Answer: C
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Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
<span>When two electrical charges, of
opposite sign and equal magnitude, are separated by a distance, a dipole is
established. The size of a dipole is measured by its dipole moment (</span>μμ). Dipole moment is measured in Debye
units, which is equal to the distance between the charges multiplied by the
charge (1 Debye equals 3.34×10−30Cm3.34×10−30Cm). The dipole moment
of a molecule can be calculated by Equation 1.11.1:
μ = qr
where
<span>
<span>μ⃗ μ→ is the dipole moment vector</span>
<span>qiqi is the magnitude of the ithith charge, and</span>
<span>r⃗ ir→i is the vector representing the position
of ithith charge.</span>
</span>
r = μ/q
<span>r = [0.838D(3.34×10−30 C⋅m/ 1D)]/ (1.6×10−19
C) *0.124
</span>
r = 1.41 x10^-10 m
Answer: 36.9 g
Explanation:
P4 + 5O2 = P4O10 Balanced equation
moles P4 present = 23.9 g x 1 mole/123.88 g = 0.193 moles
moles O2 present = 20.8 g x 1 mol/32 g = 0.65 moles O2
From balanced equation, mole ratio O2 : P4 is 5:1. Is 0.65 moles O2 5x 0.193 moles? NO. You don't have enough O2.
O2 is limiting in this reaction.
theoretical moles of P4O10 = 0.65 moles O2 x 1 mole P4O10/5 moles O2 = 0.13 moles P4O10
mass of P4O10 produced = 0.13 moles x 283.9 g = 36.9 g
Answer:
Passivation of Oxide layers of the metals.
Explanation:
Passivation is a non-electrolytic finishing process that makes most metals rust-resistant. The prosses removes free iron from the surface by using either nitric or citric acid. When this happens, it results to an inert, protective oxide layer that is very slow or less likely to chemically react with air and cause corrosion.
Passivity caused many of the metals several minutes to begin to react. Once the finishing process that makes metals less likely to react was eroded, reaction was initiated vigorously.
