Answer:
Explanation:
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In this case, since, when we see an ionic binary salt like that formed by magnesium (metal) and fluorine (nonmetal), in order to set up the correct molecular formula we need to check out the periodic table in order to identify the suitable oxidation states, since metals remain positively charged as they lose electrons and nonmetals negatively charged as they gain electrons.
In such a way, since the oxidation state of magnesium is +2 and that of fluorine is -1 we write:
Next, we need to exchange the oxidation states as subscripts without the charge to obtain:
Which is the corrected molecular formula for magnesium fluoride.
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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:
</em></u>
- 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.
I believe the answer is C. Does not touch it
Someone please correct me if I am wrong.
<span>Enthalpy is regarding the amount of heat that is given off or taken in during the process of a reaction, while entropy is about the disorderliness of a reaction.
Both are related in the equation ∆G=∆H-T∆S, where ∆G is the Gibbs free energy. So we can say that a reaction is both enthalpy and entropy driven. It's like, both of them are interlinked with each other. </span>
We know, Given mass = Molar mass * Number of moles.
A.) <span>1.25 mol CaF</span>₂
Number of moles = 1.25
Molar mass = 78
So, Mass = 78 * 1.25 = 97.5 g
B.) 3.4 mol (NH₄)₂SO₄
Number of moles = 3.4
Molar mass = 132
Mass = 3.4 * 132 = 448.8 g
Hope this helps!
