Answer:
There are 10⁻⁷ hydrogen ions in one-liter water
Step-by-step explanation:
Hi there!
a. If the pH = 7.0, then:
-log(H) = 7.0
multiply both sides of the equation by -1
log(H) = -7.0
Apply 10ˣ to both sides
10^(log (H)) = 10⁻⁷
(H) = 10⁻⁷
There are 10⁻⁷ hydrogen ions in one-liter water.
Why 10^(log (H)) = (H)?
Let´s consider this function:
y = 10^(log x)
(Apply log to both sides)
log y = log 10^(log x)
(Apply logarithmic property: log xᵃ = a · log x)
log y = log x · log 10
(log 10 = 1)
log y = log x
y = x
(since y = 10^(log x) and y = x):
10^(log x) = x
That´s why 10^(log (H)) = (H)
-3a=-3-2b
a=1+2/3b
Solution
a=1+2/3b
Answer:
(−0.103371 ; 0.063371) ;
No ;
( -0.0463642, 0.0063642)
Step-by-step explanation:
Shift 1:
Sample size, n1 = 30
Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm
Shift 2:
Sample size, n2 = 25
Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17
Mean difference ; μ1 - μ2
Zcritical at 95% confidence interval = 1.96
Using the relation :
(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)
(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)
Lower boundary :
-0.02 - 0.0833710 = −0.103371
Upper boundary :
-0.02 + 0.0833710 = 0.063371
(−0.103371 ; 0.063371)
B.)
We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.
C.)
For sample size :
n1 = 300 ; n2 = 250
(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)
Lower boundary :
-0.02 - 0.0263642 = −0.0463642
Upper boundary :
-0.02 + 0.0263642 = 0.0063642
( -0.0463642, 0.0063642)
What relationship are you talking about
