How does botulinum toxin block synaptic transmission?

kicyunya [14]

The rate of the reaction is measurable quantity that refers to the amount or how much is are chemical substances reagents used up or converted into the product over some period of time.

Rate = change in the amount/time.

This can indirectly be observed through many ways, such as the volume of gas given off if the byproduct is a gas being produced, the colour of the solution etc.

4 0

2 years ago

How did Ernest Rutherford’s experiment relate to J.J. Thomson’s work?

horsena [70]

Thomson realized that the accepted model of an atom did not account for negatively or positively charged particles. Therefore, he proposed a model of the atom which he likened to plum pudding. ... Rutherford with the assistance of Ernest Marsden and Hans Geiger performed a series of experiments using alpha particles.

8 0

2 years ago

A gas at constant temperature is enclosed in a container with a movable piston. It occupies a space of 10L at a pressure of 500

Liula [17]

Answer:

4.17L

Explanation:

V1 = 10L

V2 =?

P1 = 500torr

P2 = 1200torr

Boyle's law states that at constant temperature, the volume of a gas is inversely proportional to its pressure.

P1V1 = P2V2

V2 = ( P1 * V1 ) / P2

V2 = 4.17L

The new volume of the gas is 4.17L

8 0

2 years ago

Taking your foot off of the gas pedal when you see a police car which of the Newton’s law

larisa [96]

Answer:

newton's third law of motion

Explanation:

it states about action and reaction when you see an action you take reaction like taking off your foot from the gas pedal

3 0

1 year ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

2 years ago