
defon
"n" represents principal energy level or principal quantum number. Principal energy levels are then subdivided into sublevels. The level of principal energy level is always equal to number of sublevels. Therefore if <span>principal energy level</span> = 4, then there are also 4 sublevels present, namely: 4s, 4p, 4d and 4f sublevels.
3 0
3 years ago
If you have a 1.0 L buffer containing 0.208 M NaHSO3 and 0.134 M Na2SO3, what is the pH of the solution after addition of 50.0 m
Vlada [557]

Answer:

pH = 7.233

Explanation:

Initially, the buffer contains 0.208 moles of NaHSO₃ and 0.134 moles of Na₂SO₃.

NaHSO₃ reacts with NaOH thus:

NaHSO₃ + NaOH → Na₂SO₃ + H₂O

50.0 mL of 1.00 M NaOH are:

0.0500L × (1mol / 1L) = 0.0500moles of NaOH added. That means after the addition are produced  0.0500moles of Na₂SO₃ and consumed 0.0500moles of NaHSO₃. That means final moles of the buffer are:

NaHSO₃: 0.208 mol - 0.050 mol = <em>0.158 mol</em>

Na₂SO₃: 0.134 mol + 0.050 mol = <em>0.184 mol</em>

<em> </em>

As pKa of this buffer is 7.167, it is possible to use H-H equation to find pH, thus:

pH = pKa + log₁₀ [Na₂SO₃] / [NaHSO₃]

pH = 7.167 + log₁₀ [0.184] / [0.158]

<em>pH = 7.233</em>

6 0
3 years ago
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