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1 year ago

Which situation is the best example of regulation in an economic system?

Chemistry

1 answer:

velikii [3]1 year ago

5 0

Hope this helps!

Answer: The correct option is: "A state agency has been created to monitor the production and distribution of sports drinks."

Explanation:

The economic regulation are the provisions through which the government intervenes in the markets to set prices or quantities of production, or establish technical specifications and in general, restrictions that must be met by citizens and companies to participate in a market.

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Calculate the molality (m) of solution containing 20.5 g of ethylene glycol C,H,O,

Svetllana [295]

Answer:

Given

mass of solute (m) =20.5g

mass of solvent( ms) =155g

solution

malality =mX1000/mrxms

M=20.5x1000/62x155

M=2.1332

5 0

3 years 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

3 years ago

A heavy textbook and a QOD journal sit on the edge of your desk. Which has more gravitational potential energy?

fenix001 [56]

The textbook Would most likely have more gravitational potential energy because it is heavier. Things that are heavier have a larger gravitational pull and are pulled to the earth faster

4 0

3 years ago

A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

elena55 [62]

Answer:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$