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3 years ago

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Which hydrocarbon name adheres correctly to the IUPAC naming system? 3–ethyl–2,4–dimethylhexane 3–methethyl–2,dimethylhexane 4–d

imethylhexane–3–ethyl 4–dimenthylhexane, 3–ethyl

1 answer:

faltersainse [42]3 years ago

4 0

3–ethyl–2,4–dimethylhexane

CH₃-CH(CH₃)-CH(C₂H₅)-CH(CH₃)-CH₂-CH₃

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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 freezer compartment is covered with a 2-mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient

ruslelena [56]

Answer:

The time required to melt the frost is 3.25 hours.

Explanation:

The time required to melt the frost dependes on the latent heat of the frost and the amount of heat it is transfered by convection to the air .

The heat transferred per unit area can be expressed as:

$q=h_c*A*\Delta T\\\\q/A=h_c*\Delta T$

being hc the convective heat transfer coefficient (2 Wm^-2K^-1) and ΔT the difference of temperature (20-0=20 °C or K).

$q/A=h_c*\Delta T=2\frac{W}{m^2K}*20K=40\frac{W}{m^2}$

If we take 1 m^2 of ice, with 2 mm of thickness, we have this volume

$V=T*A = 0.002 m * 1 m^2=0.002m^3$

The mass of the frost can be estimated as

$M=\rho * V=700\frac{kg}{m^3}*0.002m^3= 1.4 kg$

Then, the amount of heat needed to melt this surface (1 m²) of frost is

$Q=L*M=334\frac{kJ}{kg}*1.4kg= 467.6kJ$

The time needed to melt the frost can be calculated as

$t=\frac{Q}{(q/A)}=\frac{467.6kJ/m2}{40W/m2} = 11.69\frac{kJ}{W}*\frac{1W*s}{1J}*\frac{1000J}{1kJ}= 11690s=3.25h$

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3 years ago

Which of the following statements about moles is correct?

jasenka [17]

The answer is A. A mole of Cu has the same number of atoms as a mole of He atoms

5 0

3 years ago

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What would be the vapor pressure at 25.0 °c of a solution of 5.00 g of glucose (c6h12o6) in 100.0 g of ethanol (c2h5oh)?

Crank

Use Raoult's Law:
Psolution = (χsolvent) (P°solvent)

24.90 = (x) (25.756)

x = 0.966765 (this is the solvent mole fraction)

χsolute = 1 - 0.966765 = 0.033235

χsolute = 0.03324 (to four sig figs)

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3 years ago

What mass of solute is needed to prepare each of the following solutions?

noname [10]

Answer:

a. 21.7824 g

b. 0.2362 g

c. 31.5273 g

Please see the answers in the picture attached below.

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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3 years ago