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C₇H₉+ HNO₃ → C₇H₆(NO₂)₃+ H₂O
Our elements: C, H, N, and O.
Balance H.
Ten on the left, eight on the right. Add a coefficient of 2 in front of H₂O.
C₇H₉+ HNO₃ → C₇H₆(NO₂)₃+ 2H₂O
Balance N.
One on the left, three on the right. Add a coefficient of 3 in front of HNO₃.
C₇H₉+ 3HNO₃ → C₇H₆(NO₂)₃+ 2H₂O
This unbalanced H, let's increase the coefficient in front of H₂O from 2 to 3 to rebalance.
C₇H₉+ 3HNO₃ → C₇H₆(NO₂)₃+ 3H₂O
Balance O.
Nine on the left, nine on the right. Already balanced.
Balance C.
Seven on the left, seven on the right. Already balanced.
Our final balanced equation:
C₇H₉+ 3HNO₃ → C₇H₆(NO₂)₃+ 3H₂O
Hope this helps!
Answer:
Q₁- The concentration of HCl = 0.075 N = 0.075 M.
Q₂- The concentration of KOH = 7.675 mN = 7.675 mM.
Q₃- The concentration of H₂SO₄ = 0.2115 N = 0.105 M.
Q₄- The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.
Explanation:
<u><em>Q₁:
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- As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.
- The normality of the NaOH and HCl = Their molarity.
∵ (NV)NaOH = (NV)HCl
∴ N of HCl = (NV)NaOH / (V)HCl = (0.15 N)(67 mL) / (134 mL) = 0.075 N.
∴ The concentration of HCl = 0.075 N = 0.075 M.
<em><u>Q₂:</u></em>
- As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.
- The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.
∵ (NV)H₂SO₄ = (NV)KOH
∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.
∴ The concentration of KOH = 7.675 mN = 7.675 mM.
<em><u>Q₃:</u></em>
- As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.
- The normality of NaOH = Molarity of NaOH = 0.5 N.
∵ (NV)H₂SO₄ = (NV)NaOH
∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄ = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.
∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.
<em><u>Q₄:</u></em>
- The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.
- The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.
- End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.
- The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.
cities have factories , more people and no water around .
Answer:
Se⁻²
Explanation:
Even though the atom is not shown, we can consult the periodic table. Selenium (Se) is in group 6 and period 4, so it has 6 valence electrons. To be stable, the atom must have a noble gas configuration with 8 valence electrons, thus, selenium must gain 2 electrons.
When an atom gain electrons, it is negatively charged, because electrons have a negative charge. So, the ion formed is Se⁻².
A) –1 B) +2 C) +6 D) +4 3.
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