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barxatty [35]
3 years ago
13

Is a match getting set on fire a chemical change?

Chemistry
1 answer:
forsale [732]3 years ago
7 0
Yes because it can not go back to its previous state
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In yeast, ethanol is produced from glucose under anaerobic conditions. A cell‑free yeast extract is placed in a solution that co
12345 [234]

Answer:

650 mmol.

Explanation:

The equation for the fermentation of one mole of glucose is:

C₆H₁₂O₆ + 2 NAD⁺ + 2 ADP + 2 P i + 2 NADH → 2 EtOH + 2 ATP + 2 NADH + 2 NAD⁺

Since NAD⁺/NADH is used and regenerated, we can eliminate it from the equation:

C₆H₁₂O₆ + 2 ADP + 2 P i  → 2 EtOH + 2 ATP

With the equation, we calculate the maximum amount of ethanol that could be obtained theoretically:

1000 mmol C₆H₁₂O₆ ------------ 2000 mmol EtOH

325 mmol C₆H₁₂O₆ ------------- x= 650 mmol EtOH

Therefore, the maximum amount of ethanol that could be produced is 650 mmol.

8 0
3 years ago
Calculate the number of C, H, and O atoms in 1.50 g of glucose, a sugar
DIA [1.3K]
Chemical formula of the glucose: C₆H₁₂O₆

We calculate the molar mass:
atomic mass (C)=12 u
atomic mass (H)=1 u
atomic mass (O)=16 u

atomic weight (C₆H₁₂O₆)=6(12 u)+12(1u)+6(16 u)=72 u+12u+96 u=180 u.
Therefore : 1 mol of glucose will be 180 g
The molar mass would be: 180 g/ mol


2) we calculate the number of moles of 1.5 g.
180 g---------------------1 mol
1.5 g----------------------  x

x=(1.5 g * 1 mol) / 180 g≈8.33*10⁻³ moles

we knows that:
1 mol = 6.022 * 10²³ particles (atoms or molecules)

3)We calculate the number of molecules:

Therefore:
1 mol-----------------------6.022*10²³ molecules of glucose
8.33*10⁻³ moles--------        x

x=(8.33*10⁻³ moles * 6.022*10²³ molecules)/1 mol≈5.0183*10²¹ molecules.

4)We calculate the number of C, H and O atoms:
A molecule of glucose have 6 atoms of C, 12 atoms of H, and 6 atoms of O,
number of atoms of C=(6 atoms/1 molecule)(5.0183*10²¹molecules)≈
3.011*10²²

number of atoms of H=(12 atoms/1 molecule)(5.0183*10²¹ molecules)≈
6.022*10²² .

number of atoms of O=(6 atoms/1 molecule)(5.0183*10²¹ molecules)≈
3.011*10²²

Answer: we have 3.011*10²² atoms of C, 6.022*10²² atoms of H, and 3.011*10²² atoms of O.
4 0
4 years ago
How much energy is required to vaporize 185 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol
Diano4ka-milaya [45]
Answer is: 73.52 kJ<span> of energy is required to vaporize butane.
</span>m(C₄H₁₀) = 185 g.
n(C₄H₁₀) = m(C₄H₁₀) ÷ M(C₄H₁₀).
n(C₄H₁₀) = 185 g ÷ 58.12 g/mol.
n(C₄H₁₀) = 3.18 mol; amount of butane.
Hvap = 23.1 kJ/mol; <span>the heat of vaporization for butane.
</span>Q = Hvap · n(C₄H₁₀).
Q = 23.1 kJ/mol · 3.18 mol; energy.
Q = 73.52 kJ.

4 0
4 years ago
Using the following thermochemical data, what is the change in enthalpy for the following reaction: 3H2(g) + 2C(s) + ½O2(g) → C2
sweet [91]
It would be -277.6 KJ/mol
7 0
3 years ago
Read 2 more answers
When is an electric charges neutral ??
dangina [55]

Answer:

Atoms of matter are electrically neutral because their nuclei contain the same number of protons as there are electrons surrounding the nuclei. Electric current and charged objects involve the separation of some of the negative charge of neutral atoms.

4 0
3 years ago
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