Answer:
VII'th group, negative ion
Explanation:
Usually fluorine is located in the seventh(VII) group as fluorine has 7 electrons in it's outer shell with chemical symbol "F"
Fluorine has 7 electrons in it's outer shell therefore, it needs one more electron to get the noble gas configuration. So more electrons means negative ion.
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I believe that it is 187.56 g/mol
The concentration of ClO₂⁻ at equilibrium if the initial concentration of HClO₂ is 0.0654.
<h3>What is concentration?</h3>
The concentration of any substance is the quantity of that substance in per square of the space or container.
The reaction is
HClO₂ + H₂O <=> H₃O⁺ + ClO₂⁻
The pH is 0.454 M
Ka = [H₃O⁺][ClO₂⁻ ] / [HClO₂]
2. 25 × 10⁻² m = [x][x] / 0.454-x]
2 + 0.011 - 0.004994 = 0
solve the quadratic equation
x = 0.0654 = [H3O+] = [ClO2-]
pH = -log (H3O+)
pH = -log(0.0654)
pH = 1.2
equilibrium concentrations of
[HClO2] = 0.454 -x = 0.454 -0.0654 = 0.3886 M
[ClO2- ] = x = 0.0654
Thus, the equilibrium concentrations is 0.0654.
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Answer:
Percentage of oxygen = 30%
Percentage of carbon = 30%
Percentage of hydrogen = 40%
Explanation:
Formula:
Percentage of element = given amount / total amount × 100
Given compound:
C₆H₈O₆
Number of atoms of carbon = 6
Number of atoms of hydrogen = 8
Number of atoms of oxygen = 6
Total number of atoms = 20
Percentage of carbon = 6/20 × 100
Percentage of carbon = 30%
Percentage of Hydrogen = 8/20 × 100
Percentage of Hydrogen = 40%
Percentage of oxygen = 6/20 × 100
Percentage of oxygen = 30%
Answer:
3.1 kg
Explanation:
Step 1: Write the balanced combustion equation
C₈H₁₈ + 12.5 O₂ ⇒ 8 CO₂ + 9 H₂O
Step 2: Calculate the moles corresponding to 1.0 kg of C₈H₁₈.
The molar mass of C₈H₁₈ is 114.23 g/mol.
1.0 × 10³ g × 1 mol/114.23 g = 8.8 mol
Step 3: Calculate the moles of CO₂ produced from 8.8 moles of C₈H₁₈
The molar ratio of C₈H₁₈ to CO₂ is 1:8. The moles of CO₂ produced are 8/1 × 8.8 mol = 70 mol.
Step 4: Calculate the mass corresponding to 70 moles of CO₂
The molar mass of CO₂ is 44.01 g/mol.
70 mol × 44.01 g/mol = 3.1 × 10³ g = 3.1 kg
