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3 years ago

How does the average reaction rate differ from an instantaneous reaction rate?

Chemistry

1 answer:

____ [38]3 years ago

7 0

Answer:

The average rate is the change in concentration over a selected period of time. It depends on when you take the measurements. The instantaneous rate is the rate at a particular time.

Explanation:

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topjm [15]

I like that game Forza horizon

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3 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago

A sample of glucose ( C6H12O6 ) of mass 8.44 grams is dissolved in 2.11 kg water. What is the freezing point of this solution? T

taurus [48]

Answer:

- 0.0413°C ≅ - 0.041°C (nearest thousands).

Explanation:

Adding solute to water causes the depression of the freezing point.

We have the relation:

ΔTf = Kf.m,

Where,

ΔTf is the change in the freezing point.

Kf is the freezing point depression constant (Kf = 1.86 °C/m).

m is the molality of the solution.

Molality is the no. of moles of solute per kg of the solution.

no. of moles of solute (glucose) = mass/molar mass = (8.44 g)/(180.156 g/mol) = 0.04685 mol.

∴ molality (m) = no. of moles of solute/kg of solvent = (0.04685 mol)/(2.11 kg) = 0.0222 m.

∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.

∴ The freezing point of the solution = the freezing point of water - ΔTf = 0.0°C - 0.0413°C = - 0.0413°C ≅ - 0.041°C (nearest thousands).

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3 years ago

Group viia elements are very active non metals give reason

Marianna [84]

Answer:

because they need only one electron

4 0

3 years ago

What is the formula for the ionic compound formed by magnesium and iodine?

MrMuchimi

The compound that is formed is: MgI2

7 0

3 years ago