When 0.243 g of mg metal is combined with enough hcl to make 100 ml of solution in a constant-pressure calorimeter, the following reaction occurs:
mg(s) + 2 hcl(aq) → mgcl2(aq) + h2(g)

if the temperature of the solution increases from 23.0 ∘c to 34.1 ∘c as a result of this reaction, calculate δh in kj/mol of mg. assume that the solution has a specific heat of 4.18 j/g∘c.

when 0.243 of metal is combined with enough to make 100 of solution in a constant-pressure calorimeter, the following reaction occurs:

if the temperature of the solution increases from 23.0 to 34.1 as a result of this reaction, calculate in of . assume that the solution has a specific heat of 4.18 .

−19.1 kj/mol
−111 kj/mol
−191 kj/mol
−464 kj/mol
−961 kj/mol

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:

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

because of all the things that happen around the us, many people get into arguments which sadly could get physical, but we should actually make a change for that.