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3 years ago

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What are the main reasons that organic chemistry is a primarily target for "green chemistry"? I. Organic solvents are often flam

mable. II. Organic solvents are often toxic. III. Organic solutes are often flammable. IV. Organic solutes are often toxic.

1 answer:

Debora [2.8K]3 years ago

8 0

Answer: Option 1 and 11

1. Organic solvents are often flammable.

II. Organic solvents are often toxic.

Explanation:

Organic solvents are often flammable and Organic solvents are often toxic are the reasons why organic chemistry is a primary target of green chemistry because green chemistry is the use of principles or approach to remove hazardous or toxic substances from chemical.profuctd or processes to make them fit or safe for use and the solvents of organic chemist are toxic, therefore green chemistry remove the toxic substances.

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How much energy, in joules, does 150.0 g of water with an initial temperature of 25 C need to absorb be raised to a final temper

satela [25.4K]

Answer:

31395 J

Explanation:

Given data:

mass of water = 150 g

Initial temperature = 25 °C

Final temperature = 75 °C

Energy absorbed = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 75 °C - 25 °C

ΔT = 50 °C

now we will put the values in formula

q = m . c . ΔT

q = 150 g × 4.186 J/g.°C × 50 °C

q = 31395 J

so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .

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3 years ago

What is required to move from one state or phase of matter to the next

tigry1 [53]

energy is required to move from one state or phase of matter to the next. Energy is used to make a liquid into a gas or a solid into a liquid.

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3 years ago

Cytochromes are critical participants in the electron transport chains used in photosynthesis and cellular respiration. How do c

goblinko [34]

Answer:

4) Each cytochrome has an iron‑containing heme group that accepts electrons and then donates the electrons to a more electronegative substance.

Explanation:

The cytochromes are <u>proteins that contain heme prosthetic groups</u>. Cytochromes <u>undergo oxidation and reduction through loss or gain of a single electron by the iron atom in the heme of the cytochrome</u>:

$Cytochrome-Fe²⁺ ⇄ cytochrome-Fe³⁺-e⁻$

The reduced form of ubiquinone (QH₂), an extraordinarily mobile transporter, transfers electrons to cytochrome reductase, a complex that contains cytochromes <em>b</em> and <em>c₁</em>, and a Fe-S center. This second complex reduces cytochrome <em>c</em>, a water-soluble membrane peripheral protein. Cytochrome <em>c</em>, like ubiquinone (Q), is a mobile electron transporter, which is transferred to cytochrome oxidase. This third complex contains the cytochromes <em>a</em>, <em>a₃</em> and two copper ions. Heme iron and a copper ion of this oxidase transfer electrons to O₂, as the last acceptor, to form water.

Each transporter "downstream" is <u>more electronegative</u><u> than its neighbor </u>"upstream"; oxygen is located in the inferior part of the chain. Thus, the <u>electrons fall in an energetic gradient</u> in the electron chain transport to a more stable localization in the <u>electronegative oxygen atom</u>.

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3 years ago

Isoelectric point. The isoelectric point (pI) is the pH at which a molecule has no net charge. The amino acid glycine has two io

timama [110]

Answer:

pI = 6.16

Explanation:

The pI is given by the average of the pKas that are involved. In this case,

Pka of carboxylic acid was given as 2.72 and that of the Amino group was given as 9.60. the average would then be ½(2.72+9.60)

= 6.16

6 0

3 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

3 years ago