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nevsk [136]
4 years ago
13

At a certain temperature, the equilibrium constant, Kc,Kc, for this reaction is 53.3. H2(g)+I2(g)↽−−⇀2HI(g)Kc=53.3 H2(g)+I2(g)↽−

−⇀2HI(g)Kc=53.3 At this temperature, 0.400 mol H20.400 mol H2 and 0.400 mol I20.400 mol I2 were placed in a 1.00 L container to react. What concentration of HIHI is present at equilibrium?
Chemistry
1 answer:
astraxan [27]4 years ago
7 0

Answer:

0.628 mol·L⁻¹

Explanation:

The balanced equation is

H₂ + I₂ ⇌ 2I ₂

Data:

 Kc = 53.3

[H₂] = 0.400 mol·L⁻¹

[I₂] = 0.400 mol·L⁻¹

1. Set up an ICE table.

\begin{array}{ccccccc}\rm \text{H}_{2}& + & \text{I}_{2} & \, \rightleftharpoons \, & \text{2HI} & & \\0.400 & & 0.400 & & 0 & & \\-x &   & -x  &   & +2x &   & \\0.400 - x &   & 0.400 - x &   & 2x & & \\\end{array}

2. Calculate the equilibrium concentrations

K_{\text{c}} = \dfrac{\text{[HI]$^{2}$}}{\text{[H$_{2}$][I$_2$]}} = \dfrac{(2x)^{2}}{(0.400 - x)^{2}} = 53.3\\\\\begin{array}{rcl}\dfrac{(2x)^{2}}{(0.400 - x)^{2}} &=& 53.3\\ \dfrac{2x }{0.400 - x} & = & 7.301\\\\2x & = & 7.301(0.400 - x)\\2x & = & 2.920 - 7.301x\\9.301x & = & 2.920\\x & = & \dfrac{2.920}{9.301}\\\\x & = & \mathbf{0.3140}\\\end{array}

[HI] = 2x mol·L⁻¹ = 2 × 0.3140 mol·L⁻¹ = 0.628 mol·L⁻¹

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(102 900 ÷ 12) + (170 × 1.27) = 8800

Step 1. Evaluate the expressions inside the parentheses (PEMDAS)

102 900 ÷ 12 = 8575

170 × 1.27 = 215.9

In multiplication and division problems, your answer can have no more significant figures than the number with the fewest significant figures.

Thus, the underlined digits are not significant, but we keep them in our calculator to avoid roundoff error.

Step 2. Do the addition (PEMDAS).

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+    215.9

= 8790.9

Everything that you add to an insignificant digit gives an insignificant digit as an answer.

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How do you find the charge of an ion
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What is the name of the compound with the formula NH HCO3 ?
svet-max [94.6K]

Answer:

Acid ammonium carbonate // Ammonium bicarbonate.

Explanation:

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How many grams of solid Na2CO3 are required to neutralize exactly 2 liters of an HCI solution of pH 2.0?
Snezhnost [94]

Answer:

The answer is 1.06g.

Explanation:

Analysis of question:

1. Identify the information in the question given.

  • volume of HCl is 2 dm3
  • pH of HCl is 2.0

2. What the question want?

  • mass of Na2CO3 is ?(unknown)
  • 3. Do calculation.
  • 1st-Write a balanced chemical equation:

Na2CO3 + 2HCl (arrow) 2NaCl + H20 + CO2

  • 2nd-Determine the molarity of HCl with the value of 2.0.

pH= -log[H+]

2.0= -log[H+]

log[H+]= -2.0

[H+]= 10 to the power of negative 2(10-2)

=0.01 mol dm-3

molarity of HCl is 0.01 mol dm-3

  • 3rd-Find the number of moles of HCl

n=MV

=0.01 mol dm-3 × 2 dm3

=0.02 mol of HCl

  • 4th-Find the second mol of it.

Based on the chemical equation,

2.0 mol of HCl reacts with 1.0 mol of Na2CO3

0.02 mol of HCl reacts with 0.01 mol of Na2CO3

<u>N</u><u>a</u>2CO3>a=<u>1</u><u> </u>mol

<u>2</u><u>H</u>Cl>b=<u>2</u><u> </u>mol

  • 5th-Find the mass of it.

mass= number of mole × molar mass

g=0.01 × [2(23)+ 12+ 3(16)]

g=0.01 × 106

# =1.06 g.

3 0
3 years ago
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