The correct answer is option A.
Presence of a co-factor is required for the activation of some enzymes. Co-factors are generally metal ions or small molecules (non-protein).Co-factors helps enzymes bind substrate tightly. For example, NAD⁺/NADH is a co-factor which is required for the activation of the enzyme dehydrogenase. NAD⁺ is the oxidized form and NADH is the reduced form.
(1) argon (3) chlorine
(2) bromine (4) sulfur
Answer: Option (2) is the correct answer.
Explanation:
STP means standard temperature and pressure.
At STP chlorine exists as a gas, sulfur exists as a solid and argon also exists as a gas.
Whereas bromine exists as a liquid at STP.
Thus, we can conclude that out of the given options bromine is the element which exists as a liquid at STP.
The element that is a liquid at standard temperature and pressure (STP) is bromine (Br). Therefore, option 2 is correct.
Bromine is the only element among the options given that is a liquid at STP. It is a reddish-brown liquid with a strong and unpleasant odor. Bromine is a halogen and exists as diatomic molecules (Br₂).
It has a boiling point of 58.8°C (137.8°F) and a melting point of -7.2°C (19.0°F). Bromine is significantly more reactive than argon but less reactive than chlorine and sulfur.
Argon (Ar) is a noble gas and exists as a colorless and odorless gas at STP. Chlorine (Cl) is a greenish-yellow gas at STP. Sulfur (S) is a solid at STP and melts at a relatively high temperature.
To learn more about the element, follow the link:
#SPJ6
32 units
128 units
256 units
Answer:
The answer is 128
Explanation:
B. newton.
C. gram.
D. nanometer.
is required?
a. 7.00 mL
b. 8.40 mL
c. 17.1 mL
d. 58.3 mL
Answer:
Explanation:
To determine the volume of the stock solution required to prepare 3.50 L of 0.200 M hydro chloric acid, we can use the formula:
M1V1 = M2V2
where:
M1 = concentration of the stock solution
V1 = volume of the stock solution
M2 = desired concentration of the diluted solution
V2 = desired volume of the diluted solution
Let's substitute the given values into the formula:
M1 = 12.0 M
V1 = ?
M2 = 0.200 M
V2 = 3.50 L
Now we can solve for V1:
12.0 M x V1 = 0.200 M x 3.50 L
V1 = (0.200 M x 3.50 L) / 12.0 M
V1 = 0.0583 L
To convert the volume from liters to milliliters, we multiply by 1000:
V1 = 0.0583 L x 1000 mL/L
V1 = 58.3 mL
Therefore, the volume of the stock solution required is 58.3 mL.
So, the correct answer is d. 58.3 mL.
To determine the volume of the stock solution required, we can use the formula:
Molarity1 x Volume1 = Molarity2 x Volume2
Where Molarity1 and Volume1 represent the initial concentration and volume, and Molarity2 and Volume2 represent the final concentration and volume.
Given:
Molarity1 = 12.0 M
Volume1 = ?
Molarity2 = 0.200 M
Volume2 = 3.50 L
Plugging in the values into the formula, we have:
12.0 M x Volume1 = 0.200 M x 3.50 L
Simplifying the equation, we can solve for Volume1:
Volume1 = (0.200 M x 3.50 L) / 12.0 M
Volume1 ≈ 0.0583 L
To convert this to milliliters, we multiply by 1000:
Volume1 ≈ 58.3 mL
Therefore, the volume of the stock solution required is approximately 58.3 mL.
The closest answer option is d. 58.3 mL.
I hope this explanation helps! Let me know if you have any further questions.