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.
The one with higher temperature is the one with NaOH as heatis given off during the neutralization reaction that occurs.
<h3>
What is volume?</h3>
Volume can be defined as the amount of space a substance or an objects occupies usually in a closed container.
Volume is measures in litres.
When water is added to dilute acid like HCl, they become more dilute.
When NaOH is added to HCl, a neutralization reaction occurs.
The student will determine the contents of the flasks by adding 10 ml of hcl to each flask. If the NaOH reacts with the Hcl, there will be an increase in temperature.
The increase in temperature is due to the heat of neutralization of the reaction between NaOH and HCl.
Learn more about volume at: brainly.com/question/1972490
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Answer:
evaporation or water runoff that's my tips
rain or thunder storm
We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
<span>
</span>
X ml - <span>25% alcohol mixture
y ml - </span><span>90% alcohol mixture
x+y = 455
0.25x ml alcohol in </span>x ml of 25% alcohol mixture
0.9y ml alcohol in y ml of 90% alcohol mixture
0.75*455= 341.25 ml alcohol in 455 ml of 75% alcohol mixture
0.25x+0.9y=341.25
System of equations:
x+y = 455 /*(-0.25) ------> -0.25x-0.25y = -0.25*455
0.25x+0.9y=341.25
-0.25x-0.25y=-113.75
0.25x+0.9y=341.25 Add both equations
0.25x+0.9y-0.25x-0.25y=341.25-113.75
0.65y =227.5
y=227.5/0.65 = 350 ml of 90% alcohol mixture
x+y=455
x+355=455
x=100 ml of 25% alcohol mixture
