Answer: Option A) 83.9g
Explanation:
KCl is the chemical formula of potassium chloride.
Given that,
Amount of moles of KCl (n) = ?
Volume of KCl solution (v) = 0.75L
Concentration of KCl solution (c) = 1.5M
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
make n the subject formula
n = c x v
n = 1.5M x 0.75L
n = 1.125 mole
Now given that,
Amount of moles of KCl (n) = 1.125
Mass of KCl in grams = ?
For molar mass of KCl, use the molar masses of:
Potassium, K = 39g;
Chlorine,Cl = 35.5g
KCl = (39g + 35.5g)
= 74.5g/mol
Since, amount of moles = mass in grams / molar mass
1.125 mole = m / 74.5g/mol
m = 1.125 mole x 74.5g/mol
m = 83.81g
Thus, 83.9 grams of KCl are needed to prepare 0.750 L of a 1.50 M solution in water
The balanced equation is Fe₂O₃ + 3 CO = 2 Fe + 3 CO₂.
Next step is to convert everything to moles.
12.6g Fe₂O₃ x (1 mol Fe₂O₃ / 159.7g Fe₂O₃) = 0.07890 mol Fe₂O₃
9.65g CO x (1 mol CO / 28.01g CO) = 0.3445 mol CO
The third step is to determine the limiting and excess reactants.
0.07890 mol Fe₂O₃ x (3 mol CO/1 mol Fe₂O₃) = 0.2367 mol CO
Therefore Fe₂O₃ is the limiting reagent while CO is in excess.
0.07890 mol Fe x (2 mol Fe(s) / 1 mol Fe₂O₃) = 0.1578 mol Fe(s)
0.1578 mol Fe x (55.84g Fe / mole Fe) = 8.812g Fe is the theoretical yield
%yield = (7.23g / 8.812g) x 100% = 82.0% is the percent yield
Answer:
Negative sign says that release of heat.
Explanation:
The expression for the calculation of the heat released or absorbed of a process is shown below as:-
Where,
is the heat released or absorbed
m is the mass
C is the specific heat capacity
is the temperature change
Thus, given that:-
Mass = 25.2 g
Specific heat = 0.444 J/g°C
So,
Negative sign says that release of heat.
Answer:
Explanation:
The atomic radius of elements are used to estimate the sizes of elements. The atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state of metals.
To solve this problem we will obtain the atomic radius values of the given elements from a standard atomic radius table;
Si 111 pm
P 98 pm
Cl 79 pm
S 87pm
pm = picometer
We see that chlorine has the least atomic radius