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2 years ago

How many atoms are in 1.14 mol so3?

2 answers:

6 0

NA = 6.022*10^23
number of atoms = number of moles * avogadros number
number of atoms = 6.022*10^23 * 1.14 moles * 4 atoms / molecules of SO3
Number of atoms = 2.75×10^24 atoms.

Lady bird [3.3K]2 years ago

3 0

1 mol ------------- 6.02x10²³ atoms
1.14 moles ------ ( atoms )

atoms = 1.14 x ( 6.02x10²³) / 1

atoms = 6.86 x 10²³ / 1

= 6.86 x 10²³ atoms

hope this helps!

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The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

Complete la siguiente comparación de similitud:

galben [10]

Answer:

metaloides

Explanation:

4 0

2 years ago

The is the variable that gets measured

ddd [48]

Answer:

what? what's the full question?

5 0

3 years ago

Nh3 is a weak base (kb = 1.8 × 10–5 m) and so the salt nh4cl acts as a weak acid. what is the ph of a solution that is 0.013 m i

ankoles [38]

First, we need to get the value of Ka:

when Ka = Kw / Kb

we have Kb = 1.8 x 10^-5

and Kw = 3.99 x 10^-16 so, by substitution:

Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11

by using the ICE table :

NH4+ + H2O →NH3 + H+
intial 0.013 0 0

change -X +X +X

Equ (0.013-X) X X

when Ka = [NH3][H+] / [NH4+]

by substitution:

2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X

∴X = 5.35 x 10^-7

∴[H+] = X = 5.35 x 10^-7

∴PH = - ㏒[H+]

= -㏒(5.35 x 10^-7)
= 6.27

6 0

2 years ago

Water vapor enters a compressor at 35 kpa and 160°c and leaves at 300 kpa with the same specific entropy as at the inlet. What a

Annette [7]

<h3>solution:</h3>

The initial entropy is obtained from the initial pressure and temperature with data from A-6 using interpolation:

s= 8.2652 kJ/kgK

The final temperature is determined from the entropy and the final pressure with data from A-6 using interpolation:

T₂ = T₁+ T₂ - T₁/ 8₂ - 8₁ ( s - s₁)

= (400 + 500 - 400/8.3271 - 8.0347 (8.2652 - 8))

= 478.83°C

The final enthalpy is determined in the same way:

h₂= h₁ + h₂ - h₁/s₂ - s₁ ( s - s₁)

= ( 3275.5+ 3486.6 - 3275.5/ 8.3271 - 8.0347) (8.265)

= 3441.91 kJ/kg

6 0

1 year ago