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Mrrafil [7]
3 years ago
9

The object of a general chemistry experiment is to determine the amount (in milliliters) of sodium hydroxide (NaOH) solution nee

ded to neutralize 1 gram of a specified acid. This will be an exact amount, but when the experiment is run in the laboratory, variation will occur as the result of experimental error. Three titrations are made using phenolphthalein as an indicator of the neutrality of the solution (pH equals 7 for a neutral solution). The three volumes of NaOH required to attain a pH of 7 in each of the three titrations are as follows: 82.00, 75.65, and 75.43 milliliters.
Required:
Use a 99% confidence interval to estimate the mean number of milliliters required to neutralize 1 gram of the acid

Mathematics
1 answer:
Bingel [31]3 years ago
7 0

Answer:

THE MEAN NUMBER OF MILLILITRES REQUIRED TO NEUTRALIZE 1g OF ACID LAY BETWEEN 68.43 AND 86.96

Step-by-step explanation:

PLEASE CHECK ATTACHMENT FOR SOLUTION AND STEP BY STEP EXPLANATION

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8x-5x+3a-9e+8

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Which shows the following expression after the negative exponents have been eliminated?
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Identify the type of sequence shown and select the appropriate response.
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Step-by-step explanation:

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4 0
3 years ago
Please answer question three and four if you can :)<br> Show full working out ty;)
nikklg [1K]

Step-by-step explanation:

3

Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

Therefore, by mid-point formula:

D = ( \frac{2 + 5}{2},  \:  \:  \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \:  \:  \frac{ 1}{2} ) \\ \therefore D= (3.5, \:  \: 0.5) \\  \& \: A=(-1,\:\:4)...(given) \\\\  now \: by \: distance \: formula \\  \\ Length  \: of \:  segment  \: AD \\  =  \sqrt{( - 1 - 3.5)^{2}  +  {(4 - 0.5)}^{2} }  \\ =  \sqrt{(4.5)^{2}  +  {(3.5)}^{2} }  \\ =  \sqrt{20.25 + 12.25 }  \\  =  \sqrt{32.5}  \\    \red{ \boxed{\therefore Length  \: of \:  segment  \: AD  = 5.7 \: units}}

4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

LHS = 3x - y-5\\=3\times 4-7-5\\= 12-12\\=0\\= RHS

Hence, point C (4, 7) lie on the straight line AB.

4(b)

Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

4 0
4 years ago
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