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Shalnov [3]
3 years ago
15

What type of change produced the rust shown above? A. chemical, because a new substance with new properties was formed B. physic

al, because a new substance with new properties was formed C. physical, because the chemical makeup of the metal did not change D. chemical, because the chemical makeup of the metal did not change
Chemistry
1 answer:
Maru [420]3 years ago
6 0
Rust or oxidation is a chemical change because it creates a new substance.

Knowing that the correct answer should be (A.)

Hope this helps!
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Identify each substance as an acid or a base.
svp [43]

Explanation:

<u>The first one is a base</u>

<u>The second one is an acid</u>

A base has a pH above 7

An acid has a pH below 7

Hope I helped!!! Have a great day!

4 0
3 years ago
Read 2 more answers
What evidence supports a conservation law? 6 CO2 → C6H12O6 6 O2 6 CO2 6 H2O light → C6H12O6 6 O2 6 H2O light → C6H12O6 6 O2 6 CO
docker41 [41]

The law of conservation has been stated that the mass and energy has neither be created nor destroyed in a chemical reaction.

The law of conservation has been evident when there has been an equal number of atoms of each element in the chemical reaction.

<h3>Conservation law</h3><h3 />

The given equation has been assessed as follows:

  • \rm 6\;CO_2\;\rightarrow\;C_6H_1_2O_6

The reactant has absence of hydrogen, while hydrogen has been present in the product. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6

The number of atoms of each reactant has been different on the product and the reactant side. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6

The reactant has the presence of carbon, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;6\;CO_2\;\rightarrow\;3\;C_6H_1_2O_6\;+\;3\;O_2

The product has the presence of hydrogen, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

Learn more about conservation law, here:

brainly.com/question/2175724

4 0
2 years ago
Ammonia gas decomposes to form nitrogen and hydrogen gases. NH3(g) → N2(g) 3H2(g) If the nitrogen gas is collected in a rigid 2
sveticcg [70]

Decomposition is a chemical reaction that breaks the reactant into two or more products. Moles of nitrogen gas (\rm N_{2}) in the cylinder is 1.63 moles.

<h3>What is the ideal gas equation?</h3>

The ideal gas equation states the relation of the hypothetical ideal gas according to the pressure, volume, temperature and moles of the gas. It is given by,

\rm PV = nRT

Where,

Pressure (P) = 2000 kPa

Volume (V) = 2L

Temperature (T) = 295 K

Gas constant (R)=  0.08206

Substituting values  in the equation:

\begin{aligned} \rm n &= \rm \dfrac{PV}{RT}\\\\&= \dfrac{2000 \times (\dfrac{1}{101.325}) \times 2}{0.08206 \times 295}\\\\&= 1.63\;\rm mol\end{aligned}

Therefore, 1.63 moles are produced.

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6 0
2 years ago
The production of iron and carbon dioxide from iron(iii) oxide and carbon monoxide is an exothermic reaction. which expression c
NARA [144]
Missing question:
A. [3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s)] / 2 
<span>B. 3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s) </span>
<span>C. 26.3 kJ/1 mol Fe2O3 (s) / 3.40 mol Fe2O3 (s) </span>
<span>D. 26.3 kJ/1 mol Fe2O3 (s) – 3.40 mol Fe2O3 (s).
</span>Answer is: B.
Chemical reaction: F<span>e</span>₂O₃<span>(s) + 3CO(g) → 2Fe(s) + 3CO</span>₂<span>(g);</span>ΔH = <span>+ 26.3 kJ.
When one mole of iron(III) oxide reacts 26,3 kJ of energy is required and for 3,2 moles of iron(III) oxide 3,2 times more energy is required.</span>
8 0
3 years ago
Read 2 more answers
Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it
lions [1.4K]

<u>Answer:</u> The equilibrium concentration of COF_2 is 0.332 M

<u>Explanation:</u>

We are given:

Initial concentration of COF_2 = 2.00 M

The given chemical equation follows:

                2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)

<u>Initial:</u>          2.00

<u>At eqllm:</u>     2.00-2x          x      x

The expression of K_c for above equation follows:

K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}

We are given:

K_c=6.30

Putting values in above expression, we get:

6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M

Hence, the equilibrium concentration of COF_2 is 0.332 M

4 0
3 years ago
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