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gulaghasi [49]
3 years ago
5

How many grams of potassium benzoate trihydrate, KC H302 - 3H20. are needed to prepare 1.5 L of a 0.120M solution of potassium b

enzoate? Mass = g
Chemistry
1 answer:
Annette [7]3 years ago
8 0

<u>Answer:</u> The mass of potassium benzoate is 8.40\times 10^{-4}g

<u>Explanation:</u>

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

We are given:

Molarity of solution = 0.120 M

Molar mass of potassium benzoate trihydrate = 214.26 g/mol

Volume of solution = 1.5 L

Putting values in above equation, we get:

0.120M=\frac{\text{Mass of potassium benzoate trihydrate}}{214.26\times 1.5}\\\\\text{Mass of potassium benzoate trihydrate}=\frac{0.120mol/L\times 1.5L}{214.26g/mol}=8.40\times 10^{-4}g

Hence, the mass of potassium benzoate is 8.40\times 10^{-4}g

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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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3 years ago
a nurse practitioner orders Medrol to be given 1.5 mg/kg of body weight. if a child weighs 72.6lb and the available stock of Med
Harrizon [31]
Answer is: 2,469 mL give to the child.
The mass m in kilograms (kg) is equal to the mass m in pounds (lb) times 0,45359237: m(child) = 72,6 · 0,045359237 = 32,93 kg.
m(Medrol) = 32,93 kg · 1,5 mg/kg.
m(Medrol) = 49,39 mg.
d(Medrol) = 20,0 mg/mL.
V(Medrol) = m(Medrol) ÷ d(Medrol).
V(Medrol) = 49,39 mg ÷ 20 mg/mL.
V(Medrol) = 2,469 mL.

8 0
3 years ago
Explain why raising the temperature of a liquid would speed up diffusion
nikitadnepr [17]
Because diffusion<span> is the process when molecules move to lower or higher concentration, so as the molecules move faster they are going to lower or high concentration faster.</span>
3 0
3 years ago
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Answer:

A

Explanation:

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5 0
3 years ago
Which aqueous solution has the highest boiling point?a. 0.20 M NaClb. 0.10 M CaCl2c. 0.1 M Ga2(SO4)3d. 0.2 M C6H12O6
abruzzese [7]

Answer:

c. 0.1 M Ga₂(SO₄)₃

Explanation:

The boiling point increasing of a solvent due the addition of a solute follows the formula:

ΔT = K*m*i

<em>Where K is boiling point increasing constant (Depends of the solute), m is molality = molarity when solvent is water, and i is Van't Hoff factor.</em>

<em />

That means the option with the higher m*i will be the solution with the highest boiling point:

a. NaCl has i = 2 (NaCl dissociates in Na⁺ and Cl⁻ ions).

m* i = 0.20*2 = 0.4

b. CaCl₂; i = 3. 3 ions.

m*i= 0.10M * 3 = 0.3

c. Ga₂(SO₄)₃ dissolves in 5 ions. i = 5

m*i = 0.10M*55 = 0.5

d. C₆H₁₂O₆ has i = 1:

m*i = 0.2M*1 = 0.2

The solution with highest boiling point is:

<h3>c. 0.1 M Ga₂(SO₄)₃</h3>
3 0
3 years ago
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