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2 years ago

Acetylcholine is released by many types of neurons. which type does not typically release ach?

Chemistry

1 answer:

rodikova [14]2 years ago

8 0

Answer:

hope it's help you ok have a good day

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What is the surface tension of H20

andrew11 [14]

Answer:

the surface tension of H20 is 72 dynes/cm at 25°C

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3 years ago

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MrRa [10]

Answer:

all the statements are true of chemical changes

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2 years ago

2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo

pogonyaev

Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

Chemical reaction involved: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: $Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2$

To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:

$\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}$

$\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)$

$\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)$

Therefore, the Gibb's free energy change at 37° C (310.15 K): ΔG = (-2.45 kJ/mol)

4 0

3 years ago

Which of the following is part of the central nervous system?

OLga [1]

The brain is part of the central nervous system.

Nerves and sensors are part of the peripheral nervous system. Only the brain and spinal cord make the central nervous system.

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3 years ago

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Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U

Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of nitrogen gas = $79mL/s$

$R_2$ = rate of effusion of sulfur dioxide gas = ?

$M_1$ = molar mass of nitrogen gas = 28 g/mole

$M_2$ = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}$

$R_2=52mL/s$

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

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3 years ago