Pollution comes from<br> many small sources.

Consider the following reaction:C2H4(g) + F2(g) -----------> C2H4F2(g) Delta H = -549 kJEstimate the carbon-fluorine bond ene

frutty [35]

Answer:

Bond energy of carbon-fluorine bond is 485 kJ/mol

Explanation:

Enthalpy change for a reaction, is given as:

$\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]$

Where $(E_{bond})_{i}$ and $(E_{bond})_{j}$ represents average bond energy in breaking "i" th bond and forming "j" th bond respectively. $n_{i}$ and $n_{j}$ are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

So, $-549kJ=(1mol\times 614kJ/mo)+(4mol\times E_{C-H})+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(4mol\times E_{C-H})-(2mol\times E_{C-F})$

or, $-549kJ=(1mol\times 614kJ/mo)+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(2mol\times E_{C-F})$

or, $E_{C-F}=485kJ/mol$

So bond energy of carbon-fluorine bond is 485 kJ/mol

8 0

3 years ago

A rotameter calibration curve (flow rate versus float position) obtained using a liquid is mistakenly used to measure a gas flow

Veronika [31]

Answer:

I would expect the gas rate determined in this manner to be too low

Explanation:

A Rotameter can be designed to respond to the sensitivity of density, velocity, to measure the flow rate of liquid or gas enclosed in a tube. Liquids are denser than gas, and since the gas rate to be determined needed to respond to the velocity head alone of the rotameter so as to bring the forces in the tube equilibrium. Knowing if there is no flow, then the float would remain at the bottom, so gas has to flow at a higher rate compared to the liquid so the float would be in a similar position making it easier to measure the flowrate. This leaves the gas rate to be determined too low.

5 0

2 years ago

A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is

aliina [53]

Answer:

$\large \boxed{\text{67 000 g}}$

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂ = 0

The formula for the heat absorbed or released by an object is

q = mCΔT, where

m = the mass of the sample

C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

q₁ + q₂ = 0

q₁ + m₂C₂ΔT₂ = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹; T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

$q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}$

(b) ΔT

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

$\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\$

$\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}$

7 0

3 years ago

Which factor below does not affect how fast a solute dissolves in a solvent?

Anit [1.1K]

Stirring this is because the three elements are factors affecting dissolving of a solvent. Eg temprature affects in hotness or coldness, Particle size affects whether it is big or small while quantity of soluble affects by the amount

3 0

3 years ago

Read 2 more answers

If a balloon is filled with a mixture of helium and oxygen, which gas will escape faster? Why?

Serjik [45]

Answer:

The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s

Explanation:

Step 1: Data given

Molar mass helium = 4.0 g/mol

Molar mass O2 = 32 g/mol

Step 2: Graham's law

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of the molecular mass : 1/(Mr)^0.5

Rate of escape for He = 1/(4.0)^0.5 = 0.5

Rate of escape for O2 = 1/(32)^0.5 = 0.177

The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s

4 0

3 years ago

Pollution comes from many small sources.