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Otrada [13]
3 years ago
15

Blood is a type of connective tissue. All of these make up blood tissue, except:

Chemistry
2 answers:
Vedmedyk [2.9K]3 years ago
7 0

Answer: Smooth Muscle Cells

aev [14]3 years ago
7 0

Answer:

The answer is D

Explanation:

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Some parts of the planet undergo very little seasonal change due to their proximity to the equator and the poles. This means that their position relative to the sun and the earth's rotation varies hardly at all.
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3 years ago
If the enantiomeric excess of a mixture is 75%, what are the % compositions of the major and minor enantiomer?
Maru [420]

Let us say that R is the major enantiomer, while S is the minor enantiomer, therefore the formula for enantiomeric excess (ee) is:

ee = (R – S) * 100%

 

Let us further say that the fraction of R is x (R = x), and therefore fraction of S is 1 – x (S = 1 – x), therefore:

75 = (x – (1 – x)) * 100

75 = 100 x – 100 + 100 x

200 x = 175

x = 0.875

 

Summary of answers:

R = major enantiomer = 0.875 or 87.5%

<span>S = minor enantiomer = (1 – 0.875) = 0.125 or 12.5%</span>

7 0
3 years ago
If the rate law for the reaction A LaTeX: \longrightarrow
Nostrana [21]

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6 0
2 years ago
Flag this question question 8 10 pts use the δh°f and δh°rxn information provided to calculate δh°f for if: δh°f (kj/mol) if7(g)
GarryVolchara [31]

\Delta H\textdegree{}_f(\text{IF} \; (g)} = -95 \;  \text{kJ} \cdot \text{mol}^{-1}

Explanation

\text{IF}_7 \; (g) + \text{I}_2 \; (s) \to \text{IF}_5 \; (g) + 2\; \text{IF} \; (g)

  • \Delta H\textdegree{}_\text{rxn} = -89\; \text{kJ} \cdot \text{mol}^{-1}
  • \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) = -941 \; \text{kJ} \cdot \text{mol}^{-1} (from the question)
  • \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) = -840 \; \text{kJ} \cdot \text{mol}^{-1} (from the question)
  • As an the most stable allotrope under standard conditions, \Delta H\textdegree{}_f (\text{I}_2) = 0\; \text{kJ} \cdot \text{mol}^{-1}

By definition,

\Delta H\textdegree{}_\text{rxn} = \Delta H\textdegree{}_f (\text{all products})  - \Delta H\textdegree{}_f (\text{all reactants})

\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + 2 \;  \Delta H\textdegree{}_f (\text{IF} \; (g) ) - \Delta H\textdegree{}_f (\text{IF}_7 \; (g) )  - \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) \\  =  \Delta H\textdegree{}_{\text{rxn}}

\begin{array}{ccc} \Delta H\textdegree{}_f  (\text{IF} \; (g) )& = & 1/2\; ( \Delta H\textdegree{}_{\text{rxn}} - \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + \Delta H\textdegree{}_f (\text{IF}_7 \; (g) )  + \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) )\\ & = & 1/2 \; (-89 - (-840) + (-941))}\\ & = & - 95 \; \text{kJ} \cdot \text{mol}^{-1} \end{array}

Note, that iodine on the reactant side is stated as a gas in the equation given in the question whereas under standard conditions it is expected to be under the solid state; the \Delta H\textdegree{} _f given in the question seemingly corresponds to the one in which the reactant iodine exists as a solid rather than as a gas. Evaluating the last expression using data from an external source

\Delta H\textdegree{}_f (\text{I}_2 \; (g) ) = \Delta H\textdegree{}_f(\text{I}_2 \; (s)) + \Delta H\textdegree{}_{\text{sublimation}}(\text{I}_2) = 62.42 \;  \text{kJ} \cdot \text{mol}^{-1} (Cox, Wagman, et al., 1984)

... yields \Delta H\textdegree{}_f  (\text{IF} \; (g) )  \approx -64 \; \text{kJ}\cdot \text{mol}^{-1}, which deviates significantly from the experimental value of -94.76    \; \text{kJ}\cdot \text{mol}^{-1} (Chase, 1998.) It is thus assumed that the \Delta H\textdegree{}_\text{rxn} value provided requires a reaction with \text{I}_2 \; (s) rather than \text{I}_2 \; (g) as a reactant.

3 0
3 years ago
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