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ki77a [65]
3 years ago
8

One of the products of a fermentation reaction is

Chemistry
1 answer:
Alex73 [517]3 years ago
5 0
There are two types of fermentation aerobic and anaerobic. Aerobic are those that need oxygen to ferment while anaerobic are those who do not need oxygen. Products of fermentation are usually water, ethanol, lactic acid and carbon dioxide. Other products are acetone and butyric acid.
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I will give brainliest!!!!!
creativ13 [48]

Answer:

1. equator

2. west to east

3. equator

4. north to south

5. hemisphears

Explanation:

Mark as brainliest please

4 0
2 years ago
A 25.00 mL sample of 0.320 M KOH is titrated with 0.750 M HBr at 25 °C.
Fudgin [204]

Answer:

a. pH = 13.50

b. pH = 13.15

Explanation:

Hello!

In this case, since the undergoing chemical reaction between KOH and HBr is:

HBr+KOH\rightarrow KBr+H_2O

As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:

pOH=-log([OH^-])=-log(0.320)=0.50

Thus, the pH is:

pH+pOH=14\\pH=14-pOH=14-0.50\\pH=13.50

Which is the same answer for a and b as they ask the same.

Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:

n_{KOH}=0.02500mL*0.320mol/L=0.00800mol\\\\n_{HBr}=0.005L*0.750mol/L=0.00375mol

It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:

n_{KOH}^{remaining}=0.00425mol

Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):

[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M

So the pOH and the pH turn out:

pOH=-log(0.142)=0.849\\pH+pOH=14\\pH=14-pOH=14-0.849\\pH=13.15

Best regards!

7 0
3 years ago
Consider an electron with charge −e−e and mass mmm orbiting in a circle around a hydrogen nucleus (a single proton) with cha
PtichkaEL [24]

Answer:

Explanation:

The net force on electron is electrostatic force between electron and proton in the nucleus .

Fc = \frac{1}{4\pi\epsilon} \times \frac{e\times e}{r^2}

This provides the centripetal force for the circular path of electron around the nucleus .

Centripetal force required = \frac{m\times v^2}{r}

So

\frac{m\times v^2}{r}=\frac{1}{4\pi\epsilon} \times \frac{e\times e}{r^2}

v^2=\frac{e^2}{4\pi \epsilon m r}

v=(\frac{e^2}{4\pi \epsilon m r})^{\frac{1}{2} }

5 0
2 years ago
How many significant figures<br> are in this number?<br> 107.051
NARA [144]

Answer:

there are 6 significant figures in 107.051

7 0
3 years ago
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How much does the mass of 12.65-g sample of copper(II) nitrate hexahydrate decrease when heated?
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