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user100 [1]
2 years ago
5

If 35.50 cm3 of a NaOH solution are required for the complete neutralization of a 25.00cm3 sample of 0.200mol dm-3 H2SO4, what i

s the concentration of the NaOH solution?
Chemistry
1 answer:
Morgarella [4.7K]2 years ago
6 0
In this question, you are given the NaOH volume but asked for concentration. 
Don't forget that for every 1 mol of NaOH there will be 1 mol OH- ion, but for every 1 mol of H2SO4 there will be 2 mol of H- ion.
To neutralize you need the same amount of OH- and H+, so the equation should be:

OH-= H+
<span>35.50cm3 * x*1= 25cm3* 0.2mol/dm3 *2
</span>x= 10/35.5 mol/dm3= 0.2816/dm3
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How many mm are equal to 4.75 x 10-2 m?
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47.5 mm

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In the compound k[co(c2o4)2(h2o)2] (where c2o42– = oxalate) the oxidation number and coordination number of cobalt are, respecti
ehidna [41]
Let's Assign Symbols to molecules like,

                    C₂O₄  =  X
and 
                    H₂O   =  Y
Then,
                                K [ Co (X)₂ (Y)₂ ]

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To neutralize, the coordination sphere must have -1 oxidation number.
So,
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As,
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Then 
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Also,
O.N of H₂O is zero as it is neutral, So,

                                    [Co - 4 + 0 ]  =  -1
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6 0
2 years ago
Read 2 more answers
If 8.00 g NH4NO3 is dissolved in 1000 g of water, the water decreases in temperature from 21.00 degrees Celsius to 20.39 degrees
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Answer:

25.7 kJ/mol

Explanation:

There are two heats involved.

heat of solution of NH₄NO₃ + heat from water = 0

q₁  +  q₂  =  0

n  =  moles of NH₄NO₃  =  8.00 g NH₄NO₃  ×  1 mol NH₄NO₃/80.0 g NH₄NO₃          

∴ n =   0.100 mol NH₄NO₃

q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln

m  =  mass of solution  =  1000.0 g + 8.00 g  =  1008.0 g

q₂  =  mcΔT  = 58.0 g  ×  4.184 J°C⁻¹  g⁻¹  × ((20.39-21)°C) = -2570.19 J

q₁  +  q₂  =  0.100 mol  ×ΔHsoln  – 2570.19 J  =  0

ΔHsoln  =  +2570.19 J  /0.100 mol  =  +25702 J/mol  =  +25.7 kJ/mol

7 0
2 years ago
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Answer:

8

Explanation:

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rate2/rate1= 8/1= 8

The rate of reaction will be 8 times faster.

7 0
2 years ago
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