<span>The pH scale goes from 0-14. 0-6.9 is acidic, 7 is neutral and 7.1-14 is basic</span>
NaOH+HCl-> NaCl+H2O
1 mole of NaOH
1 mole of HCl.
To calculate volume of NaOH
CaVa/CbVb= Na/Nb
Where Ca=2M
Cb=1M
Va=200cm³
Vb=xcm³
Substitute into the equation.
2×200/1×Vb=1/1
400/Vb=1/1
Cross multiply
Vb×1=400×1
Vb=400cm³
To calculate the mass of sodium chloride, NaCl from the neutralization rxn.
Mole of NaCl=1
Molar mass of NaCl= 23+35.5=58.5
Mass=xgrammes.
Mass of NaCl=Number of moles × Molar mass.
Substitute
Mass of NaCl= 1×58.5
=58.5g
This is what I could come up with.
The swimming pools pH is below 7, meaning it is slightly acidic. If you want to make the pH higher, you must add a base which by definition has a pH higher than 7.
D. Add base
Answer:
The reaction is endothermic.
Yes, absorbed
3.06x10¹kJ are absorbed
Explanation:
In the reaction:
2HgO(s) → 2Hg(l) + O₂(g) ΔH = 182kJ
As ΔH >0,
<em>The reaction is endothermic</em>
<em />
As the reaction is endothermic, when the reaction occurs,
<em>the heat is absorbed.</em>
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Now, based on the equation, when 2 moles of HgO (Molar mass: 216.59g/mol), 182kJ are absorbed.
72.8g are:
72.8g * (1mol / 216.59g) = 0.3361 moles HgO.
that absorb:
0.3361 moles HgO * (182kJ / 2 moles) =
<h3>3.06x10¹kJ are absorbed</h3>
2 ICl + H2 ----> I2 + 2 HCl
as given that rate is first order with respect to ICl and second order with respect to H2
The rate law will be
Rate = K [ICl] [ H2]^2
b) Given that K = 2.01 M^-2 s^-1
Concentrations are
[ICl] = 0.273 m and [H2] = 0.217 m
Therefore rate = 2.01 X (0.273)(0.217)^2 = 0.0258 M / s