The kinetic energy of a ball with a mass of 0.5kg and a velocity of 10 m/s is ?

K.E = 25 J

Explanation:

Given data:

Mass of ball = 0.5 g

Velocity of ball = 10 m/s

Kinetic energy = ?

Solution:

Formula:

K.E = 1/2 mv²

m = mass

v = velocity

Now we will put the values in formula.

K.E = 1/2 mv²

K.E = 1/2× 0.5kg × (10m/s)²

K.E = 1/2 ×0.5kg × 100m²/s²

K.E = 25 Kg.m²/s²

Kg.m²/s² = J

K.E = 25 J

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

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2