An unknown element has two isotopes: one whose mass is 68.926 amu (60.00% abundance) and the other whose mass is 70.925 amu (40.00% abundance). The average atomic mass of the element is equal to .

69.726 amu

Explanation:

From the question given above, the following data were obtained:

Isotope A:

Mass of A = 68.926 amu

Abundance of A (A%) = 60.00%

Isotope B:

Mass of B = 70.925 amu

Abundance of B (B%) = 40.00%

Average atomic mass =?

The average atomic mass of the element can be obtained as shown below:

Average atomic mass

= [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

= [(68.926 × 60)/100] + [(70.925 × 40)/100]

= 41.3556 + 28.37

= 69.726 amu

Therefore, the average atomic mass of the element is 69.726 amu

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: