<span>Since,
1000 grams of water = 1000 mL of water</span><span>
So,
At any of the given temperature:
</span>1000 mL = 10 x 100 mL
<span>
moles of NH4Cl = 53.5/53.49
= 1.0 m
= 1.0 mol/Kg
Delta T = 2 x 1.86 x 1.0
= 3.72 c
= - 3.72 °C</span>
Explanation:
Since, some of the given sample is stuck inside and behind the pipet. Hence, there will occur a decrease in the percent of acetic acid.
This is because a decrease in concentration of the acid will also lead to a decrease in the amount of sample taken for the estimation. Since. lesser is the amount or concentration present lesser will be its analyte concentration.
For example, we took 10 mg of a pickel sample but 3 mg of the sample remain stuck in the pipet. This means we actually titrating a sample less than 10 mg.
Therefore, the analyte concentration in the pickel will also be less.
Answer:
The answer is B. Urine, Feces
Explanation:
Answer:
After 2.0 minutes the concentration of N2O is 0.3325 M
Explanation:
Step 1: Data given
rate = k[N2O]
initial concentration of N2O of 0.50 M
k = 3.4 * 10^-3/s
Step 2: The balanced equation
2N2O(g) → 2 N2(g) + O2(g)
Step 3: Calculate the concentration of N2O after 2.0 minutes
We use the rate law to derive a time dependent equation.
-d[N2O]/dt = k[N2O]
ln[N2O] = -kt + ln[N2O]i
⇒ with k = 3.4 *10^-3 /s
⇒ with t = 2.0 minutes = 120s
⇒ with [N2O]i = initial conc of N2O = 0.50 M
ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)
ln[N2O] = -1.101
e^(ln[N2O]) = e^(-1.1011)
[N2O} = 0.3325 M
After 2.0 minutes the concentration of N2O is 0.3325 M
We are asked to solve for the arc length of the intercepted arc and the formula is shown below:
Arc length = 2*pi*r(central angle/360°)
r = 5 feet
central angle = 10°
Solving for the arc length, we have:
Arc length = 2*3.14*5 (10/360)
Arc length = 0.872 feet
The arc length is 0.872 feet.
