Answer:
0.7g of HCl
Explanation:
First, let us write a balanced equation for the reaction between HCl and Al(OH)3.
This is illustrated below:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:
Molar Mass of Al(OH)3 = 27 + 3(16+1)
= 27 + 3(17) = 27 + 51 = 78g/mol.
Molar Mass of HCl = 1 + 35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g
Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
From the equation above,
78g of Al(OH)3 reacted with 109.5g of HCl.
Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl
Answer:
[NaCl[ = 0.535M
Explanation:
We determine the moles of solute:
125 g . 1 mol/ 58.45 g = 2.14 moles
Molarity (mol/L) → 2.14 mol / 4L → 0.535M
Molarity is a sort of concentration that indicates the moles of solute in 1L of solution
Molybdenum in periodic table
or
Molarity definition
Answer:
4.78 %.
Explanation:
<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>
<em></em>
<em>mass % = (mass of solute/mass of solution) x 100.</em>
<em></em>
mass of MgSO₄ = 50.0 g,
mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.
mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.
<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>
Answer:
Enthalpy is the amount of heat released or absorb. The total enthalpy can be found by subtracting the products' enthalpy by the reactants' enthalpy