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Debora [2.8K]
2 years ago
11

Why neon is special from other chemical elements​

Chemistry
1 answer:
kupik [55]2 years ago
6 0
Noble gases react very unwillingly, because the outermost shell of electrons orbiting the nucleus is full, giving these gases no incentive to swap electrons with other elements. As a result, there are very few compounds made with noble gases. Like its noble gas comrades, neon is odorless and colorless.
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How many bromine atoms are present in 39.4 g of CH2Br2
strojnjashka [21]
37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2 

<span>4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms</span>
7 0
4 years ago
Which of the following is not a method of heat transfer?
abruzzese [7]
The answer is adduction lol
8 0
4 years ago
A 1.24g sample of a hydrocarbon, when completely burned in an excess of O2 yields 4.04g Co2 and 1.24g H20. Draw plausible struct
arsen [322]

Answer:

Plausible structure has been given below

Explanation:

  • Molar mass of CO_{2} is 44 g/mol and molar mass of H_{2}O is 18 g/mol
  • Number of mole = (mass/molar mass)

4.04 g of CO_{2} = \frac{4.04}{44}moles CO_{2} = 0.0918 moles of CO_{2}

1 mol of CO_{2} contains 1 mol of C atom

So, 0.0918 moles of CO_{2} contains 0.0918 moles of C atom

1.24 g of H_{2}O = \frac{1.24}{18}moles H_{2}O = 0.0689 moles of H_{2}O

1 mol of H_{2}O  contain 2 moles of H atom

So, 0.0689 moles of H_{2}O contain (2\times 0.0689)moles of H_{2}O or 0.138 moles of H_{2}O

Moles of C : moles of H = 0.0918 : 0.138 = 2 : 3

Empirical formula of hydrocarbon is C_{2}H_{3}

So, molecular formula of one of it's analog is C_{4}H_{6}

Plausible structure of C_{4}H_{6} has been given below.

4 0
3 years ago
A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa 

NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M 

CH3COONa = 0.0076 / 0.071 = 0.1070M 

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74. 

pH using Henderson - Hasselbalch equation: 

pH = pKa + log ([salt]/[acid]) 

pH = 4.74 + log ( 0.1070/0.1493) 

pH = 4.74 + log 0.717 

pH = 4.74 + (-0.14) 

pH = 4.60.

7 0
3 years ago
Someone help me pleasee?
Kamila [148]
Maybe its the second one
7 0
3 years ago
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