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nikdorinn [45]
1 year ago
15

The heat of combustion of propane, C3H8, is 2220 kJ/mol. The specific heat of copper is 0.385 J/g°C. How many grams of propane m

ust be burned to raise the temperature of a 10.0 kg block of copper from 25.0°C to 65.0°C, assuming none of the heat is lost to the surroundings?
Chemistry
1 answer:
Mamont248 [21]1 year ago
5 0

Answer:

Explanation:

q = (mass) (temp change) (specific heat)

q = (10000 g) (40 °C) (0.385 J/g⋅°C) = 154000 J = 154 kJ

154 kJ / 2220 kJ/mol = 0.069369369 mol

0.069369369 mol times 44.0962 g/mol = 3.06 g (to three sig figs)

answer choice 4

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Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from easiest to hardest. to rank
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The rate of Formation of Carbocation mainly depends on two factors'

                    1)  Stability of Carbocation:
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                     2) Ease of detaching of Leaving Group:
                                                                                   The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,

                                            R-I > R-Cl > R-F

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Platinum, which is widely used as a catalyst, has a work function φ(the minimum energy needed to eject an electron from the meta
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Answer:

A. \lambda_0=2.196\times 10^{-7}\ m

Explanation:

The work function of the Platinum = 9.05\times 10^{-19}\ J

For maximum wavelength, the light must have energy equal to the work function. So,

\psi _0=\frac {h\times c}{\lambda_0}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

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\psi _0=Work\ function

Thus,

9.05\times 10^{-19}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda_0}

\frac{9.05}{10^{19}}=\frac{19.878}{10^{26}\lambda_0}

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8 0
1 year ago
A human hair is 75 um across. How many inches is this?
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2 years ago
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A 100 mL reaction vessel initially contains 2.60×10^-2 moles of NO and 1.30×10^-2 moles of H2. At equilibrium the concentration
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Answer:

<h2>The equilibrium constant Kc for this reaction is 19.4760</h2>

Explanation:

The volume of vessel used= 100 ml

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Initial moles of H2= \frac{1.30}{10^2} moles

Concentration of NO at equilibrium= 0.161M

Concentration(in M)=\frac{moles}{volume(in litre)}

Moles of NO at equilibrium= 0.161(\frac{100}{1000})

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<u>Initial</u>          :1.3*10^-2          2.6*10^-2                0                   0        moles

<u>Equilibrium</u>:1.3*10^-2 - x     2.6*10^-2-x              x                   x/2     moles

∴\frac{2.60}{10^2}-x=\frac{1.61}{10^2}

⇒x=\frac{0.99}{10^2}

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⇒Kc=19.4760

3 0
1 year ago
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