Answer:
A flower is self-pollinated if pollen is transferred to it from any flower of the same plant and cross-pollinated if the pollen comes from a flower on a different plant.
Explanation:
This is because each flower is capable of fertilizing itself by autogamy. Autogamy means that the male part of a flower sends pollen to the female part of the same flower.
Given:
Concentration of Fluoride ions = 0.100 M
Concentration of Hydrogen Fluoride = 0.126 M
Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution
Assume: 50:50 ratio of fluoride ions and HF
12.5ml*0.1mol/L *1L/1000mL + 12.5*0.126mol/L * 1L/1000mL = 2.825x10^-3 moles F-
5ml * 0.01 mol/L *1L/1000mL = 5x10^-5 moles
Assume: Volume additive
Final concentration = 2.825x10^-3 + 5x10^-5 moles/ 30 ml * 1000ml/L =0.0958 M
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Answer:
Explanation:
By definition, <em>half neutralization</em> is the point at which half of the acid has been neutralized.
The neutralization reaction that you are studying is the acid-base reaction:
- HCl (aq) + NaOH (aq) → NaCl(aq) + H₂O (aq)
Then, since the starting molarity of the acid (HCl) is 0.2 M, you just need to find half of that concentration:
- Half molarity = M / 2 = 0.2 M / 2 = 0.1 M
So, the answer is the first choice: a. 0.1 M.
Answer:
density, yield strength, elasticity, Ductility, toughness, fatigue limit and corrosion resistance
Answer:
ΔHorxn = - 11.79 KJ
Explanation:
2 SO 2 ( g ) + O 2 ( g ) ⟶ 2 SO 3 ( g )
The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol
From the reaction above, 2 mol of SO2 reacts to produce 2 mol of SO3. Assuming ideal gas behaviour,
1 mol = 22.4l
x mol = 2.67l
Upon cross multiplication and solving for x;
x = 2.67 / 22.4 = 0.1192 mol
0.1192 mol of SO2 would react to produce 0.1192 mol of SO3.
Amount of heat is given as;
ΔHorxn = ∑mΔHof(products) − ∑nΔHof(reactants)
Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol.
ΔHorxn = 0.1192 mol * (− 395.7 kJ / mol) - 0.1192 mol * ( − 296.8 kJ / mol)
ΔHorxn = - 47.17kj + 35.38kj
ΔHorxn = - 11.79 KJ
