Ask question

Ask question

All categories

3 years ago

15

Consider the balanced chemical equation

1 answer:

Akimi4 [234]3 years ago

8 0

Answer:

A

Explanation:

The coefficents are used when determining mole ratio. So just look in front of O and B

the ratio is 3/4

you cna use this in stoich as well

You might be interested in

PLEASE!! HELP!!

german

Answer:

After 15 minutes,the volume of the balloon stops expanding

7 0

3 years ago

Plz answer quickly! :))<br><br><br> How are the shapes of covalent compounds made?

Elenna [48]

Hope this might help a little

8 0

4 years ago

Can a book sitting on a desk represent each of Newton's Laws?

pashok25 [27]

Answer:

1. Yes

Explanation:

because the book is sitting on the desk it's not in motion or movement

4 0

3 years ago

Why does water roll off of dry ice?

ziro4ka [17]

It's a combination of carbon dioxide gas from the dry ice, and tiny water droplets.) ... The dry ice fog is heavier (denser) than air, so it sinks down. But it's less dense than water, so it floats on the water. The bubbles are heavier than air, but less dense than carbon dioxide fog, so they'll float on the fog.

8 0

4 years ago

In some regions of the southwest United States, the water is very hard. For example, in Las Cruces, New Mexico, the tap water co

myrzilka [38]

Answer:

a) $46.734L$ of hard water is require to produce $45L$ of pure water.

b) Same water would not work for purifying sea water as it would require so much of energy.

Explanation:

a)

Osmotic pressure of a solution is calculated by the formula,

Π= $iMRT$

Where,

Π is the osmotic pressure

$i$ is van't hoff factor

$M$ is molar concentration (mol/L)

$R$ is gas constant

$T$ is Temperature in Kelvin

Now,

$MgCO_{3}\rightarrow Mg^{2+}+ CO_{3}^{2-}$

So, according to the above equation $i=2$

Calculation concentration at which reverse osmosis will stop

$M=\frac{ Π }{iRT}$

$M=\frac{8atm}{2\times0.082L atm mol^{-1}K^{-1} \times 300K} \\M= 0.162M$

Calculating Initial concentration that is moles of $MgCO_{3}$

$n=\frac{given mass}{molar mass} \\\\n=\frac{560 \times 10^{-6}g}{84.32g/mol}\\\\N=6.642\times10^{-6}moles$

Calculating molarity

$M= 6.642 \times 10^{-6}\times 10{^3}\\M=6.642\times10^{-3}M$

Total number of moles in total volume must remain the same,

$C_{1}V_{1}=C_{2}V_{2}$

$0.006642M \times V_{1}=0.162M \times V_{2}$

$\frac{V_{1}}{V_{2}}= \frac{0.162}{0.006.642}=25.953$

$V_{1}=25.953 V_{2}$

Also, $V_{1}=45L$

$V_{2}=\frac{45}{25.953}\\ V_{2}= 1.734L$

So, net Volume,

$V=45L+1.734L=46.734L$

So, $46.734L$ is require to produce $45L$ of pure water.

b)

Using the equation for $0.60M NaCl$ as well,

$C_{1}V_{1}=C_{2}V_{2}$

$C_{1}=0.60M\\V_{1}=45L\\C_{2}=0.162M\\V_{2}=?$

$V_{2}=\frac{0.60 \times 45}{0.167}\\V_{2}= 166.66L$

$Net water=166.66+45L=211.66L$

So, the same water would not work for purifying sea water as it would require so much of energy, as we require 211.66L of water to produce 45L of water.

7 0

3 years ago