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3 years ago

Describe step by step how we can make a field of a variable quantity?

Chemistry

1 answer:

just olya [345]3 years ago

7 0

Finding the variables in the experiment or study and measurinf

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A gas mixture contains 0.800 mol of N2, 0.200 mol of H2, and 0.150 mol of CH4. What is the mole fraction of H2 in the mixture?

Blababa [14]

Answer:

0.1739

Explanation:

0.800 mol of N2

0.200 mol of H2

0.150 mol of CH4

Total moles of the mixture = 0.8 + 0.2 + 0.150 = 1.150 mol

Mole fraction of H2 = Number of moles of H2 / Total moles

Mole Fraction = 0.2 / 1.150 = 0.1739

4 0

3 years ago

Use the periodic table to find the number of valence electrons for calcium (Ca) aluminum (Al) rubidium (Rb) oxygen (O) sulfur (S

fgiga [73]

Answer:

Calcium (Ca) => 2

Aluminium (Al) => 3

Rubidium (Rb) => 2

Oxygen (O) => 2

Sulphur (S) => 2

Iodine (I) => 1

Formulae

Calcium oxide : CaO

Aluminium iodide : AlI3

Rubidium sulphide : RbS

Aluminium oxide : Al2O3

4 0

3 years ago

Determine the ΔH for the following reaction 2NH3 + 5/2O2 = 2NO(g) + 3 H2O(g)

beks73 [17]

The enthalpy change, ΔH for the following reaction $2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$ is -452.76 kJ.

<h3>What is enthalpy change, ΔH, of a reaction?</h3>

The enthalpy change of a reaction is the heat changes that occurs when a reaction proceeds to formation of products.

Enthalpy change, ΔH = ΔH of products - ΔH of reactants

The equation of the reaction is given below

$2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$

$\Delta{H_{f}\:of\:NO = 90.25 kJ; \Delta{H_{f}\:of\:H_{2}O =-241.82kJ; \Delta{H_{f}\:of\:NH_{3} =-46.1 kJ; \Delta{H_{f}\:of\:O_{2} =0$

$\Delta{H_{f}\:of\:rxn = (90.25*2)+(-241.82*3)-( -46.1*2)= -452.76\:kJ$

Therefore, the enthalpy change, ΔH for the following reaction $2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)$ is -452.76 kJ.

Learn more about enthalpy change at: brainly.com/question/14047927

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4 0

2 years ago

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What is the boiling point of a 1.5 m aqueous solution of fructose? the boiling point elevation constant of water is 0.515Â°c/m.

Ronch [10]

Answer is: he boiling point of a 1.5 m aqueous solution of fructose is 100.7725°C.
The boiling point elevation is directly proportional to the molality of the solution according to the equation: ΔTb = Kb · b.
ΔTb - the boiling point elevation.
Kb - the ebullioscopic constant. of water.
b - molality of the solution.
Kb = 0.515°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
Tb(solution) = 100°C + 0.7725°C = 100.7725°C.

4 0

4 years ago

Which of the following correctly describes a compound?

ivann1987 [24]

Answer:they are chemically bound together and they retain their individual physical and chemical properties

Explanation:

8 0

3 years ago

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