<h3><u>Answer;</u></h3>
= 0.0000001 mol/L
= 0.0000001 mol/L
= 1:1
<h3><u>Explanation;</u></h3>
What is the concentration of H+ ions at a pH = 7?
The formula for ph is given by:
pH=−log10[H+]
In calculating for the concentration of hydrogen ion, the formula is given by:
[H+]=(10)^(-pH)
Therefore; at pH = 7
[H+]=(10)^(-7)
<u>= 0.0000001 mol/L </u>
What is the concentration of OH– ions at a pH = 7?
pH+pOH=14
7+pOH=14
pOH= 7
[OH-]=(10)^(-pOH)
[OH-]=(10)^(-7)
<u>= 0.0000001 mol/L </u>
What is the ratio of H+ ions to OH– ions at a pH = 7?
1:1
Answer:
Explanation:
Given that,
Current measure is
i=10±0.6 Amps
And also,
R=45.0±2.0 Ω
Power dissipated by
P=i²R
Then
P=(10±0.6)²(45.0±2.0)
P=10²×45
P=450Watts
Now, calculating the uncertainty
∆P=|P| • √(2(∆i/i)²+(∆R/R)²)
∆P=450√ (2×(0.6/10)²+(2/45)²)
∆P=450√(0.0072+0.001975)
∆P=450√0.009175
∆P=43.1
The uncertainty in power is 43.1
Then,
P=450 ± 43.1 Watts
Answer:
his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.
determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.
a)0.7956kg/s
b)5.437 × 10⁻³m²
Explanation:
The concepts related to the change of mass flow for both entry and exit is applied
The general formula is defined by
Where,
values are divided by inlet(1) and outlet(2) by
PART A) Applying the flow equation
PART B) For the exit area we need to arrange the equation in function of Area, that is
Answer:
The centripetal acceleration of the rock is 10.7 m/s^2, towards the centre of the circle
Explanation: