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Types of isomers possible for ch3 ch2 ch2 ch3

puteri [66]

Answer:

It has two isomers; n-butane and isobutane. Here n-butane is a straight-chain compound with four carbon atoms bonded with single covalent bonds. Explanation: Butane is an alkane with four carbon atoms so molecular formula is C4H10.

Explanation:

4 0

2 years ago

Is this correct im so confused.

inysia [295]

Answer:

the last option!!!!

Explanation:

because its right

4 0

3 years ago

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago

Although I am solid, I am so light that I can float in any liquid listed on the chart. What am I?-----

evablogger [386]

Answer:

I know that Aerogel is the lightest metal in existence, but I don't think it would help much with your answer. I mean you can give it a try?

3 0

3 years ago

**PLATO QUESTION, PLEASE ANSWER CORRECTLY, THANKS SO MUCH**

OLga [1]

None of the questions asked can be answered completely from the graph provided (GHG emissions: Direct, indirect and total Vs Year)

Reason:
1) Question A:What caused a drop in GHG emissions around 2009?. This questions in pointing towards reason for drop of GHG emission around 2009. From the graph, it can be seen that there is a drop in GHG emission around 2009. However, information for reason for this drop is not available in graph.
2) Question B: Did GHG emissions cause the melting of Arctic glaciers?. As mentioned earlier, the graph plotted provides information of GHG emissions: Vs Year. Information related to impact of GHG on environment is not available in graph.
3) Question C: How much methane was emitted by homes between 1990 and 2000?. Graph provides information of direct and indirect emission for GHG. However, it lacks information about emission from residential or industrial sources.
4) Question D: Does industrial equipment release gases other than greenhouse gases?: Present study doesnot cover type of gases emitted from industrial equipment.
5) Question E: Which types of industries were included in the study?: Present graph has not specific information related to industries.

6 0

3 years ago

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