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3 years ago

What are the physical and chemical properties of iron oxide

1 answer:

5 0

Answer: The physical properties of iron oxide are that it is an odorless, red-brown solid, and that it is insoluble. Depending on the chemical formula, one of the chemical properties is its combustibility. If the chemical formula is Fe2O3, it is not combustible. If the formula is FeO, it is readily combustible.

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A mixture of He He , N 2 N2 , and Ar Ar has a pressure of 24.1 24.1 atm at 28.0 28.0 °C. If the partial pressure of He He is 301

Karo-lina-s [1.5K]

Answer:

16.5 atm

Explanation:

A mixture of He, N₂, and Ar has a pressure of 24.1 atm at 28.0 °C. If the partial pressure of He is 3013 torr and that of Ar is 2737 mm Hg, what is the partial pressure of N₂?

The total pressure of a gaseous mixture is equal to the sum of the partial pressures.

P = pHe + pN₂ + pAr

pN₂ = P - pHe - pAr [1]

We need to express pHe and pAr in atm.

$pHe=3013torr.\frac{1atm}{760torr} =3.96atm$

$pAr=2737mmHg.\frac{1atm}{760mmHg} =3.60atm$

From [1],

pN₂ = 24.1 atm - 3.96 atm - 3.60 atm = 16.5 atm

8 0

3 years ago

Which source of energy causes most of the water evaporation on Earth’s surface?

coldgirl [10]

Which adds the most energy to the earth? Solar. So Solar will cause the most evaporation (B)

3 0

3 years ago

Read 2 more answers

Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2

PIT_PIT [208]

Answer: The standard heat for the given reaction is -138.82 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ$

Hence, the standard heat for the given reaction is -138.82 kJ

3 0

3 years ago

What are the resulting coefficients when you balance the chemical equation for the combustion of ethane, c2h6? in this reaction,

Vitek1552 [10]

Balance equation for combustion of ethane will be:

2C₂H₆(g) + 7O₂(g)--------> 4CO₂(g) + 6H₂O(g)

To balance the equation:

1. Balance the number of carbon atom on both side:

C₂H₆(g) + O₂(g)--------> CO₂(g) + H₂O(g)

1. balance the number of carbon on both side, as in reactant there are 2 but in product one,

so , multiply the CO₂, by 2 in the product.

2. Balance the number of hydrogen on both side as in reactant the number of hydrogen is 3 but in product it is 6 so, multiplythe number of H₂O by 3,

so multiply the number of H₂O by 3 in product.

3. Balance the number of oxygen on both side , as 1 and 2 step increases the number of oxygen and it becomes 7 , so to balance the number of oxygen on both side by mutiplying the number of O₂ by 7/2 in reactant .

4. Now, doubling the equation will give balance equation that is:

2C₂H₆(g) + 7O₂(g)--------> 4CO₂(g) + 6H₂O(g)

8 0

3 years ago

A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de

jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

3 0

4 years ago