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3 years ago

Define fermentation and how alcohol is produced. Describe distillation, the process used in forming " spirits".

Chemistry

1 answer:

Dimas [21]3 years ago

8 0

Answer:

Fermentation is where all alcohol is created, distillation is where the alcohol is separated and removed. In order for fermentation to occur, two things are needed: a raw material in liquid form that contains sugar, followed by the addition of yeast.

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A container of gas has a volume of 3.5 L and a pressure of 0.8 atm. Assuming the temperature remains constant, what volume of ga

strojnjashka [21]

Answer:

The correct answer is option A.

Explanation:

Initial volume of the gas = $V_1=3.5 L$

Final volume of the gas = $V_2=?$

Initial pressure of the gas = $P_1=0.8 atm$

Final volume of the gas = $P_2=0.5 atm$

Using Boyle's law:

$P_1V_1=P_2V_2$

$0.8 atm \times 3.5 L=0.5 atm\times V_2$

$V_2=5.6 L$

Hence,the correct answer is option A.

7 0

3 years ago

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An atom has a negative charge. Which of the following must have occurred?

galina1969 [7]

It should be A)It lost a neutron.

5 0

3 years ago

Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

Explain how human activity can affect the carbon cycle.

Eva8 [605]

Answers:

Human activities badly affect carbon cycle. Activities such as burning fossil fuel and deforestation have begun to effect on carbon cycle and the rise of carbon dioxide in atmosphere.

Explanation:

The carbon cycle can be affect when carbon dioxide is either released into atmosphere or remove from atmosphere. When fossil are burnt , carbon is release to the atmosphere at faster rate then it is removed. Natural gas, oil, coal, and other industrial products all are affecting carbon cycle in atmosphere.

Deforestation means permanent removal of trees from forest which cause increase in level of carbon dioxide because no trees longer to absorb carbon dioxide from atmosphere. Which result affect on carbon cycle.

4 0

3 years ago

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• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo

garik1379 [7]

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c) 0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

$H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O$

1 mole 3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

$H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O$

10 ml 17.50 ml

(x) M 0.200 M

Molarity = $\frac{0.2*17.5}{1000}$