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Ray Of Light [21]
3 years ago
5

Out of 500 silicon atoms: 460 are si-28, which has a mass of 27.98 amu 25 are si-29, which has a mass of 28.98 amu and 15 are si

-30, which has a mass of 29.97 amu. what is the average atomic mass of silicon? (round to 2 decimal places) (hint: they did not give you percentages, but you can calculate the percentages and then do the problem)
Chemistry
2 answers:
exis [7]3 years ago
7 0

Average mass can be calculated by taking average of mass of each isotopes considering their abundance.

First calculate the percentage of each isotope.

Si(28) is 460 out of 500, percentage will be \frac{460}{500}\times 100=92%

Si-29 is 25 out of 500, percentage will be \frac{25}{500}\times 100=5%

Si-30 is 15 out of 500, percentage will be \frac{15}{500}\times 100=3%

Formula for average mass used will be:

Average mass=(% of Si-28×mass of Si-28)+(% of Si-29×mass of Si-29)+(% of Si-30×mass of Si-30)

=(0.92×27.98) amu+(0.05×28.98) amu+(0.03×29.97) amu

=(25.74+1.45+0.899) amu

=28.09 amu

Thus, average atomic mass of silicon is 28.09 amu.

liberstina [14]3 years ago
6 0

Answer;

=28.09 amu

Explanation;

In this problem, they did not give us the percentages. However, since we know the number of atoms, we can easily calculate the percentages. For example:  

(460 X 100)/500 = 92%  

If we do this for all three isotopes,

(460 × 25)/500 = 5 %

(460 × 15) /500 = 3%

-We get 92%, 5%, and 3%. (We'll assume these are absolute numbers for determining our significant figures).

Now the problem is just like the previous one. First convert the percentages into decimals. Then multiply those decimals by the masses and add. Here's the solution:  

= (0.92) X (27.98 amu) + (0.05) X (28.98 amu) + (0.03) X (29.97 amu)

= 25.74 amu  +  1.449 amu  + 0.8991 amu

= 28.09 amu

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9.87 g of calcium sulfate and 12.05 g of potassium react. What is the total amount of potassium sulfate that can be produced?
blagie [28]

Answer:

                     Mass of K₂SO₄  =  12.633 g

Explanation:

                    The balance chemical equation for given reaction is as;

                                     CaSO₄ + 2 K = K₂SO₄ + Ca

To solve this reaction one should first find the limiting reagent. To do so e will calculate the moles of each reactant as,

Moles of CaSO₄  =  Given Mass / M.Mass of CaSO₄

Moles of CaSO₄  =  9.87 g / 136.14 g/mol

Moles of CaSO₄  =  0.0725 moles

Similarly for K,

Moles of K  =  Given Mass / A.Mass of K

Moles of K  =  12.05 g / 39.10 g/mol

Moles of K  =  0.308 moles

Now, according to equation,

             1 mole of CaSO₄ reacts with  =  2 moles of K

So,

         0.0725 moles of CaSO₄ will react with  =  X moles of K

Solving for X,

                     X  =  0.0725 moles × 2 moles / 1 mole

                      X  =  0.145 moles of K

As calculated above, we are provided with 0.308 moles of K while, we require only 0.145 moles of it so it means that K is in excess and CaSO₄ is the limiting reagent hence, CaSO₄ will control the yield of K₂SO₄.

So,

The amount of K₂SO₄ produced is calculated by first finding its moles as,

According to equation,

             1 mole of CaSO₄ produced  =  1 mole of K₂SO₄

So,

         0.0725 moles of CaSO₄ will produce  =  X moles of K₂SO₄

Solving for X,

                     X  =  0.0725 moles × 1 mole / 1 mole

                      X  =  0.0725 moles of K₂SO₄

Now convert moles of K₂SO₄ to mass as,

Mass of K₂SO₄  =  Moles × M.Mass

Mass of K₂SO₄  =  0.0725 × 174.26 g/mol

Mass of K₂SO₄  =  12.633 g

7 0
4 years ago
When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3,a yellow precipitate forms and Ag+ becomes negligibly small. Whi
jok3333 [9.3K]

Answer:

Option A

Explanation:

Number of millimoles of Na3PO4 = 1 × 100 = 100

Number of millimoles of AgNO3 = 1 × 100 = 100

When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion

When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-

As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible

Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-

And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-

∴ Increasing order of concentration will be  PO43- < NO3- < Na+

4 0
4 years ago
Read 2 more answers
Can someone help me?!!!!!
tangare [24]

Answer:

Subtract all the variables and do the opposite operation when added.

Explanation:

4 0
3 years ago
80.0 grams of NaOH is dissolved in water to make a 6.00 L solution what is the molarity of the solution
Elanso [62]

Answer:

0.33 mol/ L

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer will help you. Have a nice day !

3 0
3 years ago
If 18.8 mLmL of 0.800 M HClM HCl solution are needed to neutralize 5.00 mLmL of a household ammonia solution, what is the molar
lidiya [134]

<u>Answer:</u> The molar concentration of ammonia is 3.008 M

<u>Explanation:</u>

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NH_3

We are given:

n_1=1\\M_1=0.800M\\V_1=18.8mL\\n_2=1\\M_2=?M\\V_2=5.00mL

Putting values in above equation, we get:

1\times 0.800\times 18.8=1\times M_2\times 5.00\\\\x=\frac{1\times 0.800\times 18.8}{1\times 5.00}=3.008M

Hence, the molar concentration of ammonia is 3.008 M

3 0
3 years ago
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