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Alinara [238K]
2 years ago
5

suppose you find a website that describes rusting of a nail as a change to only the physical properties of iron. would it be acc

urate
Chemistry
1 answer:
Kaylis [27]2 years ago
5 0

Answer:

The website is inaccurate.

Explanation:

Remember, a chemical change is characterized by many things, but most commonly:

  1. A change in color, temperature or state.
  2. An irreversible change
  3. One or more new substances are formed

Rusting (Iron oxide) is a redox reaction (Oxidation-Reduction Reaction) where  iron molecules come in contact with water and oxygen molecules. The iron molecules lose electrons, becoming oxidized, while oxygen gains electrons, becoming reduced.

Is rust reversable? (No)

Did the colour of the iron on the surface change? (Yes)

Did a new substance form? (Yes, Read above).

Hope I helped! Have a good day!

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The density of copper is 8.96g/mL. The mass of 7.00 of copper is
Molodets [167]
7.00 what? Moles? mL?
3 0
2 years ago
When 45 g of an alloy, at 25oC, are dropped into 100.0 g of water, the alloy absorbs 956 J of heat. If the final temperature of
taurus [48]

Answer:

The answer to the question is

The specific heat capacity of the alloy = 1.77 J/(g·°C)

Explanation:

To solve this, we list out the given variables thus

Mass of alloy = 45 g

Initial temperature of the alloy = 25 °C

Final temperature of the alloy = 37 °C

Heat absorbed by the alloy = 956 J

Thus we have

ΔH = m·c·(T₂ - T₁) where  ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C  and m = mass of the alloy = 45 g

∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c  or

c = 956 J/(540 g·°C) = 1.77 J/(g·°C)

The specific heat capacity of the alloy is 1.77 J/(g·°C)

3 0
3 years ago
Iday!' Read the following phrases from the article. 1. Full-scale war 2. Braces for a full invasion 3. Hostile act 4. Defenses o
maria [59]

Answer:

c.

Explanation:

7 0
1 year ago
What is the concentration of no3- ions in a solution prepared by dissolving 15.0 g of ca(no3)2 in enough water to produce 300. m
r-ruslan [8.4K]

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻

M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol

15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2


Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol                               2 mol
0.100 mol                           0.200 mol

We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M

Concentration of NO3⁻ is 0.667 M.



4 0
3 years ago
Pentane is a small liquid hydrocarbon, between propane and gasoline in size at C5H12. The density of pentane is 0.626 g/mL and i
Goshia [24]

Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and 3.073\times 10^4kJ/L respectively.

Explanation :

Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).

As we are given that:

\Delta H^o_{comb}=-3535kJ/mol

Fuel value = \frac{\Delta H^o_{comb}}{\text{Molar mass of pentane}}

Molar mass of pentane = 72 g/mol

Fuel value = \frac{3535kJ/mol}{72g/mol}

Fuel value = 49.09 kJ/g

Now we have to calculate the fuel density of pentane.

Fuel density = Fuel value × Density

Fuel density = (49.09 kJ/g) × (0.626g/mL)

Fuel density = 30.73 kJ/mL = 3.073\times 10^4kJ/L

Thus, the fuel density of pentane is 3.073\times 10^4kJ/L

4 0
3 years ago
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