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2 years ago

5

The compound MgSO4 , Epsom salt, is used to soothe sore feet and muscles. How many grams will you need to prepare a bath contain

ing 6.20 moles of Epsom salt?

1 answer:

natima [27]2 years ago

3 0

Answer:

the value of the experiment with you and your family are doing well and that's what I was thinking took a break

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Which applies to latitudes father from the equator

konstantin123 [22]

Higher elevations...

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3 years ago

Which of the following is an example of a physical change?

MatroZZZ [7]

Answer:

A

Explanation:

because you can turn the melted iron back to the solid it was before

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3 years ago

The density of a block of wood with a volume of 50 cubic centimeters and a mass of 100 grams is?

german

2 g/cm³ is the density of the block of wood

Explanation:

To determine density of the block of wood we use the following formula:

density = mass / volume

density of the wood block = 100 g / 50 cm³

density of the wood block = 2 g/cm³

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density

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7 0

3 years ago

Which of these following terms are made of matter and which or not made of matter: air,light,radio,magnetic field,car,radio wave

viva [34]

Matter: radio, car, flashlight, textbook, human

Not matter: everything besides the matter items

3 0

3 years ago

Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar

kakasveta [241]

Answer:

The answer is " $\bold{0.525\ \ atm^{-1}}$ "

Explanation:

Given equation:

$2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$

Given value:

$\Delta rH =-198.2 \ \ \frac{KJ}{mol}$

$Kp=1100 \ K$

$\Delta x = 2-(2+1)\\\\$

$= 2-(2+1)\\\\= 2-(3)\\\\= -1$

$\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right$

calculating the total pressure on equilibrium= $(1-2x)+(0.5-x)+2x \ atm\\\\$

$= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\$

$\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\$

calculating the pressure in $So_2$ :

$= (1-2 \times 0.15)$

$= 1-0.30 \\\\ =0.70 \ atm$

calculating the pressure in $O_2$ :

$= (0.5- 0.15)\\\\= 0.35 \ atm \\$

calculating the pressure in $So_3$ :

$= (2 \times 0.15)\\\\= (.30) \ atm \\\\$

Calculating the Kp at 1100 K:

$= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\$

4 0

3 years ago