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2 years ago

What is the IUPAC name for this product?

Chemistry

1 answer:

mihalych1998 [28]2 years ago

7 0

The IUPAC name for the given product is 2 chloro Butane.

<h3>What is IUPAC nomenclature?</h3>

IUPAC stands for 'International Union of Pure and Applied Chemistry', which givers some rule for designing the name of compounds of chemistry.

In the given product total four carbon atoms are present and between all of them single bonds are present.
In the second carbon atom, chlorine group is present.
During the nomenclature process, first we write down the name of the attached group which is followed by the alkane chain.

Hence name of the product is 2 chloro Butane.

To know more about IUPAC nomenclature, visit the below link:

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The molarity of a 2 liter aqueous solution that contains 222.2 grams of dissolved calcium chloride ( CaCl2), expressed with two

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Answer:

The answer to your question is 1 M

Explanation:

Data

Molarity = ?

mass of CaCl₂ = 222.2 g

Volume = 2 l

Process

1.- Calculate the molar mass of CaCl₂

CaCl₂ = 40 + (35.5 x 2) = 40 + 71 = 111 g

2.- Calculate the moles of CaCl₂

111g of CaCl₂ ---------------- 1 mol

222.2 f of CaCl₂ ---------------- x

x = (222.2 x 1) / 111

x = 222.2 / 111

x = 2 moles

3.- Calculate the Molarity

Molarity = moles / Volume

-Substitution

Molarity = 2/2

-Result

Molarity = 1

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3 years ago

0.0145 moles of helium gas are introduced into a balloon so that the volume of the balloon is 2.54 liters. An additional amount

olga_2 [115]

Answer:

4.43L is final volume of the ballon

Explanation:

Avogadro's law of ideal gases states that <em>equal volumes of gases, at the same temperature and pressure, have the same number of molecules</em>.

The formula is:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

Where V and n are volume and moles of the gas in initial and final conditions.

If the initial conditions are 0.0145 moles and 2.54L and final amount of moles is 0.0253moles, final volume is:

$\frac{2.54L}{0.0145mol} =\frac{V_2}{0.0253mol}$

V₂ = <em>4.43L is final volume of the ballon</em>

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3 years ago

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3 years ago

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A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What

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Answer:

Approximately $1.854\; \rm mol\cdot L^{-1}$ .

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

<h3>Formula mass of strontium hydroxide</h3>

Look up the relative atomic mass of $\rm Sr$ , $\rm O$ , and $\rm H$ on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

$\rm Sr$ : $87.62$ .
$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Calculate the formula mass of $\rm Sr(OH)_2$ :

$M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}$ .

<h3>Number of moles of strontium hydroxide in the solution</h3>

$M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}$ means that each mole of $\rm Sr(OH)_2$ formula units have a mass of $121.634\; \rm g$ .

The question states that there are $10.60\; \rm g$ of $\rm Sr(OH)_2$ in this solution.

How many moles of $\rm Sr(OH)_2$ formula units would that be?

$\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}$ .

<h3>Molarity of this strontium hydroxide solution</h3>

There are $8.71467\times 10^{-2}\; \rm mol$ of $\rm Sr(OH)_2$ formula units in this $47\; \rm mL$ solution. Convert the unit of volume to liter:

$V = 47\; \rm mL = 0.047\; \rm L$ .

The molarity of a solution measures its molar concentration. For this solution:

$\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}$ .

(Rounded to four significant figures.)

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What are the functional groups in oxybenzone?

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Explanation:

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