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3 years ago

Balance equation P2O5 + H20 - - - - > H3PO4 Help me.....

Chemistry

1 answer:

Archy [21]3 years ago

8 0

Answer:

Search by reactants (P 2O 5, H 2O) and by products (H 3PO 4)

H2O + P2O5 → H3PO4

H2O + HNO3 + P2O5 → H3PO4 + N2O5

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For which of the following aqueous solutions would one expect to have the largest van’t Hoff factor (i)? a. 0.050 m NaCl b. 0.50

ira [324]

Answer:

The van't hoff factor of 0.500m K₂SO₄ will be highest.

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Van't Hoff factor was introduced for better understanding of colligative property of a solution.

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a) For NaCl the van't Hoff factor is 2

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Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.

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3 years ago

Write isotopic symbols in the form AZX for each isotope. a. the copper isotope with 34 neutrons b. the copper isotope with 36 ne

krek1111 [17]

Explanation:

The form AZX denoted that;

Z = The atomic number or proton number (symbol Z) of a chemical element is the number of protons found in the nucleus of every atom of that element.

A = Mass Number ( Sum of Neutrons and Protons)

X represents the element.

A. the copper isotope with 34 neutrons

X = Cu

Atomic Number of Copper, Z = 29

Mass Number, A = 29 + 34 = 63

63.29.Cu

B. the copper isotope with 36 neutrons

X = Cu

Atomic Number of Copper, Z = 29

Mass Number, A = 29 + 36 = 65

65.29.Cu

C. the potassium isotope with 21 neutrons

X = K

Atomic Number of Potassium, Z = 19

Mass Number, A = 19 + 21 = 40

40.19.K

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X = Ar

Atomic Number of Argon, Z = 18

Mass Number, A = 18 + 22 = 40

40.18.Ar

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2 years ago

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .

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3 years ago