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2 years ago

Does iron oxide have a chemical reaction? If so, can you describe it?

Chemistry

1 answer:

SVEN [57.7K]2 years ago

5 0

Answer:

The oxygen and water in air react with the metal to form the hydrated oxide. The familiar red form of rust is (Fe2O3), but iron has other oxidation states, so it can form other colors of rust

Explanation:

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Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T

shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

C = $\frac{P}{K_{h}}$

where, pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at $25^{o}C$ are as follows.

$p(O_{2})$ = 0.21 atm = 0.21 bar

$p(N_{2})$ = 0.78 atm = 0.78 bar

$K_{h}$ for $O_{2}$ = $7.9 \times 10^{2} bar/mol$

$K_{h}$ for $N_{2}$ = $1.6 \times 10^{3} bar /mol$

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

$C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L$

$C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L$

Now, we will calculate the number of moles as follows.

$n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3$ mol

$n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3$ mol

As the molar mass of $O_2$ = 32 g/mol

Hence, mass of oxygen will be as follows.

Mass of $O_2 = 32 \times 1.995 \times 10^-3 \times 1000$

= 63.84 mg

As the molar mass of $N_{2}$ = 28

Mass of $N_{2} = 28 \times 3.66 \times 10^-3$ = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg and nitrogen is 102.5 mg.

5 0

3 years ago

Read 2 more answers

8.310x10^2 – 7.210x10^1<br><br>[?]x10^[?]

alisha [4.7K]

Answer:

$7.589\times 10^{2}$

Explanation:

The expression can be solved mathematically as follows:

1) $8.310\times 10^{2}-7.210\times 10^{1}$ Given

2) $8.310\times 10^{1+1} - 7.210\times 10^{1}$ Definition of sum

3) $(8.310\times 10^{1})\times 10^{1}-7.210\times 10^{1}$ $a^{m+n} = a^{m}\cdot a^{n}$ /Associative property

4) $(8.310\times 10^{1}-7.210)\times 10^{1}$ Distributive property

5) $(83.100-7.210)\times 10^{1}$ Multiplication

6) $75.89\times 10^{1}$ Subtraction.

7) $(7.589\times 10^{1})\times 10^{1}$ Multiplication/Associative property

8) $7.589\times (10^{1}\times 10^{1})$ Associative property

9) $7.589\times 10^{2}$ $a^{m+n} = a^{m}\cdot a^{n}$ /Result

7 0

2 years ago

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

4 0

2 years ago

Write a balanced net ionic equation for the following reaction: SrCl2(aq) + H2CO3(aq) → SrCO3(s) + HCl (aq)

max2010maxim [7]

We will balance the equation in the following order: metals, amethals, carbon, hydrogen and oxygen (the most common order).
The metal present in the equation is Sr, which is already balanced (there are 1 on each side of the equation).
The amethal present in the equation is Cl. There is 2 Cl in the left side and only one in the right side. So, we will multiply the quantity of the molecule that contains Cl by 2. Doing this, we'll obtain:

$SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}$

Looking at the equation, we can see that it is now fully balanced. Hence, a balanced equation of the reaction is:

$SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}$

5 0

2 years ago

What is the molar mass of fluorine gas, F2?

Paladinen [302]

Answer:

37.99681 g/mol

Explanation:

Hope it helps!

6 0

3 years ago

Read 2 more answers