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nadya68 [22]
3 years ago
9

5

Chemistry
2 answers:
brilliants [131]3 years ago
5 0

Answer:

A. move very little, so it cannot easily change its shape.

sineoko [7]3 years ago
4 0

Answer:

this answer is b no u can it very easy

Explanation:

because solid and liquid can compress easily

plz support me

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The aldol condensation involves nucleophilic attack of an enolate to a carbonyl. Use the pop-up menus to identify the total numb
Sindrei [870]

Aldol condensation involves the reaction of an acid or base with a carbonyl group producing a nucleophile that attacks another carbonyl compound  to yield a β-hydroxyaldehyde or β-hydroxyketone compound.

<h3>What is aldol condensation?</h3>

The aldol condensation is a reaction in organic chemistry in which there is a reaction between an acid or base and a carbonyl group which then serves as the nucleophile that attcks a second carbonyl to yield a β-hydroxyaldehyde or β-hydroxyketone compound.

The aldol condensation may be acid catalysed or base catlysed. The question is incomplete hence the complete mechanimsms can not be decuced.

Learn more about aldol condensation: brainly.com/question/9415260

3 0
2 years ago
The energy acquired in the light-dependent reactions is used in the light-independent reactions to build glucose molecules. How
jeyben [28]

Answer: through energy carriers, ATP and NADPH

Explanation:in the light dependent stage,energy from a light photon is used to create ATP through ADP and an inorganic phosphate.

It does this by the transfer of energetic electron from one electron carrier to another.NADPH is also formed.

In the light independent reaction,ATP and NADP are used to reduce carbon dioxide to 3-phosphoglycerate

8 0
3 years ago
A synthesis reaction takes place when carbon monoxide (CO) and hydrogen gas (H2) react to form methanol (CH3OH). How many grams
kakasveta [241]

Answer:

A. 3.2grams

Explanation:

First we need to get the chemical equation for this reaction:

CO + H₂ → CH₃OH

We then need to balance the equation:

CO + <u>2</u>H₂ → CH₃OH

Our next step is to first convert our given into moles. We do this by figuring out first how many grams of each substance there are in 1 mole by adding up the atomic mass of each element in each substance.

             Carbon(1)       Oxygen(1)

CO =       12.011(1)      +   15.999(1)   = 28.01 g/mole

            Hydrogen(2)

H₂ =        1.008(2)        =  2.016g/mole

We can then use this to determine how many moles of each reactant we have given the mass.

CO = 2.8g\\2.8g\times\dfrac{1mole}{28.01g}=0.1moles\\\\H_{2}=0.50g\\\\0.50g\times\dfrac{1mole}{2.016g}=0.469moles

So here we have our new given:

0.1 moles of CO

0.496 moles of H₂

We then need to determine how much product we produce with our given.

According to our chemical equation we can assume that:

For every 1 mole of CO we can produce 1 mole of CH₃OH

Fore every 2 moles of H₂ we can produce 1 mole of CH₃OH

Using this ratio we can determine how much product each reactant will produce by using the ratios:

0.1 moles of CO\times\dfrac{1moleofCH_{3}OH}{1moleofCO}=0.1moles of CH_{3}OH\\\\ 0.496molesofH_{2}\times\dfrac{1moleofCH_{3}OH}{2molesofH_{2}}=0.248molesofCH_{3}OH

Now notice that they are not equal. The reactant that produces the least amount of product will be our limiting reactant. The limiting reactant determines the amount of product that is produced, because once it is completely used up, the reaction stops. So in this case, the amount of CH₃OH that is produced is 0.10 moles.

Now we need to figure out how many this is in grams. To do that we need to find out how many grams of CH₃OH there are in 1 mole of CH₃OH.

                   Carbon (1)             Hydrogen(4)          Oxygen(1)

CH₃OH =      12.001(1)     +           1.008(4)       +       15.999(1)

            =      12.001         +           4.032          +        15.999     = 32.032g/mole

We then use this for converting:

0.10 moles\times\dfrac{32.032g}{1mole}=3.2032g

So the reaction will produce 3.2032g of CH₃OH or 3.2g of CH₃OH

3 0
3 years ago
What is the of 0.068 M of OH^ - ion?
Marina CMI [18]

Answer:

reeeeeeeeeeeeee

Explanation:

4 0
3 years ago
A reactant decomposes with a half-life of 29.5 s when its initial concentration is 0.229 M. When the initial concentration is 0.
Dmitrij [34]

Answer :

The order of reaction is, 0 (zero order reaction).

The value of rate constant is, 0.00388Ms^{-1}

Explanation :

Half life : It is defined as the time in which the concentration of a reactant is reduced to half of its original value.

The general expression of half-life for nth order is:

t_{1/2}\propto \frac{1}{[A_o]^{n-1}}

or,

\frac{t_{1/2}_1}{t_{1/2}_2}=\frac{[A_2]^{n-1}}{[A_1]^{n-1}}

or,

n=\left(\frac{\log\frac{(t_{1/2})_1}{(t_{1/2})_2}}{\log\frac{(A)_2}{(A)_1}}\right )+1           .............(1)

where,

t_{1/2} = half-life of the reaction

n = order of reaction

[A] = concentration

As we are given:

Initial concentration of A = 0.229 M

Final concentration of A = 0.639 M

Initial half-life of the reaction = 29.5 s

Final half-life of the reaction = 82.3 s

Now put all the given values in the above formula 1, we get:

n=\left (\frac{\log \frac{29.5}{82.3}}{\log\frac{0.639}{0.229}}\right )+1

n=0.000196\approx 0

Thus, the order of reaction is, 0 (zero order reaction).

Now we have to determine the rate constant.

To calculate the rate constant for zero order the expression will be:

t_{1/2}=\frac{[A_o]}{2k}

When,

t_{1/2} = 29.5 s

[A_o] = 0.229 M

29.5s=\frac{0.229M}{2k}

k=0.00388Ms^{-1}

Thus, the value of rate constant is, 0.00388Ms^{-1}

4 0
3 years ago
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