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myrzilka [38]
2 years ago
11

A study is conducted to examine the effect of NaCl concentration on enzyme activity with the following results. Activity was tes

ted at various salt concentrations. Negative controls were conducted without enzyme. Which of the following is the best interpretation of the 0.27 p value when comparing the activity at 0.1 M to that 1t 0.05M
NaC1 Con (M) 0 0.01 0.02 0.05 0.1 0.2
Avg. Activity (nm/mg/min) 15.5 14 35.5 175.5 195 50.5
Standard Deviation 1.5 1.5 3.6 17 20 5
T-test vs 0.1M <0.05 <0.05 <0.05 0.27 NA <0.05
T-test vs Neg. Ctrl 0.56 0.6 <0.05 <0.05 <0.05 <0.05

a. The activity at 0.1 M NaCl is statistically significantly greater than that at 0.05M NaCl
b. There is a 27% difference between the average enzyme activities at 0.1 and 0.05 M NaCl
c. The variability among the replicates at 0.1 M and 0.05 M NaCl is 27%
d. The activity at 0.1 M NaCl is not statistically significantly greater than that at 0.05M NaCl
Chemistry
1 answer:
Marina86 [1]2 years ago
3 0

Answer:

The answer is "Option D".

Explanation:

The behavior of 0.1M NaCl also isn't substantially larger objectively than those of 0.05M NaCl because a p-value above 0.05 (p>0.05) indicates no ability to tell differential and is a strong proof in favor of a null hypothesis.

The other wrong choices can be defined as follows:

  • Option A as it's just the reverse of the correct answer to the null.
  • Options B and C because p worth tests to support nor oppose the null hypothesis.
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2.54 mL of ethanol to mol ethanol (Hint: the density of ethanol is 0.789<br> g/mL.)
Katarina [22]

0.04350179862

Explanation:

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3 0
3 years ago
How many moles of gold, Au, are in 3.60 x 10^-5 g of gold?
zlopas [31]
<h3>Answer:</h3>

1.83 × 10⁻⁷ mol Au

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.60 × 10⁻⁵ g Au (Gold)

<u>Step 2: Identify Conversions</u>

Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})
  2. Multiply:                            \displaystyle 1.82769 \cdot 10^{-7} \ mol \ Au

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au

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2 years ago
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