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monitta
3 years ago
6

This chemical equation represents the burning of methane, but the equation is incomplete. What is the missing coefficient in bot

h the reactants and the products? CH4 + ____O2 → CO2 + ____H2O
Chemistry
2 answers:
liraira [26]3 years ago
8 0

<u>Answer:</u> The missing coefficients are 2 and 2.

<u>Explanation:</u>

The given reaction is a combustion reaction which is defined as the reaction where a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

Every equation follow law of conservation of mass which states that the total number of individual atoms present on the reactant side is always equal to the total number of individual atoms present on the product side.

For the given chemical equation:

CH_4+O_2\rightarrow CO_2+H_2O

<u>On reactant side:</u>

Number of carbon atoms = 1

Number of hydrogen atoms = 4

Number of oxygen atoms = 2

<u>On product side:</u>

Number of carbon atoms = 1

Number of hydrogen atoms = 2

Number of oxygen atoms = 3

As, the number of hydrogen and oxygen atoms are not balanced, therefore we add some coefficients to make them balanced.

Hence, the balanced chemical equation follows:

CH_4+2O_2\rightarrow CO_2+2H_2O

Thus, the missing coefficients are 2 and 2.

Zolol [24]3 years ago
6 0
CH4+(x)O2=CO2 +(Y)H2O
C=1 +H=4 +O=? = C=1 +O=2+? +H=?
H=4>>Y=2
C=1 +H=4 +O=? = C=1 +O=(2+2) +H=4
C=1 +H=4 +O=4 = C=1 +O=4 +H=4
O=4>>X=2
CH4+(2)O2 =CO2 +(2)H2O

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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

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

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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)

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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

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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 )

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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

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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

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

we substitute

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

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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

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