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2 years ago

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Complete combustion of 1.11 0 g of a gaseous hydrocarbon yields 3.613 g of carbon dioxide and one 1.109 g of water. 80.288 g sam

ple of the hydrocarbon occupies a volume of 131 milliliters at 24.8°C and 753 mm Hg. What is the molecular formula of the hydrocarbon?

1 answer:

Mamont248 [21]2 years ago

4 0

Answer:

C4H6

Explanation:

we know that

n = P×V / RT

= (753 ) x (0.131 ) / ((62.36367) x (24.8 + 273.15) ) = 0.005309

molecular weight = (0.288) / (0.005309) = 54.25 g/mol

molecular weight of CO_2 in terms of C

(3.613 ) / (44.00964 ) x (12.01078) = 0.986033g C

Molecular weight of H_2O in terms of H

(1.109 ) / (18.01532 ) x (2 / 1) x (1.007947) = 0.124096g H

Number of atoms of C

(0.986033) / (1.110) x (54.25) / (12.01078) = 4.012

(0.124096) / (1.110) x (54.25 ) / (1.007947) = 6.017

Round to the nearest whole numbers to find the molecular formula: C4H6

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algol13

the answer should be C.

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An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char

ArbitrLikvidat [17]

Answer:

Part a)

$W = 1.58 \times 10^{-20} J$

Part b)

$v = 1.86 \times 10^5 m/s$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = 4.00 \times 10^{-12} C/m^2$

now the electric field is given as

$E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}$

$E = 0.225 N/C$

Now acceleration of an electron due to this electric field is given as

$a = \frac{eE}{m}$

$a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}$

$a = 3.97 \times 10^{10}$

Now work done on the electron due to this electric field

$W = F.d$

$d = 0.470 - 0.03$

$d = 0.44 m$

So work done is given as

$W = (ma)(0.44)$

$W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)$

$W = 1.58 \times 10^{-20} J$

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

$W = \frac{1}{2}mv^2 - 0$

$1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$v = 1.86 \times 10^5 m/s$

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3 years ago

Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou

kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

i = current flow = 3 A
t = time interval for which the current flow = $4\ h = 4\times 3600\ s = 14400\ s$
V = terminal voltage of the battery
R = rate of energy = 9 cents/kWh

<u>Assume:</u>

Q = charge transported as a result of charging
E = energy expended
C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

$\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\$

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

$E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\$

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

$\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents$

Hence, 0.108V cents is the charging cost of the battery.

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Resistivity of nichrome is high.

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sergiy2304 [10]

The electron charge is equal to $-e=-1.6\cdot 10^{-19}C$ . The atomic nucleus of the problem has a charge of $+40 e=40\cdot (1.6\cdot 10^{-19}C)=6.4\cdot 10^{-18}C$ . The distance between the nucleus and the electron is $r=10^{-9}m$ , so we can calculate the electrostatic (Coulomb) force between the two:
$F=k_e \frac{(-e)(+40e) }{r^2} =8.99\cdot 10^9 Nm^2C^{-2} \frac{(-1.6\cdot 10^{-19}C)(6.4\cdot 10^{-18}C)}{(10^{-9}m)^2} =$
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which is attractive, since the two charges have opposite sign.

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