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Savatey [412]
3 years ago
13

How many moles of carbon are in 7.00 x 1023 atoms of carbon?

Chemistry
1 answer:
Andru [333]3 years ago
8 0

Answer:

<h2>1.16 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{7 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{7}{6.02}  \\  = 1.162790....

We have the final answer as

<h3>1.16 moles</h3>

Hope this helps you

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The specific heat capacity of a certain type of cooking oil is 1.75 J/(g⋅∘C). What is the amount of heat exchanged when the temp
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Answer:

Q = -811440 J

Explanation:

Given data:

Mass of oil = 2.76 Kg (2.76× 1000 = 2760 g)

Initial temperature = 191 °C

Final temperature = 23°C

Specific heat capacity of oil = 1.75 J/g.°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 23°C - 191 °C

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3 years ago
Calculate the number of moles of Cl2 produced at equilibrium when 3.98 mol of PCl5 is heated at 283.9 deg celsius in a vessel ha
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Answer:

1.274 moles

Explanation:

The equation for the reaction can be represented as follows:

                       PCl_5(g)           ⇄       PCl_3(g)     +         Cl_2(g)

K = 0.060

K = \frac{[PCl_3][Cl_2]}{[PCl_5]}

Concentration of   PCl_5(g)  = \frac{numbers of moles}{volume}

Concentration of   PCl_5(g)  = \frac{3.98}{10.0}

Concentration of   PCl_5(g)  = 0.398 moles

If we construct an ICE table for the above equation; we have:

                       PCl_5(g)           ⇄       PCl_3(g)     +         Cl_2(g)

Initial                0.398                          0                          0

Change            - x                               + x                        + x

Equilibrium    (0.398 - x)                      x                          x

K = \frac{[PCl_3][Cl_2]}{[PCl_5]}

K = \frac{[x][x]}{[0.398-x]}

K = \frac{x^2}{0.398-x}

0.060 = \frac{x^2}{0.398-x}

0.06(0.398-x) = x²

0.02388 - 0.060x = x²

x² + 0.060x - 0.02388 = 0               (quadratic equation)

a = 1;       b= 0.06;      c= -0.02388

Using quadratic formula;

=  \frac{-b+/-\sqrt{b^2-4ac} }{2a}

= \frac{-0.06+/-\sqrt{(0.06)^2-4(1)(-0.02388)} }{(2*1)}

= \frac{-0.060+/-\sqrt{0.0036+0.09552} }{2}

= \frac{-0.06+/-\sqrt{0.09912} }{2}

= \frac{-0.06+/-0.3148}{2}

= \frac{-0.060+0.3148}{2}   or \frac{-0.060-0.3148}{2}

= \frac{0.2548}{2}  or \frac{-0.3748}{2}

= 0.1274 or -0.1874

We go by the positive value which says:

[x] = 0.1274 M

number of moles = 0.1274 × 10.0

= 1.274 moles

∴ the number of moles  of Cl₂ produced at equilibrium = 1.274 moles

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