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sukhopar [10]
3 years ago
10

One would not expect yeast to undergo fermentation in the presence of

Chemistry
1 answer:
Hunter-Best [27]3 years ago
3 0

Answer:

Oxygen.

Explanation:

Hello,

Yeast fermentation in the food industry must be carried out under anaerobic conditions as long as when oxygen is present respiration occurs rather than fermentation.

Best regards.

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A researcher claims that an ancient scroll originated from greek scholars in about 500 bce. a measure of its carbon-14 decay rat
Naddik [55]
The amount of substance present in a certain object with a given half-life in terms of h can be expressed through the equation,

     A(t) = (A(o))(0.5)^(t/h)

where A(t) is the amount of substance after t years and A(o) is the original amount. In this item we are given that A(t)/A(o) is equal to 0.89. Substituting the known values,

     0.89 = (0.5)(t / 5730 years)

The value of t from the equation is 963.34 years.

<em>Answer: 963 years</em>
8 0
3 years ago
A can contains a gas with a volume of 86 mL at 30oC. What is the volume in the can if it is heated to 65oC?
Ber [7]

Answer:

New volume of gas = 95.93 ml (Approx)

Explanation:

Given:

Old volume of gas = 86 ml

Old temperature = 30°C = 30 + 273 = 303 K

New temperature = 65°C = 65 + 273 = 338 K

Find:

New volume of gas

Computation:

V1T2 = V2T1

(86)(338) = (V2)(303)

New volume of gas = 95.93 ml (Approx)

8 0
2 years ago
ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a
Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{20\text{ min}}

k=3.465\times 10^{-2}\text{ min}^{-1}

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 3.465\times 10^{-2}\text{ min}^{-1}

t = time passed by the sample  = 80 min

a = initial amount of the reactant  = 1.6 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}

a-x=0.100M

Therefore, the concentration of A after 80 min is, 0.100 M

3 0
2 years ago
What is the unit for volume?<br> Atm<br> Liters<br> Mol
Mrrafil [7]

Liters.

Hope this helps.

5 0
3 years ago
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