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angellove1707
07.03.2021 •
Chemistry
To completely neutralise 200cm3 of 0.5mol/dm3 sodium hydroxide (NaOH), a student adds 100cm3 of 0.5mol/dm3 sulfuric acid (H2SO4). The temperature of the solution goes up by 4.5°C.
The equation for the reaction is
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)
a. Calculate the amount, in moles, of NaOH in the sodium hydroxide solution.
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Ответ:
0.1 mol
Explanation:
Step 1: Given data
Volume of the solution of sodium hydroxide (V): 200 cm³Molar concentration of sodium hydroxide (C): 0.5 mol/dm³Step 2: Convert "V" to dm³
We will use the conversion factor 1 dm³ = 1000 cm³.
200 cm³ × 1 dm³/1000 cm³ = 0.200 dm³
Step 3: Calculate the moles (n) of NaOH
The molarity of the NaOH solution is equal to the moles of NaOH divided by the volume of solution.
C = n/V
n = C × V
n = 0.5 mol/dm³ × 0.200 dm³ = 0.1 mol
Ответ:
c an Arrhenius acid increases H+ in the solution