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3 years ago

Identify the acid, base, and salt in the following reaction HNO3 + KOH —> H2O + KNO3

Chemistry

1 answer:

Yuri [45]3 years ago

7 0

Explanation:

acid ,HNO3

base,KOH

salt ,KNO3

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What is the specific heat of ethanol if 4650 j are required to raise the temperature of 250.0g of ethanol from 22.0oc to 29.6 de

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$C=2.45\frac{J}{g\°C}$

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3 years ago

The value of Kc for the reaction between water vapor and dichlorine monoxide, H2O(g) 1 Cl2O(g) 4 2 HOCl(g) is 0.0900 at 25°C. De

max2010maxim [7]

Answer:

[HOCl] = 0.001 127 mol·L⁻¹; [H₂O] = [Cl₂O] = 0.003 76 mol·L⁻¹

Explanation:

The balanced equation is

H₂O + Cl₂O ⇌ 2HOCl

Data:

Kc = 0.0900

[H₂O] = 0.004 32 mol·L⁻¹

[Cl₂O] = 0.004 32 mol

1. Set up an ICE table.

$\begin{array}{ccccccc}\rm \text{H$_{2}$O}& + & \text{Cl$_{2}$O} & \, \rightleftharpoons \, & \text{2HOCl} & & \\0.00432 & & 0.00432 & & 0 & & \\-x &&-x&&+2x&&\\0.00432-x &&0.00432 - x& & 2x&&\\\end{array}$

2. Calculate the equilibrium concentrations

$K_{\text{c}} = \dfrac{\text{[HOCl]$^{2}$}}{\text{[H$_{2}$O][Cl$_2$O]}} = \dfrac{(2x)^{2}}{(0.00432 - x)^{2}} = 0.0900\\\\\begin{array}{rcl}\dfrac{4x^{2}}{(0.00432 - x)^{2}} &=& 0.0900\\ \dfrac{2x }{0.00432 - x} & = & 0.300\\2x & = & 0.300(0.00432 - x)\\2x & = & 0.001296 - 0.300x\\2.300x & = & 0.001296\\x & = & \mathbf{5.63\times 10^{-4}}\\\end{array}$

[HOCl] = 2x mol·L⁻¹ = 2 × 5.63 × 10⁻⁴ mol·L⁻¹ =0.001 127 mol·L⁻¹

[H₂O] = [Cl₂O] = (0.004 32 - 0.000 563) mol·L⁻¹ = 0.003 76 mol·L⁻¹

Check:

$\begin{array}{rcl}\dfrac{0.001127^{2}}{0.00376^{2}} & = & 0.0900\\\\\dfrac{1.270 \times 10^{-6}}{1.411 \times 10^{-5}} & = & 0.0900\\0.0900 & = & 0.0900\\\end{array}$