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3 years ago

Major product of free radical of chlorination of propene

Chemistry

1 answer:

tia_tia [17]3 years ago

6 0

Answer:

For example, propene reacts with aqueous chlorine to form 1-chloro-2-propanol as the major product.

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What type of climate conditions are needed to form coal deposits.

guajiro [1.7K]

Answer:

Subbituminous coal can form at temperatures as low as 35 to 80 °C (95 to 176 °F) while anthracite requires a temperature of at least 180 to 245 °C (356 to 473 °F).

Sub-types: Cannel coal

Child material class: Lignite

Explanation:

6 0

3 years ago

In the laboratory you dissolve 19.4 g of potassium acetate in a volumetric flask and add water to a total volume of 125 mL. What

aleksandrvk [35]

The molarity of the potassium acetate solution given the data is 1.584 M

<h3>What is molarity? </h3>

This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

<h3>How to determine the mole of CH₃COOK</h3>

Mass of CH₃COOK = 19.4 g
Molar mass of CH₃COOK = 98 g/mol
Mole of CH₃COOK =?

Mole = mass / molar mass

Mole of CH₃COOK = 19.4 / 98

Mole of CH₃COOK = 0.198 mole

<h3>How to determine the molarity of CH₃COOK</h3>

Mole of CH₃COOK = 0.198 mole
Volume = 125 mL = 125 / 1000 = 0.125 L
Molarity of CH₃COOK = ?

Molarity = mole / Volume

Molarity of CH₃COOK = 0.198 / 0.125

Molarity of CH₃COOK = 1.584 M

Learn more about molarity:

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5 0

2 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

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