Answer:
The equation for the reaction of one sodium bicarbonate ( NaHCO3 ) molecule with one citric acid (C6H8O7) molecule is the following:
Sodium Bicarbonate + Citric Acid ⇒ Water + Carbon Dioxide + Sodium Citrate
NaHCO3 + C6H8O7 ⇒ 3 CO2 + 3 H2O + Na3C6H5O7
Explanation:
The reaction is in balance, that is, the whole H2CO3 is not finished, but a little bit of this acid is left in the solution. Therefore, when sodium bicarbonate is added to the solution with citric acid, sodium citrate salt (C6H5O7Na3) and carbonic acid (H2CO3) are formed, which is rapidly broken down into water (H2O) and carbonic oxide (CO2).
C6H8O7 + NaHCO3 ⇒ C6H5O7Na3 + 3 H2CO3
C6H5O7Na3 + 3 H2CO3 ⇔ C6H5O7Na3 + 3 H2O + 3 CO2
Answer is: Benzene is trigonal (or triangular) planar.
VSEPR theory (The Valence Shell Electron Pair Repulsion Theory) uses the AXE notation (m and n are integers, m + n = number of regions of electron density).
For benzene molecule (C₆H₆):
m = 3; the number of atoms bonded to the central atom.
n = 0; the number of lone pairs on the central atom.
Answer:
Mass = 6.183 g
Solution:
Step 1: Calculate number of moles of Boric acid using following formula,
Molarity = Moles ÷ Volume
Solving for Moles,
Moles = Molarity × Volume
Putting Values,
Moles = 0.05 mol.L⁻¹ × 2.0 L
Moles = 0.1 mol
Step 2: Calculate Mass of Boric Acid using following formula,
Moles = Mass ÷ M.mass
Solving for Mass,
Mass = Moles × M.mass
Putting values,
Mass = 0.1 mol × 61.83 g.mol⁻¹
Mass = 6.183 g
Flask used to prepare this solution is called as Volumetric flask. Take 2 L volumetric flask, add 6.183 g of Boric acid and fill it to the mark with distilled water.
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Answer:</h3>
1 x 10^13 stadiums
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Explanation:</h3>
We are given that;
1 stadium holds = 1 × 10^5 people
Number of iron atoms is 1 × 10^18 atoms
Assuming the stadium would carry an equivalent number of atoms as people.
Then, 1 stadium will carry 1 × 10^5 atoms
Therefore,
To calculate the number of stadiums that can hold 1 × 10^18 atoms we divide the total number of atoms by the number of atoms per stadium.
Number of stadiums = Total number of atoms ÷ Number of atoms per stadium
= 1 × 10^18 atoms ÷ 1 × 10^5 atoms/stadium
= 1 × 10^13 Stadiums
Thus, 1 × 10^18 atoms would occupy 1 × 10^13 stadiums
