Explanation:
Let us assume that the ratio for the given reaction is 1:1.
Therefore, we will calculate the moles of 
as follows.
Moles of solution = molarity × volume (L)
= 0.0440 M × 0.014 L
= 0.000616 moles
Moles of excess EDTA = 0.000616 moles
Also, the initial moles of EDTA will be calculated as follows.
Total initial moles of EDTA = 0.0600 M × 0.025 L
= 0.0015
Therefore, moles of EDTA reacted with 
will be as follows.
= 0.0015 - 0.000616
= 0.00088 moles
Since, we have supposed a 1 : 1 ratio between and EDTA
.
So, moles of = 0.00088 moles
Now, we will calculate the molarity of as follows.
Molarity of solution = 
= 
= 0.015 M
Thus, we can conclude that the original concentration of the solution is 0.015 M.
Answer: option c. magnitude and direction
Explanation:
1) Force is a vector: it has magnitude and direction.
2) Magnitude measures the "intensity" of the force. It is measured in newtons (N), in the SI (international system of units). One newton is the force exerted to confere an acceleration of 1 m/s² to a mass of 1 kg.
3) Telling the magnitude of the force is not enough information to understand what the force is and to predict its effect.
It is necessary to tell the direction in which the force is applied.
It is not the same a force of 10 N that pulls that the same magnitude pushing. And it is not the same a hhorizontal force of 100 N to move an object, than the same magnitude applied at an agle.
That is why the force must be measured and reported as a magnitude and a direction.
4) Examples of forces correctly reported are:
i) 100 N vertically upward
ii) 1000 N 20° to the east of the north.
iii) 0.2 N with an elevation angle of 50°.
Then, scientists must measure the magnitude and the direction of the force.
<span>Balancing the choices, we have as follows:
A. 3AgBr + GaPO4 → Ag3PO4 +GaBr3
B. 3H2SO4 + 2B(OH)3 → B2(SO4)3 + 6H2O
C. Fe + 2AgNO3 → Fe(NO3)2 + 2Ag
D. C2H4O2 +2O2 → 2CO2 + 2H2O
Therefore, the correct answers are option A and B. Hope this answers the question. Have a nice day.</span>
Answer: - 986.6 kj/mol
Explanation:
1) Equation given:
CaO(s) + H₂O (l) → Ca (OH)₂ (s) δh⁰ = −65.2 kj/mol
2) Standard enthalpies of formation given:
CaO, δhf⁰ = −635.6 kj/mol
H₂O, δhf⁰ = −285.8 kj/mol
3) Calculate the standard enthalpy of formation of Ca(OH)₂.
δh⁰ = ∑δh⁰f of products - ∑ δh⁰f of reactants
Using the mole coefficients of the balanced chemical equation:
δh⁰ = δh⁰f Ca(OH)₂ - δh⁰f CaO - δh⁰f H₂O
⇒ δh⁰f Ca(OH)₂ = δh⁰ + δh⁰f CaO + δh⁰f H₂O
⇒ δh⁰f Ca(OH)₂ = - 65.2 kj/mol − 635.6 kj/mol) − 285.8 kj/mol) = - 986.6 kj/mol.
