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.
Answer:
"Freezing point and ability to react with oxygen" are chemical properties
Explanation:
The change of liquid into solid is the freezing point. The melting point is more than the freezing point in certain cases of mixtures for certain organic compounds like fats. As soon as the mixture freezes it becomes solid and which results in change in the composition from the liquid and solid in this way the it drastically reduces the freezing point. The melting point gets higher due to the pressure. This happens due to the release of heat of which results in the rise of temperature to the freezing point
.Also the reaction of elements with oxygen which leads to formation of new substance is also an chemical property
Answer:
the answer is 6
Explanation:
there is 3 hydrogen molecules in NH3 and there's 2 molecules of NH3 so in total, there are 6 hydrogen molecules on the products side.
<span>Well if you're looking for grams, all you need to do is cancel out units.
(ml)(g/ml)=g because the ml cancels out.
Thus, multiply: (85.32ml)(1.03g/ml)=...I'll let you solve this. :)
Good luck! Hope that helped. When in doubt, look at the units.</span>
Answer:
Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt
The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1
E decreseas 3/2 as fast as G increases = 0.30 M/s
Explanation:
Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt
When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:
Given data = d[D]/dt = 0.10 M/s
-d[D] / 2dt = d[H]/dt
d[H]/dt = 0.05 M/s
The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1
When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:
d[G] / 2dt = -d[H]/3dt
E decreseas 3/2 as fast as G increases = 0.30 M/s
