Below left is a cylinder containing water. an ball with a mass of 21g and a volume of 15 cm3 is lowered into the water. sketch the object and the new water level in the cylinder on the right.

You didn’t show the cylinder containing water, so I created one that you can use as a model (see image).

The water level was originally at 37 mL.

Then you added the ball, and it displaced its volume of water.

The new volume reading is 52 mL, so

Volume of ball = volume of displaced water = 52 mL – 37 mL = 15 mL.

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:

h