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3 years ago

Which type of fermentation produces CO2 bubbles in baking?

Chemistry

1 answer:

Anika [276]3 years ago

6 0

Alcoholic fermentation fermentation produces CO2 bubbles in baking.

<u>Explanation:</u>

The other name given for the Alcoholic Fermentation is Ethanol fermentation. In this process of fermentation, ethanol and carbon dioxide are the resultant by-products. These are formed by the conversion of fructose,sucrose and glucose to cellular energy. This type of fermentation do not require oxygen for the process to take place. Hence, these are known to be an anaerobic process

This type of fermentation has its application like ethanol fuel production, cooking of bread, etc. A dough rises of the Ethanol fermentation. this is because, the sugars that are present in a dough are absorbed by yeast . this produces ethanol and carbon dioxide. During baking process,bubbles are formed by this carbon dioxide.

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A uniform cylinder of mass 4.3 kg and radius 0.4 m rolls down a ramp inclined at an angle 0.15 radians to the horizontal. what i

tester [92]

The cylinder's acceleration is

$a=\frac{gsin}{1+\frac{I}{mrL} }$

$= 2g\frac{sin}{3}$ θ

$= \frac{2(9.8) sin (0.15 rad)}{3} \\=0.9763 \frac{m}{sec r}$

<h3>How do you determine a cylinder's acceleration?</h3>

The cylinder's complete radius, $R$ , from the center marks the contact point, therefore the torque created by friction is given by $fR = I\alpha$ , where is the rotating acceleration. This rotational acceleration's corresponding linear acceleration, $\alpha$ , is equal to $R$ . The cylinder, which has its mass concentrated in its center, triumphs in the competition, followed by the disc and the hoop, with their respective final velocities being roughly $1.4:1.2:1$ . We can also see that our findings are unaffected by the cylinders' masses.

To learn more about Acceleration of cylinder, visit

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5 0

2 years ago

How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?

Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

$C_{6}H_{8}O_{6}$

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

$C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g$

To know the percent of each element, we need to to the following:

$C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)$

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

$C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}$

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

$C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol$

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

5 0

3 years ago

Balance the following reaction:

miv72 [106K]

The balanced reaction is as below

3A₂B + 2DC₃→ 6 AC + D₂B₃

The number that must be to the left of AC is 6

Explanation

According to the law of mass conservation , the number of atoms in reactant side must be equal to number to the number of atoms in product side.

Therefore the equation above is balance since it obey the law of mass conservation.

For example there is 6 atoms of A in reactant side and 6 in product side.

5 0

3 years ago

Read 2 more answers

HELPPP PLS

photoshop1234 [79]

I think it could be C maybeee though

6 0

3 years ago

Type the correct answer in each box.

Arturiano [62]

Answer:

5SiO2 + 2CaC2 = 5Si + 2CaO + 4CO2

Explanation:

balancing equations is a lot of trial and error. My strategy to approaching this equation was to get the O's balanced. After trying several combonations I found that I needed 10 O's on each side of the equation for the other elements to match up. After I balanced the O's, I balanced my C's to 4 on each side. Then I balanced my Ca's to have 2 on each side. And last but not least I balanced my Si to have 5 on each side.

8 0

3 years ago