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Vlad1618 [11]
4 years ago
5

Calculate the change in enthalpy for the following reaction, using the bond enthalpy data provided below. (Note: you may round t

he bond energies to the nearest 100 kJ/mol during your calculations.) 2 N2H2(g) 3 O2(g)
Chemistry
1 answer:
aliina [53]4 years ago
3 0

Answer:

-514 kJ/mol

Explanation:

The bond enthalpy which is also known as bond energy can be defined as the amount of energy needed to split one mole of the stated bond. The change in enthalpy of a given reaction can be estimated by subtracting the sum of the bond energies of the reactants from the sum of the bond energies of the products.

For the given chemical reaction, the change in enthalpy of the reaction is:

ΔH_{rxn} = [2(409) + 4(388) + 3(496) - 4(630) - 4(463)] kJ/mol = 818 + 1552 + 1488 - 2520 - 1852 = -514 kJ/mol

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A cube of an unknown metal measures 0.200 cm on one side. The mass of the cube is 52 mg. Which of the following is most likely t
Margarita [4]

Answer:

Option E. Zirconium

Explanation:

From the question given above, the following data were obtained:

Length of side (L) of cube = 0.2 cm

Mass (m) of cube = 52 mg

Name of the unknown metal =?

Next, we shall determine the volume of the cube. This can be obtained as follow:

Length of side (L) of cube = 0.2 cm

Volume (V) of the cube =?

V = L³

V = 0.2³

V = 0.008 cm³

Next, we shall convert 52 mg to g. This can be obtained as follow:

1000 mg = 1 g

Therefore,

52 mg = 52 mg × 1 g / 1000 mg

52 mg = 0.052 g

Thus, 52 mg is equivalent to 0.052 g.

Next, we shall determine the density of the unknown metal. This can be obtained as follow:

Mass = 0.052 g.

Volume = 0.008 cm³

Density =?

Density = mass / volume

Density = 0.052 / 0.008

Density of the unknown metal = 6.5 g/cm³

Comparing the density of the unknown metal i.e 6.5 g/cm³ with those given in table in the above, we can conclude that the unknown metal is zirconium

7 0
3 years ago
(chem) which is more concentrated: 45.0 grams of HCOOH dissolved in 189 mL of water or 1.5 moles of CH↓2COOH dissolved in twice
Liula [17]

Answer:

CH3COOH would be more concentrated

Explanation:

The higher the concentration value, the more concentrated it is.

The relationship between concentration, moles and volume is given by the equation;

Concentration = No of moles / Volume

5.0 grams of HCOOH dissolved in 189 mL of water

Number of moles = Mass / Molar mass = 5 / 46.03 = 0.1086 mol

Concentration = 0.1086 / 0.189 = 0.5746 mol/L

1.5 moles of CH3COOH dissolved in twice as much water

Volume = 2 * 189 = 378 ml = 0.378 L

Concentration = 1.5 / 0.378 = 3.9683 mol/L

Comparing both concentration values;

CH3COOH would be more concentrated

6 0
3 years ago
4 NH3 + 6 NO → 5 N2 + 6 H2O How many moles of NH3 are necessary to produce 0.824 mol N2?
GrogVix [38]
4 mol NH₃ → 5 mol N₂
x mol NH₃ → 0.824 mol N₂

x=0.824*4/5=0.6592 mol
6 0
4 years ago
What is the theoretical yield of NaBr
dolphi86 [110]

The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles

<h3>Balanced equation </h3>

2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

<h3>How to determine the theoretical yield of NaBr</h3>

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

Therefore,

2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr

Therefore,

Thus, the theoretical yield of NaBr is 7.08 moles

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

7 0
2 years ago
An
Pavel [41]
I think it is c but I’m not sure
6 0
3 years ago
Read 2 more answers
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