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3 years ago

if you take a dried out white rose and place the stem in food coloring, will the rose turn the color you put the stem in

Chemistry

1 answer:

cricket20 [7]3 years ago

7 0

Answer:

Yes, this is true. The reason is that the flower transpires and sucks the water in and distributes it as much as it can. You can also flip it upside down and hang it with petals down , allowing the liquid to enter the flower and then retaining color for longer periods of time and having more color.

Explanation:

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What components did you collect to convert your two molecules of pyruvate to acetyl-coa?

GenaCL600 [577]

You need to have NAD+ as a source of oxidation for the pyruvate, as well as a supply of coenzyme A. CO2 is released by the pyruvate as a carboxyl group is removed

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3 years ago

Question # 10

satela [25.4K]

The answer should be D, cells can only create an identical copy of the original cell.

3 0

3 years ago

How many atoms of hydrogen are in 0.500 mol of ch3oh molecules?

cestrela7 [59]

In 1 mol of CH3OH, you have 4 H-atoms (because 3 H-atoms are attached to the C-atom, and one H-atom in the OH group). That means in 0.500 mol of CH3OH, you have 2 H-atoms since it is halved. And then we have Avogadro's constant: 6.02 * 1023.

The question asks for how many hydrogen atoms there are in 0.500 mol CH3OH. Using the numbers that we have (Avogadro's constant and no. of H-atoms), the answer of the question will be something like:

H-atoms in CH3OH = 2 * 6.02 * 1023 = ~1.2 * 1024

8 0

3 years ago

The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa

Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

$\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

$\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%$

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

$\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%$

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

8 0

3 years ago

It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high

lara [203]

Answer:

Biphenyl

Explanation:

The reaction of bromo benzene with magnesium-ether solution yields a Grignard reagent.

The byproduct of this reaction is biphenyl. It is formed when two unreacted bromobenzene molecules are coupled together.

Hence, It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of biphenyl by-product formed.

4 0

3 years ago