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Zepler [3.9K]
3 years ago
5

What can the increase in the size of the "cousins be compared to on the periodic table?​

Chemistry
1 answer:
stira [4]3 years ago
3 0

Atomic radius

Explanation:

The increase in size on the periodic table can be compared to the atomic radius on the table.

Elements in the same group on the periodic table can be regarded as a family.

Atomic radius is the property used to quantify the size of an atom. It is taken as half of the inter-nuclear distance between two covalently bonded atoms of a non-metal or the distance between the nuclei of two metals.

  • Across a nucleus, it decreases rapidly due to pull of the nucleus.
  • Atomic radius increases down a group on the periodic table.

Learn more:

Periodic table brainly.com/question/2014634

#learnwithBrainly

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What mass of sodium chloride (NaCl) forms when 7.5 g of sodium carbonate (Na2CO3) reacts with a dilute solution of hydrochloric
Gre4nikov [31]

 The mass of NaCl  formed  is 8.307  grams


<u><em> calculation</em></u>

step 1: write the equation  for reaction

Na₂CO₃  + 2HCl → 2 NaCl  +CO₂ +H₂O

Step 2: find the  moles of Na₂CO₃

moles = mass/molar mass

 The  molar mass of Na₂CO₃  is = (23 x2) + 12 + ( 16 x3) = 106 g/mol

moles  = 7.5 g/106 g/mol =0.071 moles

Step 3: use the  mole ratio to determine the  mole of NaCl

Na₂CO₃:NaCl  is  1:2  therefore the moles of NaCl =0.07  x2 =0.142 moles


Step 4:  calculate mass  of NaCl

mass= moles x molar mass

the molar  mass of NaCl= 23 +35.5 =58.5 g/mol

mass  = 0.142  moles x 58.5 g/mol =8.307  grams

8 0
3 years ago
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Rank the ions in order of increasing basicity: CH3NH-, CH3CH2-, CH30-
Alexxx [7]

In order to find out the ranking of ions basicity, check the pKa values of each ions. The principle that you need to remember is that the stronger the acid the weaker the corresponding conjugate base. The pKa dictates acid value of the compound. The answer would be CH3NH, CH3O-, and CH3CH2-. 

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tino4ka555 [31]
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8 0
3 years ago
On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?
enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

HNO_2\ (aq) \leftrightharpoons  H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\  K_a = 4.0\times \ 10^{-4}

H^{+}=?

Formula:

Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}

Let

[H^{+}] = [NO_2^{-}] = x at equilibrium

x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \  M\\\\

therefore,

[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M

Calculating the % ionization:

= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\

6 0
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