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sveticcg [70]
2 years ago
15

Sewage and industrial pollutants dumped into a body of water can reduce the dissolved oxygen concentration and adversely affect

aquatic species. In one study, weekly readings are taken from the same location in a river over a two-month period. The week 3 measurement in the data set is suspected of being an outlier.
True or False: According to the Q test, the value at week 3 can be rejected at the 95% confidence level.
Week Dissolved O2 (ppm)
1 4.9
2 5.1
3 5.6
4 4.3
5 4.7
6 4.9
7 4.5
8 5.1
Chemistry
1 answer:
MArishka [77]2 years ago
7 0

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.

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A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of
jek_recluse [69]

<u>Answer:</u> The volume when the pressure and temperature has changed is 1.6\times 10^2mL

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

Let us assume:

P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K

Putting values in above equation, we get:

\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL

Hence, the volume when the pressure and temperature has changed is 1.6\times 10^2mL

5 0
3 years ago
The Lewis diagram for BH2 is:
Luda [366]

Answer:

Electron pair geometry- trigonal planar

There is one lone pair around the boron atom

The geometry of BH2 is bent

Explanation:

The valence shell electron pair repulsion theory offers a frame work for determining the shape of molecules based on the number of electron pairs of the valence shell of the central atom in the molecule.

In BH2-, the central atom is boron. There is a lone pair on boron. Owing to the lone pair on boron, the molecular geometry of BH2 is bent.

4 0
2 years ago
30.5 g of sodium metal reacts with chlorine gas to produce sodium chloride. How much (in grams) chlorine gas must react with thi
Lena [83]

Answer:

Mass of chlorine = 47.22 g

Explanation:

Given data:

Mass of sodium = 30.5 g

Mass of chlorine= ?

Solution:

Chemical equation:

 2Na + Cl₂      →      2NaCl  

Number of moles of Na:

Number of moles = mass/molar mass

Number of moles = 30.5g/ 23 g/mol

Number of moles = 1.33 mol

Now we will compare the moles of Cl ₂ with Na from balance chemical equation.

                    Na            :              Cl ₂

                     2              :               1

                     1.33          :            1/2×1.33 = 0.665 mol

Mass of chlorine gas:

Mass = number of moles × molar mass

Mass = 0.665 mol × 71 g/mol

Mass = 47.22 g

6 0
3 years ago
What do the symbols in the parentheses indicate
beks73 [17]

Answer:

(C) the physical state of each reactant and product

Explanation:

Hope this helps

8 0
3 years ago
A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to . He'll do this by add
erma4kov [3.2K]

Answer:

0.295 L

Explanation:

It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:

" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "

Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.

To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂

Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.

  • 47.2 mL * 150 mM = 24.0 mM * V₂
  • V₂ = 295 mL

And <u>converting into L </u>becomes:

  • 295 mL * \frac{1 L}{1000mL} = 0.295 L

6 0
3 years ago
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