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strojnjashka [21]

4 years ago

9

Is calcium carbonate chemical change ?

1 answer:

N76 [4]4 years ago

7 0

Calcium oxide is known as lime and is one of the top 10 chemicals produced annually by thermal decomposition of limestone. .Over time, this reacts with carbon dioxide in the air to form crystals of calcium carbonate, which lock the sand grains together to form a hard rock-like material.

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If 35.0 g of CH₃OH (MM = 32.0 g/mol) are dissolved in 500.0 mL of solution, what is the concentration of CH₃OH in the solution?

Stels [109]

Answer:

mlllllllllllllllllll lodsssss

6 0

3 years ago

Suppose a 500.mL flask is filled with 1.9mol of NO3 and 1.6mol of NO. The following reaction becomes possible: NO3gNOg 2NO2g The

never [62]

There is an error in the first sentence of the question; the right format is:

Suppose a 500.mL flask is filled with 1.9mol of NO3 and 1.6mol of NO2.

It should be NO2 and not NO.

Answer:

The equilibrium molarity of NO = 0.21695 m

Explanation:

Given that :

the volume = 500 mL = 0.500 m

number of moles of $NO_3 = 1.9 \ mol$

number of moles of $NO_2 = 1.6 \ mol$

Then we can calculate for their respectively concentrations as :

$[NO_3] = \frac{number \ of \ moles}{volume}$

$[NO_3] = \frac{1.9}{0.500}$

$[NO_3] = 3.8 \ M$

$[NO_2] = \frac{number \ of \ moles}{volume}$

$[NO_2] = \frac{}{} \frac{1.6}{0.500}$

$[NO_2] = 3.2 \ M$

The chemical reaction can be written as:

$NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}$

The ICE table is as follows;

$NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}$

Initial 3.8 - 3.2

Change +x x -2x

Equilibrium 3.8+x +x 3.2 - 2x

$K_c=\frac{[NO_2]^2}{[NO_3][NO]}$

$where \ K_c = 8.33$

$8.33 = \frac{(3.2-2x)^2}{(3.8+2x)x} \\ \\ 8.33 = \frac{(3.2-2x)^2}{(3.8x+2x^2)}$

$8.33(3.8x + 2x^2) = (3.2-2x)^2 \\ \\ 31.654x + 16.66x^2 = (3.2-2x)(3.2-2x) \\ \\ 31.654x + 16.66x^2 = 10.24 - 12.8x +4x^2 \\ \\ 10.24 - 44.454x -12.66 x^2 = 0 \\ \\ 12.66x^2 +44.454x -10.24 = 0$

Using quadratic formula;

$\frac{-b\pm \sqrt{(b)^2-4ac} }{2a}$

= $\frac{-(44.454) + \sqrt{(44.454)^2-4(12.66)(-10.24)} }{2(12.66)} \ \ OR \ \ \frac{-(44.454) - \sqrt{(44.454)^2-4(12.66)(-10.24)} }{2(12.66)}$

= 0.21695 OR -3.7283

Going by the positive value;

x = 0.21695

$[NO_3] = 3.8 +x = 3.8 + 0.21695$

= 4.01695 m

[NO] = x = 0.21695 m

$[NO_2] = 3.2 +x = 3.2 + 0.21695$

= 3.41695 m

3 0

3 years ago

A 300-gallon anaerobic digester will be loaded daily with a feedstock that contains two parts dairy manure and one-part water by

Mars2501 [29]

Answer:

The volume of feedstock needed to mantain an organic load rate of 2 kgVS/day is 0.055 m3/day of feedstock.

The HRT is 20.6 days.

Explanation:

First, we calculate how many kg is 1 m3 of feedstock. We know the density, so we can calculate the mass:

$M=\rho*V=37.5\frac{lb}{ft^3}*1m^3*(\frac{3.281ft}{1m}) ^3=1324.5lb=600.7 kg$

If the VS are 6% in weight,

$M_{vs}=0.06*M=0.06*600.7\,kg/m^3=36,0kgVS/m3$

The volume per day needed to feed 2 kg of VS/day is:

$V=\frac{2kg}{36kg/m^3}= 0.055m3/day=5.5litres/day$

The HRT depends on the volume of the tank and the flow. Its equation is

$HRT=\frac{V}{Q}=\frac{300gal}{0.055 m^3/day}*\frac{1m^3}{264.172gal}\\ \\HRT=20.6\,days$

6 0

3 years ago

g Hydrogen iodide, HI, decomposes at moderate temperatures according to the equation The amount of I2 in the reaction mixture ca

atroni [7]

Answer:

Equilibrium constant, Kc = 0.023

Explanation:

Equation for the decomposition of Hydrogen iodide is given below:

2HI ----> H₂ + I₂

Initially, the number of moles of the reactant and the products are given as follows:

n(HI) = 2 * 3.800 moles = 7.600 moles

nH₂) = 0.000 moles

n(I₂) = 0.000 moles

At equilibrium, the equation becomes: 2HI <----> H₂ + I₂

Number of moles of the reactant and the products are given as follows:

n(HI) = 7.600 - (0.886 + 0.886) moles = 5.828 moles

nH₂) = 2 * 0.443 = 0.886 moles

n(I₂) = 2 * 0.443 = 0.886 moles

Equilibrium constant, Kc = [H₂] [I₂] / [HI]²

Equilibrium constant, Kc = (0.886) * (0.886) / (5.828)²

Equilibrium constant, Kc = 0.023

5 0

3 years ago

A parachute on a racing dragster opens and changes the speed of the car from 85 m/s to 45 m/s in a period of 4.5 seconds. What i

LiRa [457]

Answer:

- 8.89 $m/s^{2}$

Explanation:

$\frac{(Vf-Vo)}{time}$ = $\frac{(45-85 m/s)}{4.5s}$ = - 8.89 $m/s^{2}$

7 0

3 years ago

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