Which of the following types of waste should not be recycled at home?

Let’s say you have 3 UV active spots in the crude material and co-spot TLC plate. One has the same Rf value as the starting mate

Troyanec [42]

Answer:

Answer: (1R,2S) / (1S, 2R) , (1R,2R) / (1S, 2S)

Explanation:

Sodium borohydride reduction of benzoin will give four possible stereo isomers out of which are (1R,2S) - (1S, 2R) isomers and (1R,2R) - (1S, 2S) isomers which are known as enantiomers.

In general enantiomers show single spot in the TLC as they do not show any difference in Rf value (i.e) (1R,2S) - (1S, 2R) isomers show only one spot although they are two compounds and also (1R,2R) - (1S, 2S) isomers also show one spot. That is the reason why you are observing two spots in the TLC ( of reaction mixture) other than starting materilal.

8 0

2 years ago

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20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago

40 liters is equal to?<br> A. 40 cm3<br> B. 40 m3<br> C. 4,000 ml

9966 [12]

B I think sorry if I’m wrong

5 0

3 years ago

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How many molecules are in 0.55 moles of Cu(NO3)2?

xeze [42]

<h3>Answer:</h3>

3.3 × 10²³ molecules Cu(NO₃)₂

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.55 mol Cu(NO₃)₂

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 0.55 \ mol \ Cu(NO_3)_2(\frac{6.022 \cdot 10^{23} \ molecules \ Cu(NO_3)_2}{1 \ mol \ Cu(NO_3)_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 3.3121 \cdot 10^{23} \ molecules \ Cu(NO_3)_2$