Answer:
₁₅P = 1s² 2s² 2p⁶ 3s² 3p³
Explanation:
Phosphorus is the second element of group (v) with atomic number 15 and the electrons per shell are arrange as thus; 2, 8, 5.
The electronic configuration of phosphorus is
₁₅P = 1s² 2s² 2p⁶ 3s² 3p³
Or
₁₅P [Ne] 3s² 3p³
In its first shell I.e 1s shell, it has 2 electrons filling it to s-orbital requirement. In its second shell, electrons fill it in 2s² 2p⁶ filling the second orbital and attaining an octet configuration. The third shell contains 5 electrons filling 3s² 3p³ orbital.
Answer:
The unknown NaOH base has a concentration of 0.636M
Explanation:
<u>Step 1:</u> the balanced equation
NaOH + HCl → NaCl + H2O
This means for 1 mole of NaOH consumed there is 1 mole of HCl needed to produce 1 mole of NaCl and 1 mole of H2O
<u>Step 2</u>: Calculate moles of HCl used
Number of moles = Concentration * volume = 0.5M * 25*10^-3 L =0.0125 moles
<u>Step 3</u>: Calculate moles of NaOH
Since the mole ratio for HCl and NaOH is 1:1 this means we have 0.0125 moles of NaOH for 0.0125 moles of HCl
<u>Step 4:</u> Calculate Concentration of the unknown NaOH base
Concentration = Number of moles / Volume
Volume of NaOH = 24.64-5 =19.64 mL = 0.01964 L
Concentration = 0.0125/0.01964 = 0.636 M
The unknown NaOH base has a concentration of 0.636M
The uncertainties of the delta measurements and the uncertainty of the atomic weight derivedfrom the best measurement of isotopic abundances constrain the number of significant figures in theatomic-weight values of the upper and lower bounds. For carbon, the fifth digit after the decimal pointis uncertain because of the uncertainty value of 0.000 027. Therefore, the number of significant digitsin the atomic-weight value is reduced to four figures after the decimal point. The Commission may rec-ommend additional conservatism and reduce the number of significant figures further. For the lowerbound of carbon, 12.009 635 is truncated to 12.0096. For an upper bound, the trailing digit is increasedto ensure the atomic-weight interval encompasses the atomic-weight values of all normal materials. Inthe case of carbon, the upper bound is adjusted from 12.011 532 to 12.0116 to express four digits afterthe decimal point. The lower and upper bounds are evaluated so that the number of significant digits ineach is identical. If a value ends with a zero, it may need to be included in the value to express therequired number of digits. The following are examples of lower and upper atomic-weight bounds foroxygen that could be published by the Commission in its various tables.
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Extracted from" Atomic weights of the elements 2009 (IUPAC Technical Report)"
Compounds are substances made of at least 2 different elements. Molecules have at least 2 atoms, doesn't have to be different elements. Compounds and molecules both work.
Answer:
Percent composition of the solution is 26 % of sucrose and 74 % of water
Explanation:
Percent composition is the mass of solute, either of solvent in 100 g of solution.
Mass of solution = Mass of solvent + Mass of solute
Mass of solute = 35 g
Mass of solvent = 100 g
As we know, water density = 1g/mL
So 1g/mL . 100 mL = 100 g
35 g + 100 g = 135 g → Mass of solution
(Mass of solute / Mass of solution) . 100 =
(35 g / 135 g) . 100 = 26 %
(Mass of solvent / Mass of solution) . 100 =
(100 g / 135 g) . 100 = 74 %
