All categories

3 years ago

You give a sample containing .90 moles of a substance

1 answer:

3 0

1 mole Cu ------------------ 6.02x10²³ atoms
0.90 moles Cu ------------ ??

0.90 x ( 6.02x10²³ ) / 1 =

= 5.41x10²³ atoms

You might be interested in

The doctor ordered 750ml of a 15 percent solution. the stock solution is 50 percent. how much stock solution and diluent are nee

garik1379 [7]

The doctor required a 15% 750mL solution. To lower the concentration of the solution, you have to add more diluting liquid.

750mL * 0.15 = 112.5mL. This 112.5mL signifies the pure chemical or compound in the solution. This 112.5mL can be taken from the stock solution.

112.5 * 2 = 250mL of stock solution would be needed and 500 mL of the diluting agent should be added. Adding the two together would result to a 750mL solution of 15% concentration.

3 0

3 years ago

1. How many moles of iron is needed to react with sulfur in order to produce 8.6 moles

AleksAgata [21]

Answer:

17.2 moles of Iron are required

Explanation:

Based on the chemical reaction:

2Fe + 3S → Fe₂S₃

2 moles of Fe react with 3 moles of S to produce 1 mole of Iron (III) sulfide

Assuming and excess of sulfur, if we want to obtain 8.6 moles of Fe₂S₃ are required:

8.6 moles Fe₂S₃ * (2 moles Fe / 1mol Fe₂S₃) =

<h3>17.2 moles of Iron are required</h3>

7 0

3 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$