Determine the concentration of h 3 o + to the correct number of significant figures in a solution with each ph. describe how these calculations show the relationship between the number of digits to the right of the decimal place in ph and the number of significant figures in concentration. ph = 2.50 ph = 2.51 ph = 2.52

To calculate the concentration of H3O+

1- when PH = 2.5 :

The explanation:

we are going to use this formula:

PH = - ㏒[H3O+]

and we here have PH = 2.5 so, by substitution:

2.5 = -㏒ [H3O+]

∴ [H3O+] = 0.00316 ≈ 0.003

2- when PH = 2.51 :

The explanation:

we are going to use this formula:

PH = - ㏒[H3O+]

when here we have PH = 2.51 so, by substitution:

2.51 = - ㏒[H3O+]

∴[H3O+] = 0.00309 ≈ 0.003

3- when PH = 2.52 :

The explanation:

we are going to use this formula:

PH = - ㏒[H3O+]

when here we have PH =2.52 so, by substitution:

2.52 = -㏒[H3O+]

∴[H3O+] = 0.00301 ≈ 0.003

we can see that when the the number of digits to the right of the decimal place in ph increase, the number of significant figures in concentration decrease.

electron microscopy and scanning tunneling microscopes gives us a greater understanding of atoms by being able to see the atom itself, and lets us see some atomic reactions.

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