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morpeh [17]
2 years ago
14

What are the complete Ionic and Net Ionic equations for the following

Chemistry
1 answer:
zhenek [66]2 years ago
5 0

Answer:

Check photo

Explanation:

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Ascorbic acid (vitamin
svlad2 [7]

Let us see the structure of ascorbic acid


As shown there is no COOH group however the OH group can lose a proton and forms conjugate base

The conjugate base formed is stabilized due to resonance

More the stability of conjugate base more the strength of acid

Hence ascorbic acid behaves as an acid

8 0
3 years ago
C2F4 effuses through a barrier at a rate of 4.6x10-6 mol/hour, while an unknown gas effuses at a rate of 5.8x10-6 mol/hour. What
umka21 [38]
The  molar mass  of  the unknown  compound  is   calculated   as   follows

let the unknown  gas be represented by   letter  Y

Rate of C2F4/  rate of  Y  = sqrt of   molar  mass of gas Y/ molar mass of  C2F4

 =  (4.6  x10^-6/ 5.8  x10^-6)  = sqrt  of  Y/ 100

remove  the  square  root  sign  by  squaring  in both  side

(4.6  x  10^-6 / 5.8  x10^-6)^2 =  Y/100

= 0.629 =Y/100

multiply  both side  by  100

Y=  62.9 is  the molar  mass of unknown  gas



5 0
2 years ago
A reaction produces 74.10 g Ca(OH)2 after 56.08 g CaO is added to 36.04 g H2O. How should the difference in the masses of reacta
melomori [17]
Cao +  H2O  ---->Ca(OH)2
Calculate   the  number  of  each   reactant  and  the  moles  of  the  product
that  is
moles = mass/molar mass
The  moles  of  CaO=  56.08g/  56.08g/mol(molar  mass  of  Cao)=  1mole
the  moles  of  water=  36.04 g/18  g/mol=  2.002moles
The   moles  of Ca (OH)2=74.10g/74.093g/mol= 1mole

 The  mass  of differences  of  reactant  and  product  can   be  therefore 
 explained  as 
 1  mole   of  Cao  reacted  completely   with   1  mole   H2O  to  produce  1 mole  of  Ca(OH)2. The  mass  of  water   was  in  excess  while  that  of  CaO  was  limited

3 0
3 years ago
At 400 K oxalic acid decomposes according to the reaction:H2C2O4(g)→CO2(g)+HCOOH(g)In three separate experiments, the intial pre
padilas [110]

Answer:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

Explanation:

For the reaction:

H₂C₂O₄(g) → CO₂(g) + HCOOH(g)

At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:

H₂C₂O₄(g) = P₀ - x

CO₂(g) = x

HCOOH(g) = x

P at t=20000 is:

P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x

For 1st point:

x = 92,8-65,8 = 27

Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8

2nd point:

x = 130-92,1 = 37,9

H₂C₂O₄(g): 92,1 - 37,9 = 54,2

3rd point:

x = 157-111 = 46

H₂C₂O₄(g): 111-46 = 65

Now, as the rate law is :

v = k P[H₂C₂O₄]

Based on integrated rate law, k is:

(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k

1st point:

k = 2,64x10⁻⁵

2nd point:

k = 2,65x10⁻⁵

3rd point:

k = 2,68x10⁻⁵

The averrage of this values is:

k = 2,66x10⁻⁵

That means law is:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

I hope it helps!

4 0
2 years ago
Calculate the wavelength of a photon of light having a frequency of 3.50 x 1015 Hz.
Aliun [14]

Answer:

Wavelength, \lambda=8.57\times 10^{-8}\ m

Explanation:

Given that,

Frequency, f=3.5\times 10^{15}\ Hz

We need to find the wavelength of a photon of light. The relation between frequency and wavelength is as follows :

c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{3.5\times 10^{15}}\\\\\lambda=8.57\times 10^{-8}\ m

So, the wavelength of the light is 8.57\times 10^{-8}\ m.

4 0
2 years ago
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