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myrzilka [38]
3 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]3 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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Answer:

\large \boxed{\text{392 u}}

Explanation:

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The formula for the freezing point depression by a nonelectrolyte is

\Delta T_{f} = -K_{f}b\\b = -\dfrac{\Delta T_{f}}{ K_{f}} = -\dfrac{-0.465 \, ^{\circ}\text{C}}{\text{1.86 $\, ^{\circ}$C$\cdot$kg$\cdot$mol}^{-1}} = \text{0.250 mol/kg}

2. Calculate the moles of solute

\begin{array}{rcl}b & = & \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}\\\\\text{moles of solute} & = & b \times {\text{kilograms of solvent}}\\n & = &\text{0.250 mol/kg} \times \text{1.00 kg}\\ & = & \text{0.250 mol}\\\end{array}

3. Calculate the molecular mass

\begin{array}{rcl}\text{Moles} & = &\dfrac{\text{mass}}{\text{molar mass}}\\\\\text{0.250 mol} & = & \dfrac{\text{98.0 g}}{MM}\\\\MM & = & \dfrac{\text{98.0 g }}{\text{0.250 mol}}\\\\& = &\textbf{392 g/mol}\\\end{array}\\\text{The molecular mass of the solute is $\large \boxed{\textbf{392 u}}$}

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olga55 [171]

Answer:

The answer to your question is: first option.

Explanation:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + energy.  This option is correct, 1 molecule of glucose reacts with 6 molecules of oxygen and produces 6 molecules of carbon dioxide, six molecules of water and energy.

C5H12O6 + 6 O2 → 5 CO2 + 6 H2O + energy. This option is incorrect because glucose has 6 carbons not five.

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A nuclear fission reaction has mass difference between the products and the reactants of 3.86 g. Calculate the amount of energy
iris [78.8K]

<u>Answer:</u> The amount of energy released by the reaction is 3.47\times 10^{14}J

<u>Explanation:</u>

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

where,

\Delta m = difference between the masses of products and reactants = 3.86 g = 0.00386 kg    (Conversion factor: 1 kg = 1000 g)

c = speed of light = 3\times 10^8m/s

Putting values in above equation, we get:

E=(3.86\times 10^{-3}kg)\times (3\times 10^8m/s)^2

E=3.47\times 10^{14}J

Hence, the amount of energy released by the reaction is 3.47\times 10^{14}J

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