Answer:
Explanation:
a )
pH = - log[ H⁺]
8.26 = - log[ H⁺]
[ H⁺] = 10⁻⁸°²⁶ mole / l
= 5.49 x 10⁻⁹ moles / l
[ H⁺] [OH⁻] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ / 5.49 x 10⁻⁹
= .182 x 10⁻⁵ moles / l
b )
10.25 = - log[ H⁺]
[ H⁺] = 10⁻¹⁰°²⁵ mole / l
= 5.62 x 10⁻¹¹ moles / l
[ H⁺] [OH⁻] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ / 5.62 x 10⁻¹¹
= .178 x 10⁻³ moles / l
c )
4.65 = - log[ H⁺]
[ H⁺] = 10⁻⁴°⁶⁵ mole / l
= 2.24 x 10⁻⁵ moles / l
[ H⁺] [OH⁻] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ / 2.24 x 10⁻⁵
= .4464 x 10⁻⁹ moles / l
Answer:
This metal has a specific heat of 0.9845J/ g °C
Explanation:
Step 1: Given data
q = m*ΔT *Cp
⇒with m = mass of the substance
⇒with ΔT = change in temp = final temperature T2 - initial temperature T1
⇒with Cp = specific heat (Cpwater = 4.184J/g °C) (Cpmetam = TO BE DETERMINED)
Step 2: Calculate specific heat
For this situation : we get for q = m*ΔT *Cp
q(lost, metal) = q(gained, water)
- mass of metal(ΔT)(Cpmetal) = mass of water (ΔT) (Cpwater)
-5 * (15-100)(Cpmetal) = 20* (15-10) * (4.184J/g °C =
-5 * (-85)(Cpmetal) = 418.4
Cpmetal = 418.4 / (-5*-85) = 0.9845 J/g °C
This metal has a specific heat of 0.9845J/ g °C
Answer:
When the magma reaches it boiling point and it starts to rise
Explanation:
Answer:
I believe is the npp i kinda forgot how to do these lol
