Answer:
The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.
Explanation:
The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.
What I’m seeing on quizlet says what you’re describing is a ball-and-stick model.
Answer:
1.6g/mL
Explanation:
Density equation is D=m/v
Density = g/mL
m=mass of sample in grams
v = volume of sample in mL
The volume of a square can be calculated by V=l*w*h.
In this case it is 5cm*5cm*5cm = 125cm^3
Since we know that 1cm^3 ~ 1mL we can convert the volume to mL as so:
125cm^3 (1mL/(1cm^3)) = 125mL
Then simply plug into the density equation:
D=200g/125mL = 1.6g/mL
On the other hand ammonia is a very dangerous chemical which has a pungent smell and effect the eyes of the user. Thus it kept always in the fume exhaust hood for storing and dispensing function.
The pH of ammonia buffer contains ammonium hydroxide (NH₄OH) and a salt of ammonia with a strong acid like (HCl) which produces, ammonium chloride (NH₄Cl) mixture. The evaporation rate of ammonia is so high at room temperature thus on opening of the buffer solution the ammonia get evaporated very fast and the concentration of ammonia decreases which affect the pH of the buffer solution.
Thus the reason to put ammonia buffer in fume hood is explained.
Answer:
FALSE
Since 0.385 < 0.526, the value for week 3 is accepted.
Explanation:
Qexp = (|Xq - Xₙ₋₁|)/w
where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data
First, the data are arranged in decreasing order, from highest to lowest:
3. 5.6
2. 5.1
8. 5.1
1. 4.9
6. 4.9
5. 4.7
7. 4.5
4. 4.3
Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3
Qexp = (|5.6 - 5.1|)/1.3 = 0.385
From tables, at 95% confidence level, for n = 8, Qcrit = 0.526
Since 0.385 < 0.526, the value for week 3 is accepted.
