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2 years ago

Why is regulation of the cell cycle important?

1 answer:

3 0

Adult human being =100 trillion cells
All cells originated from single cell essential for: propagation
Replacement of dead cells
Increasing number of cells for specific functions. Cell cycle regulation is disrupted in cancer

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Help out please?????

Alexxandr [17]

Answer:

A non- metal

Explanation:

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5 0

2 years ago

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Which pair of reactants will undergo aldol condensation to produce methylvinyl ketone?

xenn [34]

When formaldehyde and acetone then react with each other( aldol condensation) then it will be formed methyl vinyl ketone.

In organic chemistry, an aldol condensation would be a condensation reaction in which an enol and enolate ion combines with a carbonyl chemical to produce a -hydroxy aldehyde or -hydroxy ketone, that is then dehydrated to produce a conjugated enone.

In aldol condensation, when formaldehyde and acetone then react with each other then it will be formed methyl vinyl ketone.

It can be written as

$CH_{2} O + CH_{3} COCH_{3}$ → $HOCH_{2} -CH_{2} -CO-CH_{3}$

When it will be heated then it gives methyl vinyl ketones.

$HOCH_{2} -CH_{2} -CO-CH_{3}$ → $CH_{2} =CH-CO-CH_{3}$

So, the pair of reactants will be formaldehyde and acetone

To know more about aldol condensation

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1 year ago

After learning about climate change in her science class, selena wants to make a few changes at home so she can affect climate c

ivanzaharov [21]

Selena can change from burning fossil fuel in her home to using solar energy.

<h3>What causes climate changes?</h3>

Climate change is caused by the release of the greenhouse gases which is gotten through the use and burning of fossil fuels such as coal.

The use of solar panels and solar energy by Selena in her house can help adjust the climate changes by decreasing the amount of greenhouse gases released into the atmosphere.

Learn more about greenhouse gases here:

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8 0

1 year ago

What is meant by the term sensitivity in Analytical Chemistry? (a) The term refers to the limit at which the absorption instrume

ivolga24 [154]

The term sensitivity in Analytical Chemistry is "the slope of the calibration curve or a function of analyte concentration or amount".

Answer: Option B

Explanation:

In a sample, the little amounts of substances can be accurately evaluated by a method is termed as "Analytical sensitivity". This detect a target analyte like an antibody or antigen, process is considered as potential of a test to and generally demonstrated as the analyte's minimum detectable concentration.

The acceptable diagnostic sensitivity is not guaranteed by high analytical sensitivity. The percentage of individuals who have a given disarray who are identified by the method as positive for the disarray is known as "Diagnostic sensitivity".

6 0

2 years ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

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2 years ago