All categories

3 years ago

How do I convert/scale a recipe using a t chart

Chemistry

1 answer:

Rina8888 [55]3 years ago

6 0

Answer:

Obtain the conversion factor by dividing the required yield (from Step 2) by the old yield (from Step 1). That is, conversion factor = (required yield)/(recipe yield) or conversion factor = what you NEED ÷ what you HAVE.

Explanation:

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You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.220 M sodium benzoate

ddd [48]

Answer:

34.5 mL

Explanation:

Given, we ned to prepared buffer solution with pH = 4.00 , volume = 100.0 mL .

[C6H5COOH] = 0.100 M , [C6H5COO-] = 0.120 M , pKa = 4.20

First we need to calculate the ratio of the conjugate base and acid

We know, Henderson Hasselbalch equation

pH = pKa + log [C6H5COO-] /[C6H5COOH]

4.00 = 4.20 + log [C6H5COO-] /[C6H5COOH]

log [C6H5COO-] /[C6H5COOH] = 4.00-4.20

= - 0.20

Antilog from both side

[C6H5COO-] /[C6H5COOH] = 0.631

Now we are given the molarity of each acid and its conjugate base and we need to calculate for volume

Sum of volume = 100 mL = 0.100 L

volume of benzoic acid = x

volume of benzoate = 0.10 -x ,

so Volume of acid + volume of conjugate base = 0.100 L

[C6H5COO-] /[C6H5COOH] = 0.631

[C6H5COO-] = 0.631 * [C6H5COOH]

0.120 (0.1-x) = 0.631 *0.100x

So, x = 0.065

So, volume of benzoic acid = x = 0.0655 L

= 65.5 mL

So, volume of sodium benzoate = 0.1 -x

= 0.1-0.0655

= 0.0345 L

= 34.5 mL

So we need to mix 65.5 mL of 0.100 M benzoic acid and 34.5 mL of 0.120 M sodium benzoate form 100.0 mL of a pH=4.00 buffer solution.

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3 years ago

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How do most cells release energy?

Shkiper50 [21]

Answer:

Explanation:

Cells use oxygen to release energy stored in sugars such as glucose. I'm not sure if that will support the reason why I picked (C.)

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3 years ago

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Elements can be described by various properties, and identified by their boiling and melting points. For example, gold melts at

aleksandr82 [10.1K]

Answer:

Explanation:

Melting and boiling point variations are not clear (do not have uniform pattern) in periodic table. But we can see, some elements have higher melting and boiling points and some have less. Here we study melting and boiling points of s, p, d blocks elements. IVAth group elements (C,Si) show high melting and boiling points because they have covalent gigantic lattice structures.

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3 years ago

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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

3 0

3 years ago

Why parafin oil use in melting

MatroZZZ [7]

Answer:

Paraffin oil is used for determination of boiling point and melting point for the following reasons: It has a very high boiling point and so it can be used to maintain high temperatures in the boiling and melting point apparatus without loss of the substance.

4 0

3 years ago