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Helga [31]
2 years ago
14

How many kilograms of water must be added to 6.07 grams of oxalic acid (H2C2O4) to make a 0.025 m solution?

Chemistry
1 answer:
uranmaximum [27]2 years ago
4 0
This dilution problem uses the equation
M
a
V
a
=
M
b
V
b

M
a
= 6.77M - the initial molarity (concentration)
V
a
= 15.00 mL - the initial volume
M
b
= 1.50 M - the desired molarity (concentration)
V
b
= (15.00 + x mL) - the volume of the desired solution
(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x
59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.
I hope this was helpful.
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The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How
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First convert grams to moles using molar mass of butane that is 58.1 g

3.50g C4H10 x (1 mol C4H10)/(58.1g C4H10) = 0.06024 mol C4H10 <span>

<span>Now convert moles to molecules by using Avogadro’s number

0.06024 mol C4H10 x (6.022x10^23 molecules C4H10)/(1 mol C4H10) = 3.627x10^22 molecules C4H10 

And there are 4 carbon atoms in 1 molecule of butane, so use the following ratio: 

3.627 x 10^22 molecules C4H10 x (4 atoms C)/(1 molecule C4H10) 
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Answer:

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Explanation:

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Consider the following reaction:
vfiekz [6]

Answer:

Kc = 4.76

Explanation:

To find the concentrations of CO and H₂ at equilibrium, you have to set up an ICE (Initial, Change, Equilibrium) table.

    CO (g) + 2H₂(g) ⇌ CH₃OH (g)

I   0.32 M   0.53 M          0

C    -x            -2x            +x

E  0.32-x    0.53-2x        0.16 M

Since you know the concentration of CH₃OH at equilibrium, it would be equal to x since 0 + x = 0.16. So,

[CH₃OH] = 0.16 M

[CO] = 0.32 - 0. 16 = 0.16 M

[H₂] = 0.53 - 2(0.16) = 0.21 M

Now that you have all the concentrations at equilibrium, you can calculate the equilibrium constant.

Kc  = products ÷ reactants

     = [CH₃OH] ÷ [CO][H₂]²

     =  0.16 ÷ (0.16)(0.21)

Kc =   4.76

The equilibrium constant at this temperature is 4.76.

Hope this helps.

     

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