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rjkz [21]
3 years ago
14

Five students performed a Kjeldahl nitrogen analysis of a protein sample. The following weight % nitrogen values were determined

: 15.5, 13.8, 15.7, 15.4, 15.8. What is the value of G_calculated for the Grubbs test? Should the outlier be rejected with 95% confidence?
Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

G_calculated = 1.756

The outlier should be rejected, as G_cal > G_tab (= 1.463) at 95 % confidence.

Explanation:

The Grubb's test is used for identifying an outlier in data, which is from the same population. For this, a statistical term, G, is calculated for the suspected outlier. If the calculated value is greater than the tabulated G value then the suspected value is rejected. This term is given as,

G_calculated = | suspect value - mean| / s

Here,  suspect value is 13.8, mean is to be taken of all the data (including suspected value). s is the standard deviation of the sample data.

s is calculated from the following formula:

s = (Σ(xi - x)²/(N-1))^1/2

Here, x is the mean, which is 15.24, xi is individual value and N is the total number of data (5).

From the above formula, s is found to be

Standard Deviation, s = 0.820

Now for G value,

G_calculated = | 13.8 - 15.24| / (0.820)

G_ calculated = 1.756

The tabulated G value at 95 % confidence and N -1 (5 - 1 = 4) degree of freedom is, 1.463.

As calculated G (1.756) is greater than the tabulated G (1.463), the value 13.8 is considered an outlier at 95 % confidence.  

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<u>Explanation:</u>

The equation relating the pre-exponential factor and activation energy follows:

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T = temperature = ?

Putting values in above equation, we get:

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How many moles of ammonium nitrate will be produced 110.0g of ammonium carbonate
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Answer:

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Explanation:

The Balance chemical equation is as follow:

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To solve this problem we will do following steps:

Finding moles of Ammonium Carbonate:

As we know,

                        Moles  =  Mass / M.Mass

So,

                        Moles  =  110 g  /  96.08 g/mol

                        Moles  =  1.144 moles

Calculating moles of Ammonium Nitrate:

According to balance chemical equation;

                      1 mole of (NH₄)₂CO₃ produces  =  2 moles of NH₄NO₃

So,

          1.144 moles of (NH₄)₂CO₃ will produce  =  X moles of NH₄NO₃

Solving for X,

                     X =  2 moles × 1.144 moles ÷ 1 mole

                     X  =  2.288 moles of NH₄NO₃

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