The stages of the protozoan life cycle are:

Which elements are present in all organic compounds? 1. Hydrogen and oxygen 2. Nitrogen and Carbon 3. Nitrogen and oxygen 4. Hyd

cupoosta [38]

Answer:

my answer is num:3 Hydrogen and oxygen.

8 0

2 years ago

Draw the structure of two alkenes that would yield 1-methylcyclohexanol when treated with Hg(OAc)2 in water, then NaBH4:Draw the

disa [49]

Answer:

Two possible compounds are shown below- one with an exocyclic double bond and another one with an endocyclic double bond

Explanation:

Reaction of alkene with $Hg(OAc)_{2}$ gives a complex of mercurous ion.

Then water molecule attacks this complex through $S_{N}2$ type reaction at more substituted position.

$NaBH_{4}$ cleaves the resultant C-Hg bond and forms a C-H bond.

Two possible structures of an alkene is possible to yield 1-methylcyclohexanol which are shown below.

3 0

3 years ago

Write the name of the alkyne next to the drawing of the molecule. (2 points)

azamat

1) 1-Pentyne

2) But-2-yne

3) Ethyne

4) 1-Propyne

5) 1-Butyne

6) Pent-2-yne

<h3>What is alkyne?</h3>

Molecules that contain a triple bond between two carbon atoms are known as alkynes.

1) The number of carbon on the longest chain = 5, (two carbon atoms between the triple bond) therefore the prefix is pen-

The functional group is the triple bond on the first carbon with the suffix -yne

The name of the compound is 1-Pentyne

2) There are 4 carbon atoms on the longest chain, therefore, the prefix is but-

The location of the triple bond functional group (suffix; -yne) is on the second carbon atom

Therefore, the name of the compound is But-2-yne

3) The triple bond functional group is between the only two carbon atoms, therefore, the name is Ethyne

4) The number of carbon atoms = 3, with a prefix prop-, and the functional group on the first carbon atom on the longest chain is the triple bond, with suffix -yne

Therefore, the name of the compound is 1-Propyne

5) The number of carbon atoms on the longest chain = 4, the prefix is but-

The location of the triple bond functional group (suffix -yne) is on the first carbon atom

Therefore, the name is 1-Butyne

6) The number of carbon atoms on the longest chain = 5, the prefix of the compound is pen-

The location of the triple bond functional group (suffix; -yne) is on the second carbon atom, therefore;

The name of the compound is Pent-2-yne

Learn more about alkyne here:

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6 0

2 years ago

What is pollution ? briefly

Marrrta [24]

Answer:

pollution there are many kind of pollution..... water pollution is when people dumb and trash into the ocean and air pollution is when factories pump coal and other substance to make light but also cause harmful gas into the atmosphere.

Explanation:

7 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