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ira [324]
3 years ago
7

In the gaseous state, chlorine exists as a diatomic molecule Cl2 (Molar mass = 70.9 g/mol). Calculate the number of moles of chl

orine present in 140 g of chlorine gas. Express the quantity in moles to three significant figures.
Chemistry
1 answer:
kap26 [50]3 years ago
8 0

Cl is stable as a diatomic molecule where the 2 Cl atoms are held together by a covalent bond

molar mass of the diatomic molecule is 70.9 g/mol

therefore 70.9 g of Cl₂ is - 1 mol

then 140 g of Cl₂ is - 1/70.9 x 140 = 1.97 mol

there are 1.97 mol of Cl₂ present

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(CH3)2-CH-CH2-O(CH3)3IUPAC NAME
bazaltina [42]

Answer:

1-(tert-butoxy)-2-methylpropane

Note: there is a mistake in formula, the correct formula is (CH₃)₂-CH-CH₂-O-C(CH₃)₃ not (CH₃)₂-CH-CH₂-O(CH₃)₃, because oxygen is a divalent compound.

Explanation:

<em>Structural formula is attached</em>

IUPAC naming rules

1. start numbering the chain from the functional group. In this compound we        start from oxygen side.

2. Here we can see that at position 1 there is an oxy group along with a tertiary carbon having three methyl groups. So we write it as 1-tert-butoxy. Which means that there is a methoxy group at position 1 along with a tertiary carbon.

3. At position 2 we can see that there is a methyl group attached to the main chain, so we write it as 2-methyl.

4. Now we count the total number of carbons in the main chain. As we can see that there are 3 carbons in the remaining or parent chain, so we write it as propane

5. So the IUPAC name of the compound will be 1-(tert-butoxy)-2-methylpropane

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3 years ago
What are the uses of an enema? Select all that apply. what are the uses of an enema? select all that apply. ?
shutvik [7]

An enema administration is a technique used to stimulate stool evacuation The process helps push waste out of the rectum when you cannot do so on your own

7 0
3 years ago
Why do the deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer?
rodikova [14]

Answer:

The deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer because Organic compounds that are neither acids or bases do not react with either NaOH or HCl and, therefore remain more soluble in the organic solvent and are not extracted.

Explanation:

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3 years ago
I WIll upvote for answer<br> Balance the following reaction Fe^2 O ^3 +HCI-&gt;FeCI^3+H^2O
kotykmax [81]
Fe^2 O^3 + 6HCl --> 2FeCl^3 + 3H^2 O
7 0
3 years ago
5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0
3 years ago
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