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elena-14-01-66 [18.8K]
3 years ago
13

Question 12 A chemistry student must write down in her lab notebook the concentration of a solution of sodium thiosulfate. The c

oncentration of a solution equals the mass of what's dissolved divided by the total volume of the solution. Here's how the student prepared the solution: The label on the graduated cylinder says: empty weight: 8.2g She put some solid sodium thiosulfate into the graduated cylinder and weighed it. With the sodium thiosulfate added, the cylinder weighed 76.148g . She added water to the graduated cylinder and dissolved the sodium thiosulfate completely. Then she read the total volume of the solution from the markings on the graduated cylinder. The total volume of the solution was 172.7mL . What concentration should the student write down in her lab notebook? Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
o-na [289]3 years ago
3 0

Weight of sodium thiosulfate   = 76.148 - 8.2

= 67.948 g.

Concentration of the solution = 67.948 / 172.7

= 0.393 g / mL.  to the nearest thousandth . (answer).

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9. The times of first sprinkler activation (in seconds) for a series of tests of fire-prevention sprinkler systems that use aque
Nezavi [6.7K]

This question is incomplete, the complete question is;

The times of first sprinkler activation (in seconds) for a series of tests of fire-prevention sprinkler systems that use aqueous film-forming foam is as follows; 27 41 22 27 23 35 30 33 24 27 28 22 24

( see " use of AFFF in sprinkler systems," Fire technology, 1975: 5)

The system has been designed so that the true average activation time is supposed to be at most 25 seconds.

Does the data indicate the design specifications have not been met?

Test the relevant hypothesis at significance level 0.05 using the P-value approach  

     

Answer:

since p-value (0.042299) is lesser than the level of significance ( 0.05)

We Reject Null Hypothesis

Hence, There is no sufficient evidence that the true average activation time is at most 25 seconds

Explanation:

Given the data in the question;

lets consider Null and Alternative hypothesis;

Null hypothesis H₀ : There is sufficient evidence that the true average activation time is at most 25 seconds

Alternative hypothesis H₁ : There is no sufficient evidence that the true average activation time is at most 25 seconds

i.e

Null hypothesis H₀ : μ ≤ 25

Alternative hypothesis H₁ :  μ > 25

level of significance σ = 0.05

first we determine the sample mean;

x^{bar} = \frac{1}{n}∑x_{i}

where n is sample size and ∑x_{i} is summation of all the sample;

=  \frac{1}{13}( 27 + 41 + 22 + 27 + 23 + 35 + 30 + 33 + 24 + 27 + 28 + 22 + 24 )

=   \frac{1}{13}( 363

sample mean x^{bar} = 27.9231

next we find the standard deviation

s = √( \frac{1}{n-1}∑(x_{i}-x^{bar})²

x                    (x_{i}-x^{bar})                       (x_{i}-x^{bar})²

27                   -0.9231                          0.8521

41                    13.0769                        171.0053

22                  -5.9231                          35.0831  

27                  -0.9231                          0.8521

23                  -4.9231                          24.2369

35                  7.0769                          50.0825

30                  2.0769                          4.3135

33                  5.0769                          25.7749

24                  -3.9231                          15.3907

27                  -0.9231                          0.8521

28                  0.0769                          0.0059

22                 -5.9231                          35.0831  

24                 -3.9231                          15.3907

sum                                                    378.9229

so ∑(x_{i}-x^{bar})² = 378.9229

∴

s = √( \frac{1}{13-1} ×378.9229 )

s = √31.5769

standard deviation s = 5.6193

now, the Test statistics

t = ( x^{bar} - μ ) / \frac{s}{\sqrt{n} }

we substitute

t = ( 27.9231 - 25 ) / \frac{5.6193}{\sqrt{13} }

t = 2.9231 / 1.5585

t = 1.88

now degree of freedom df = n - 1 = 13 - 1 = 12

next we calculate p-value

p-value = 0.042299 ( using Execl's ( = TDIST(1.88,12,1)))

Here x=1.88, df=12, one tail

now we compare the p-value with the level of significance

since p-value (0.042299) is lesser than the level of significance ( 0.05)

We Reject Null Hypothesis

Hence, There is no sufficient evidence that the true average activation time is at most 25 seconds

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A solution with a hydrogen ion concentration of 3.25 × 10-2 m is ________ and has a hydroxide concentration of _______
Romashka-Z-Leto [24]

To know the acidity of a solution, we calculate the pH value. The formula for pH is given as:

<span>pH = - log [H+]            where H+ must be in Molar</span>

We are given that H+ = 3.25 × 10-2 M

Therefore the pH is:

pH = - log [3.25 × 10-2] 

pH = 1.488

Since pH is way below 7, therefore the solution is acidic.

 

To find for the OH- concentration, we must remember that the product of H+ and OH- is equivalent to 10^-14. Therefore,

[H+]*[OH-] = 10^-14 <span>
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Answer:

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