The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol / 0.250 mol/L = 40.0 mL
40.0 mL of 0.250 M HCl is required
<span>1)false a in chemical equilibrium concentration of reactant is equal to concentration of product
2)as here they said heat is added in product side means its endothermic reaction and in endothermic reaction on increasing temp. equilibrium shift towards forward direction so its true
3) B)as here mole are equal in reactant and product side that is 2 and if we increase pressure equilibrium shift in dat direction where no. of moles are less and here mole are equal so it will remain unaffected</span>
I think it's balanced ?
There are two H's in the both side
And one P in the both side
the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol) is 1.36g
The reaction is 3 MnO2 + 4 Al ------ 2Al2o3+ Mn
3 mole of manganese oxide give 2 moles of aluminum oxide so by the reaction n( MnO2)/3 =n(al203)2
the formula is n= mass/M so, now substituting values
m (Al2O3)= m(MnO2) X 2 X M (Al2O3) / M(MnO2 X3
so, by substituting values, 2 X101.96 X1.74g / 3 X 86.94 =1.36g
so mass of aluminum oxide obtained = 1.36g
To learn more about Mass:
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Answer:
12,375 cm³
Explanation:
v=l×w×h
Since there are 3 books the equation would then be
v=3(l×w×h)
v=3(25×15×11)
v=3(4,125)
v=12,375 cm³
