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Zinaida [17]
3 years ago
10

What type of change occurs when rust on a nail a. chemical change b. physical change

Chemistry
1 answer:
Korvikt [17]3 years ago
5 0

Answer:

Chemical change

Explanation:

The rusting of iron/a nail is a chemical change

Iron (Fe) and Oxygen (O) combine to create the compound Iron Oxide (Fe2O3), which is rust.

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Varvara68 [4.7K]
The first answer is B and the second answer is B
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3 years ago
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Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica
Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = d_{ideal} = \dfrac{P\times M}{RT}

here:

P = pressure of the O₂ gas = 310 bar

= 310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂  gas = 32 g/mol

∴

d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}

d_{ideal} = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol²    &

b = 0.0319  L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}

310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}

310V^3 -44.389V^2+1.382V-0.044=0

After solving;

V = 0.1152 L

∴

d_{Van \ der \ Waal} = \dfrac{32}{0.1152}

d_{Van \ der \ Waal} = 277.77  g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

5 0
3 years ago
Temperature and silica content determine the _______ of magma. A. color B. elasticity C. viscosity D. texture
Rudik [331]
Hello.

The answer is 

<span> C. viscosity.

</span>Temperature and silica content determine the viscosity of magma. 

The viscosity of amagma<span> is largely controlled by the </span>temperature.
<span>
Have a nice day.</span>
3 0
3 years ago
The chemical reaction for the formation of syngas is: CH4 + H2O -&gt; CO + 3 H2 What is the rate for the formation of hydrogen,
grin007 [14]

Answer :  The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

CH_4+H_2O\rightarrow CO+3H_2

The expression for rate of reaction :

\text{Rate of disappearance of }CH_4=-\frac{d[CH_4]}{dt}

\text{Rate of disappearance of }H_2O=-\frac{d[H_2O]}{dt}

\text{Rate of formation of }CO=+\frac{d[CO]}{dt}

\text{Rate of formation of }H_2=+\frac{1}{3}\frac{d[H_2]}{dt}

The rate of reaction expression is:

\text{Rate of reaction}=-\frac{d[CH_4]}{dt}=-\frac{d[H_2O]}{dt}=+\frac{d[CO]}{dt}=+\frac{1}{3}\frac{d[H_2]}{dt}

As we are given that:

+\frac{d[CO]}{dt}=0.35M/s

Now we to determine the rate for the formation of hydrogen.

+\frac{1}{3}\frac{d[H_2]}{dt}=+\frac{d[CO]}{dt}

+\frac{1}{3}\frac{d[H_2]}{dt}=0.35M/s

\frac{d[H_2]}{dt}=3\times 0.35M/s

\frac{d[H_2]}{dt}=1.05M/s

Thus, the rate for the formation of hydrogen is, 1.05 M/s

6 0
3 years ago
Maurels
lora16 [44]
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Explanation:

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2 years ago
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