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harkovskaia [24]
3 years ago
10

Rahul burnt 3kg of fire wood got only 500 g of ashes and says that Law of conservation of mass is not valid do you agree or disa

gree
Chemistry
1 answer:
N76 [4]3 years ago
3 0

Answer:

disagree

Explanation:

Cellulose (main component of wood beside lignin) is composed from carbon, hydrogen and oxygen with the following molecular formula  (C₆H₁₀O₅)ₓ (where x is the degree of condensation of beta-glucose units which are the basic monosaccharide of the cellulose polysaccharide).

Now when you burn cellulose you obtain carbon (ashes), carbon dioxide (CO₂) and water  (H₂O). Of course, beside the main products we may have other by-products but we consider them in very low quantity.

The difference between 3 kg of starting wood and 500 g of ashes is representing the quantity of carbon dioxide (CO₂)  and water (H₂O) formed.

We have to say that CO₂ and H₂O are in gaseous state and escape in the atmosphere.

The law of conservation of mass is always valid.

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A cloud that forms on the ground is called fog. Some clouds you see in the sky might be from airplanes. These are called contrails. High level cirrus clouds may travel at speeds up to 100 mph.
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3 years ago
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Enter the oxidation number of one atom of each element in each reactant and product.
Anton [14]

Answer:

1. -4

2. +1

3. 0

4. +4

5. -2

6. +1

7. -2

reduced = H

oxidized = O

Explanation:

Know oxidation rules.

- Hope this helped! Please let me know if you would like to learn this. I could show you the rules and help you work through them.

7 0
3 years ago
Using the atomic masses and relative abundance of the isotopes of nitrogen given below, determine the average atomic mass of nit
mylen [45]
The formula for average atomic mass is :

mass of isotope A * % of isotope A + mass of isotope B * % of isotope B + ....

Now,
Here,
Average atomic mass of nitrogen = (14.003 * 99.63%) + (15.000 * 0.37%)
                                                 = (14.003 * 0.9963) + ( 15.000 * 0.0037)
                                                 = 13.951 + 0.056
                                                 = 14.007 a.m.u.                                                 
3 0
3 years ago
In performing the "STEP TEST" for 3 minutes, what was your score ?
slava [35]

Answer:

78

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That's my score when I did a 3-minute Step test

8 0
3 years ago
At a particular temperature, K = 4.1 ✕ 10−6 for the following reaction. 2 CO2(g) 2 CO(g) + O2(g) If 2.3 moles of CO2 is initiall
mestny [16]

Answer:

concentration of [O_2] = 0.0124 = 12.4 ×10⁻³ M

concentration of [CO] = 0.0248 = 2.48 ×10⁻² M

concentration of [CO_2] = 0.4442 M

Explanation:

Equation for the reaction:

2CO_2_{(g)                ⇄          2CO_{(g)       +       O_2_{(g)

Concentration of   CO_2_{(g) = \frac{2.3}{4.9}  = 0.469

For our ICE Table; we have:

                       2CO_2_{(g)                ⇄          2CO_{(g)       +       O_2_{(g)

Initial                 0.469                              0                           0

Change              - 2x                                +2x                      +x

Equilibrium       (0.469-2x)                       2x                         x

K = \frac{[CO]^2[O]}{[CO_2]^2}

K = \frac{[2x]^2[x]}{[0.469-2x]^2}

4.1*10^{-6}=\frac{2x^3}{(0.469-2x)^2}

Since the value pf K is very small, only little small of  reactant goes into product; so (0.469-2x)² = (0.469)²

4.1*10^{-6} = \frac{2x^3}{(0.938)}

2x^3 =3.8458*10^{-6

x^3 =\frac{3.8458*10^{-6}}{2}

x^3=1.9229*10^{-6

x=\sqrt[3]{1.9929*10^{-6}}

x = 0.0124

∴ at equilibrium; concentration of  [O_2] = 0.0124 = 12.4 ×10⁻³ M

concentration of [CO] = 2x  = 2 ( 0.0124)

= 0.0248

= 2.48 ×10⁻² M

concentration of [CO_2] = 0.469-2x

= 0.469-2(0.0124)

= 0.469 - 0.0248

= 0.4442 M

3 0
3 years ago
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