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lesantik [10]
3 years ago
10

How many moles of ethylene (C2H4) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin? C2H4(g) + 3O2(g) y

ields 2CO2(g) + 2H2O(g) 0.21 moles 0.32 moles 0.63 moles 0.84 moles
Chemistry
1 answer:
7nadin3 [17]3 years ago
5 0
1) Convert 12.9 liters of Oxygen to mol at the given conditions:

PV = nRT ⇒ n = PV/RT

n = [1.2atm*12.9 l] / [0.082 atm l /K mol * 297K]

n = 0.636 mol of O2

2) use the stoichiometry derived from the balanced chemical equation

 1mol C2H4 / 3  mol O2 =  x mol C2H4 / 0.636 mol O2

x = 0.636 / 3 mol O2 = 0.212 mol O2.

Answer: 0.212 mol O2
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Heat flow is the ________of energy from a waemer object to a cooler one. whats the blank??​
Romashka-Z-Leto [24]

Answer:

transfer

Explanation:

as heat flows from one object to another, the first object loses the heat(energy), while the second object gains heat(energy).

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3 years ago
50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the
trapecia [35]

Answer:

-21 kJ·mol⁻¹  

Explanation:

Data:

                    H₃O⁺ +  OH⁻ ⟶ 2H₂O

       V/mL:    50         50  

c/mol·dm⁻³:   1.0         1.0

     

ΔT = 4.5 °C  

       C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

     nΔH   +         mCΔT       + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 +   50×4.5  = 0

0.050ΔH +          1883        +      225    = 0

                                  0.050ΔH + 2108 = 0

                                              0.050ΔH = -2108

                                                        ΔH = -2108/0.0500

                                                              = -42 000 J/mol

                                                              = -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

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Any element above 92 on the
Fynjy0 [20]
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explanation

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