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3 years ago

HELP PLEASE! I will make you brainlist IF YOU CAN ANSWER ALL OF THE QUESTIONS! (Read the directions)

Chemistry

1 answer:

Oliga [24]3 years ago

3 0

Explanation:

I would love to help but im not able to find where the questions are can you help with that?

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How much heat is required to vaporize 25g of water at 100*c

ElenaW [278]

Answer:

Heat required = 13,325 calories or 55.75 KJ.

Explanation:

To convert a water to steam at 100 degree celsius to vapor, we have to give latent heat of vaporization to water

Which equals ,

Q = mL,

Where, m is the mass of water present

L = specific latent heat of vaporization

Here , m= 25 gram

L equals to 533 calories (or 2230 Joules)

So, Q = 25×533 = 13,325 Calories

Or , Q = 55,750 Joules = 55.75 KJ

so, Heat required = 13,325 calories or 55.75 KJ.

4 0

3 years ago

Which of the following sets of quantum numbers is NOT allowed? a. n = 5, l= 4, ml= –2, ms = +1/2 b. n = 2, l = 1, ml= 0, ms = +1

klemol [59]

Answer:

Explanation:

the n value must always be greater than the l value

8 0

3 years ago

A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to

melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0

3 years ago

The gas phase decomposition of sulfuryl chloride at 600 K SO2Cl2(g)SO2(g) + Cl2(g) is first order in SO2Cl2. During one experime

krok68 [10]

Answer: The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample = 559 min

a = let initial amount of the reactant = $2.83\times 10^{-3}$

a - x = amount left after decay process = $3.06\times 10^{-4}$

$559min=\frac{2.303}{k}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=\frac{2.303}{559}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=3.96\times 10^{-3}min^{-1}$

The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

6 0

3 years ago

Isopentyl acetate is an ester that smells like bananas. As such, it is often referred to as banana oil. Benzocaine, also an este

miskamm [114]

The answer is false

7 0

3 years ago

HELP PLEASE! *I will make you brainlist IF YOU CAN ANSWER ALL OF THE QUESTIONS!* (Read the directions)

HELP PLEASE! I will make you brainlist IF YOU CAN ANSWER ALL OF THE QUESTIONS! (Read the directions)