(a) what volume of 0.115 m hclo4 solution is needed to neutralize 50.00 ml of 0.0875 m naoh? (b) what volume of 0.128 m hcl is needed to neutralize 2.87 g of mg1oh22? (c) if 25.8 ml of an agno3 solution is needed to precipitate all the cl- ions in a 785-mg sample of kcl (forming agcl), what is the molarity of the agno3 solution? (d) if 45.3 ml of a 0.108 m hcl solution is needed to neutralize a solution of koh, how many grams of koh must be present in the solution?

a) HClO4 Volume= 38.04mL

b) HCl Volume= 0.19 litters = 190mL

c) AgNO3 Molarity= 0.41 mol/l = 0.41 M

d) KOHmass = 0.275g

Explanation:

In both, the neutralization reactions (a, b and d) and in the single exchange reactions (d) the relationship between the moles of the reagents indicates how much is needed to complete the reaction.

The number of moles of a substance can be calculated as:

moles= Molarity* volume (liters)      >For solutions

or

moles= mass(g)/MM (g/mol)      > For pure substance

MM= Molecular Mass

a) moles of NaOH = moles of HClO4

NaOH Molarity*NaOH Volume= HClO4 Molarity* HClO4 Volume

HClO4 Volume= (NaOH Molarity* NaOH Volume)/HClO4 Molarity

HClO4 Volume= (0.0875M*50.00mL)/0.115M

HClO4 Volume= 38.04mL

b) moles of Mg(OH)2 =2 moles of HCl

Mg(OH)2  mass/ Mg(OH)2 MM = 2*HCl Molarity* HCl Volume

HCl Volume= (Mg(OH)2  mass/ Mg(OH)2 MM )/ 2*HCl Molarity

HCl Volume= (2.87g/58.32g/mol)/ 2*0.128mol/l

HCl Volume= 0.19 l =190ml

c) moles of KCl = moles of AgNO3

KCl mass/ KCl MM = AgNO3 Molarity* AgNO3 Volume

AgNO3 Molarity= (KCl mass/ KCl MM )/AgNO3 Volume

AgNO3 Molarity= (0.785g/74.55 g/mol))/0.0258 l

AgNO3 Molarity= 0.41 mol/l = 0.41 M

d) moles of KOH= moles of HCl

KOHmass/ KOH MM = HCl Molarity* HCl Volume

KOHmass = HCl Molarity* HCl Volume*KOH MM  

KOHmass = 0.108 (mol/l) *0.0453 l*56.11( g/mol)

KOHmass = 0.275g

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:

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j