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il63 [147K]
3 years ago
13

Plzzz help me! I will do anything ;) How many molecules are there in 45g of caffeine

Chemistry
1 answer:
babymother [125]3 years ago
6 0

Answer:

45 IS THE ANSWER

Explanation:

MARK AS BRAINLIEST PLSSS

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4 0
3 years ago
Read 2 more answers
A sample of pure NO2NO2 is heated to 335 ∘C∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2N
Alborosie

Answer:

k_c = 1. 1 × 10⁻²

Explanation:

Given that:

Temperature = 335 ° C = (335+ 273)K = 608

Pressure = 0.750 atm

Volume = 1 Litre

number of moles of NO2 = ???

Rate Constant =0.0821 L atm /K/mol

Using the Ideal gas equation

PV = nRT

n = \frac{PV}{RT}

n = \frac{0.75*1}{0.0821*608}

n = 0.015

n = 1.5 × 10⁻² mole

Density = 0.525 g/L

The equation for the reaction can be illustrated as:

                     2NO2(g)         ⇌          2NO(g)         +         O2(g)

For the ICE table; we have:

 

Initial                 x                                   0                              0

Change            -2y                               + 2y                          +y

Equilibrium        (x - 2y)                        2y                             y

Total moles at equilibrium = (x-2y)+2y+y

= x + y moles

However,

1.5 × 10⁻² mole of the mixture has a mass of 0.525 g

i.e x + y moles = 1.5 × 10⁻² mole

Now, molar mass of 1 mole of NO2 = 46g/mol

Since number of moles = \frac{mass}{molar mass}

mass of (x-2y) moles = 46 × (x-2y) g

Molar mass of NO = 30 g/mol

Also, mass of NO = 2y × 30 = 60y

Molar mass of O2 = 32 g/mol

Mass of O2 = y × 32 = 32y

Total mass = ( 46x - 90y)+60y+32y = 0.525

46x = 0.525

x = \frac{0.525}{46}

x = 0.0114

x = 1.14 × 10⁻²

x + y moles = 1.5 × 10⁻²

y =  1.5 × 10⁻² -  1.14 × 10⁻²

y = 0.0036

y = 3.6 × 10⁻³

At equilibrium

[NO2] = ( 1.14 - 2(0.36))× 10⁻² = 4.2 × 10⁻³ M

[NO] = 2 ( 3.6 × 10⁻³)  = 7.2 × 10⁻³ M

[O2] = 3.6 × 10⁻³ M

k_c = \frac{[NO]^2[O_2]}{[NO_2]^2}

k_c = \frac{(7.2*10^{-3})^2(3.6*10^{-3})}{(4.2*10^-3)^2}

k_c = 0.011

k_c = 1. 1 × 10⁻²

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3 years ago
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A.) Liquid

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C.) Gas

5 0
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