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3 years ago

Why do we need to study chemistry?

1 answer:

3 0

Chemistry is very important to study. Chemistry explains the natural world around us and also explains certain chemical processes that happen here on earth.

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A bacterium is in a nasty environment. What sort of reproduction would it use in this situation? Why? How else would it protect

lana [24]

The bacteria in nasty environment undergoes multiple fission.

<h3><u>Explanation</u>:</h3>

The bacteria is a unicellular prokaryotic organisms that are found in each and every places of the world. They can survive in extremes of temperatures and pH. They can save themselves through special processes in the extreme climates.

The bacteria undergoes multiple fission in these climates. They cover themselves up with a strong and tough capsule inside which they undergo several Binary fissions. This leads to the formation of multiple cells enclosed with a capsule.

With the return of the favourable climate, the capsule rupture and these newly formed cells come out.

7 0

2 years ago

The rate constant for a certain reaction is measured at two different temperatures:

Talja [164]

Answer: The activation energy Ea for this reaction is 22689.8 J/mol

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

$ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

$k_1$ = rate constant at temperature $T_1$ = $2.3\times 10^8$

$k_2$ = rate constant at temperature $T_2$ = $4.8\times 10^8$

$E_a$ = activation energy = ?

R= gas constant = 8.314 J/kmol

$T_1$ = temperature = $280.0^0C=(273+280)=553K$

$T_2$ = temperature = $376.0^0C=(273+376)=649K$

Putting in the values ::

$ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]$

$E_a=22689.8J/mol$

The activation energy Ea for this reaction is 22689.8 J/mol

3 0

2 years ago

PLEASE ANSWER THIS QUESTION ASAP.

Natali5045456 [20]

Equal is the answer

Hope this helps

6 0

2 years ago

Read 2 more answers

How many atoms are found in 3.45g of CO2?

USPshnik [31]

<u>Answer:</u> The number of carbon and oxygen atoms in the given amount of carbon dioxide is $4.72\times 10^{22}$ and $9.44\times 10^{22}$ respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of carbon dioxide gas = 3.45 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in above equation, we get:

$\text{Moles of carbon dioxide gas}=\frac{3.45g}{44g/mol}=0.0784mol$

1 mole of carbon dioxide gas contains 1 mole of carbon and 2 moles of oxygen atoms.

According to mole concept:

1 mole of a compound contains $6.022\time 10^{23}$ number of molecules

So, 0.0784 moles of carbon dioxide gas will contain $1\times 0.0784\times 6.022\times 10^{23}=4.72\times 10^{22}$ number of carbon atoms and $2\times 0.0784\times 6.022\times 10^{23}=9.44\times 10^{22}$ number of oxygen atoms

Hence, the number of carbon and oxygen atoms in the given amount of carbon dioxide is $4.72\times 10^{22}$ and $9.44\times 10^{22}$ respectively

3 0

3 years ago

Draw the major product formed when the structure shown below undergoes substitution in ch3ch2oh with heat. (use a wavy bond (sin

muminat

Answer:

Explanation:

The missing image is attached below.

The objective of this question is to draw the major product formed from the diagram attached below.

From the diagram attached, we will see the reaction of a tertiary alkyl halide together with a weak nucleophile (ch3ch2oh) undergoing a nucleophilic substitution (SN₁) mechanism to yield a racemic mixture(i.e., compound that is not optically active but contains an equal amount of dextrorotatory and levorotatory stereoisomers) as a product.

7 0

2 years ago