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2 years ago

What is the mass of 2.5 moles of hydrogen fluoride gas HF

Chemistry

1 answer:

Nady [450]2 years ago

7 0

Answer:

You will get 5.0 g of hydrogen.

Explanation:

As with any stoichiometry problem, we start with the balanced equation.

2HF

→

SnF

Moles of H

2.5

mol Sn

1 mol H

mol Sn

2.5 mol H

Mass of H

2.5

mol H

2.016 g H

mol H

5.0 g H

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How many cycles of the synthesis pathway are needed to produce myristic acid, c13h27cooh?

Alenkasestr [34]

The primary cycle of fatty acid synthesis produces a four-carbon acyl group. Every following cycle adds a three-carbon malonyl group, which stretches off one molecule of CO2 (one carbon atom) in a condensation response. Thus, the net number of carbon atoms additional by each following cycle is two. For the reason that the first cycle makes a four-carbon chain, and each added cycle extends the chain by two carbon atoms, the second cycle fallouts in a six-carbon chain, the third cycle fallouts in an eight-carbon chain, and so on. And so, the synthesis of the 14-carbon myristic acid necessitates six repetitions.

7 0

2 years ago

1. Complete the following Data Table as you conduct the lab

miskamm [114]

Answer:

Do to half of the mnairals this can not be made into a lab there is an error

Explanation:

4 0

2 years ago

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15. An apparatus consists of a 4.0 dm3

Solnce55 [7]

Answer:

i· Partial pressure of nitrogen gas is 219.429kPa and partial pressure of argon gas is 33.714kPa ·

ii· Total pressure of the gas mixture is 253.143kPa·

Explanation:

<h3>solution for i :</h3>

Assuming that the given gases to be ideal,

so by ideal gas equation

PV=nRT

where,

P is the pressure of the gas

V is the volume occupied by the gas

n is the number of moles of the gas

R is the ideal gas constant

T is the temperature of the gas

Actually partial pressure of each gas is the pressure of the gas exerted when it occupies complete volume

As the temperature of both gases are same, the mixing process is an isothermal process

PV=constant

Initially for nitrogen gas PV=803×4=3212

let P $x_{1}$ be the partial pressure of the nitrogen gas

(P $x_{1}$ )×14=3212

∴P $x_{1}$ =219.429kPa

∴Partial pressure of nitrogen gas is 219.429kPa

Initially for argon gas PV=47.2×10=472

let P $x_{2}$ be the partial pressure of the argon gas

(P $x_{2}$ )×14=472

∴P $x_{2}$ =33.714kPa

∴Partial pressure of argon gas is 33.714kPa

solution for ii :

Total pressure of the gas mixture will be the sum of the partial pressures of each gas as

Partial pressure of the gas=(total pressure of the mixture)×(mole fraction of the gas)

∴Total pressure=P $x_{1}$ +P $x_{2}$

=219.429+33.719

Total pressure =253.143kPa

6 0

2 years ago

Read 2 more answers

A. The measured pH of a 0.100 M HCl solution at 25 degrees Celsius is 1.092. From this information, calculate the activity coeff

ra1l [238]

Answer:

activity coefficient $\mathbf{\gamma =0.809}$

activity coefficient $\mathbf{\gamma = 0.791}$

The change in pH in part A = 0.092

The change in pH in part B = 0.102

Explanation:

From the given information:

pH of HCl solution = 1.092

Activity of the pH solution [a] = $10^{-1.092}$

[a] = 0.0809 M

Recall that [a] = $\gamma$ × C

where;

$\gamma$ = activity coefficient

C = concentration

Making the activity coefficient the subject of the formula, we have:

$\gamma = \dfrac{[a]}{C}$

$\gamma = \dfrac{0.0809 \ M}{0.100 \ M}$

$\mathbf{\gamma =0.809}$

The pH of a solution of HCl and KCl = 2.102

[a] = $10^{-2.102}$

[a] = 0.00791 M

activity coefficient $\gamma = \dfrac{0.00791 \ M}{0.01 \ M}$

$\mathbf{\gamma = 0.791}$

C. The change in pH in part A = 1.091 - 1.0 = 0.092

The change in pH in part B = 2.102 -2.00 = 0.102

3 0

2 years ago

A gas sample occupies 3.50 liters of volume at 20.°c. what volume will this gas occupy at 100.°c (reported to three significant

VLD [36.1K]

Explanation:

According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.

That is, $Volume \propto Temperature$

Hence, it is given that $V_{1}$ is 3.50 liters, $T_{1}$ is 20 degree celsius, and $T_{2}$ is 100 degree celsius.

Therefore, calculate $V_{2}$ as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

$\frac{3.50 liter}{20^{o}C} = \frac{V_{2}}{100^{o}C}$

$V_{2}$ = 17.5 liter

Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.

6 0

2 years ago

Read 2 more answers