Answer:
We need 1.1 grams of Mg
Explanation:
Step 1: Data given
Volume of water = 78 mL
Initial temperature = 29 °C
Final temperature = 78 °C
The standard heats of formation
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Step 2: The equation
The heat is produced by the following reaction:
Mg(s)+2H2O(l)→Mg(OH)2(s)+H2(g)
Step 3: Calculate the mass of Mg needed
Using the standard heats of formation:
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Mg(s) + 2 H2O(l) → Mg(OH)2(s) + H2(g)
−924.54 kJ − (2 * −285.8 kJ) = −352.94 kJ/mol Mg
(4.184 J/g·°C) * (78 g) * (78 - 29)°C = 15991.248 J required
(15991.248 J) / (352940 J/mol Mg) * (24.3 g Mg/mol) = 1.1 g Mg
We need 1.1 grams of Mg
A. The negative ionic radius is larger than the neutral atomic radius.
<span><span>The column of a fractional distillation apparatus should be
aligned vertical to get better separation because
the column is essentially longer. It will prevent channeling in the
column. </span>Therefore column efficiency will be higher if it is as
vertical as possible instead of tilted.<span> Distillation is a process used to separate a mixture of two (or
more) components. In a typical </span>fractional
distillation, a liquid mixture
is heated in the distilling flask and the
process begins.</span>
Answer:
The answer to your question is 25.9 g of KCl
Explanation:
Data
Grams of KCl = ?
Volume = 0.75 l
Molarity = 1 M
Formula

Solve for number of moles

Substitution
Number of moles = 1 x 0.75
Simplification
Number of moles = 0.75 moles
Molecular mass KCl = 39 + 35.5 = 34.5
Use proportions to find the grams of KCl
34.5 g of KCl ---------------- 1 mol
x ---------------- 0.75 moles
x = (0.75 x 34.5) / 1
x = 25.9 g of KCl
Answer:
2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.
Explanation:
Suppose x is the number of litres added to the 10% mixture than the quantity of new mixture is given as below
litres
litres
Also the quantity of alcohol is given as
Now the equation is as

So 2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.