Quick help I really need help

Pls help check dhe paper

Semmy [17]

A= 16.5x100=1650
b=120 divided by 100= 1.2lb
that’s all I know.

7 0

2 years ago

Read 2 more answers

A molar solution is a solution that contains one mole of solute in

BartSMP [9]

Answer:

1 liter of the solution

Explanation:

Molarity or concentration of a solution refers to the number of moles of solute contained in a volume of 1 liter or 1 dm³ solution.
Therefore, the concentration of a solution is calculated by dividing the number of moles of a solute by the volume of the solution.
Concentration is given in mol/L or mol/dm³.
Hence, a molar solution is a solution that contains 1 mole of a solute in 1 liter of the solution. This is equivalent to 1M or 1 mol/L.

7 0

3 years ago

You are looking at two samples of iron. One sample is very small and came from the red soil on the planet Mars. The other sample

aivan3 [116]

The two samples don’t contain different atoms so it will be false

3 0

3 years ago

What is the molar mass of C2H6O

Neporo4naja [7]

46.0684g I think because you have to concert grams to moles or moles to grams

5 0

3 years ago

If 13.2 kg of Al 2 O 3 ( s ) , 50.4 kg of NaOH ( l ) , and 50.4 kg of HF ( g ) react completely, how many kilograms of cryolite

Elodia [21]

Answer : The mass of cryolite produced will be, 54.38 kg

Solution : Given,

Mass of $Al_2O_3$ = 13.2 kg = 13200 g

Mass of $NaOH$ = 50.4 kg = 50400 g

Mass of $HF$ = 50.4 kg = 50400 g

Molar mass of $Al_2O_3$ = 101.9 g/mole

Molar mass of $NaOH$ = 40 g/mole

Molar mass of $HF$ = 20 g/mole

Molar mass of $Na_3AlF_6$ = 209.9 g/mole

First we have to calculate the moles of $Al_2O_3,NaOH$ and $HF$ .

$\text{ Moles of }Al_2O_3=\frac{\text{ Mass of }Al_2O_3}{\text{ Molar mass of }Al_2O_3}=\frac{13200g}{101.9g/mole}=129.54moles$

$\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{50400g}{40g/mole}=1260moles$

$\text{ Moles of }HF=\frac{\text{ Mass of }HF}{\text{ Molar mass of }HF}=\frac{50400g}{20g/mole}=2520moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Al_2O_3+6NaOH+12HF\rightarrow 2Na_3AlF_6+9H_2O$

From the balanced reaction we conclude that

The mole ratio of $Al_2O_3,NaOH$ and $HF$ is, 1 : 6 : 12

And, the ratio of given moles of $Al_2O_3,NaOH$ and $HF$ is, 129.54 : 1260 : 2520

From this we conclude that, $NaOH,HF$ is an excess reagent because the given moles are greater than the required moles and $Al_2O_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Na_3AlF_6$

From the reaction, we conclude that

As, 1 mole of $Al_2O_3$ react to give 2 mole of $Na_3AlF_6$

So, 129.54 moles of $Al_2O_3$ react to give $129.54\times 2=259.08$ moles of $Na_3AlF_6$

Now we have to calculate the mass of $Na_3AlF_6$

$\text{ Mass of }Na_3AlF_6=\text{ Moles of }Na_3AlF_6\times \text{ Molar mass of }Na_3AlF_6$

$\text{ Mass of }Na_3AlF_6=(259.08moles)\times (209.9g/mole)=54380.892g=54.38kg$

Thus, the mass of cryolite produced will be, 54.38 kg

5 0

3 years ago