Answer:
92.04%
Explanation:
Given:
Mass of CO₂ obtained = 53.0 grams
Mass of calcium carbonate heated = 1.31 grams
Now,
the molar mass of the calcium carbonate = 100.08 grams
The number of moles heated in the problem = Mass / Molar mass
= (1.31 grams) / (100.08 grams/moles)
= 0.013088 moles
now,
1 mol of calcium carbonate yields 1 mol of CO₂
thus,
0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂
now,
Theoretical mass of 0.013088 moles of CO₂ will be
= Number of moles × Molar mass of CO₂
= 0.013088 × 44 = 0.5758 grams
Thus, the percent yield for this reaction = 
or
the percent yield for this reaction = 
or
the percent yield for this reaction = 92.04%
Molar mass of LiBr (mm )= 86.845 g/mol
Molarity ( M ) = 4 M
Mass of solute ( m ) = 100 g
Volume ( V ) = in liters ?
V = m / mm * M
V = 100 / 86.845 * 4
V = 100 / 347.38
V = 0.2875 L
hope this helps!.
<span>plagioclase feldspars have striations and potassium feldspars don't have striations</span>
1 mole Hg ---------------- 6.02x10²³ atoms
?? ------------------------- 1.30 x10⁷ atoms
1.30x10⁷ x 1 / 6.02x10²³ =
= 1.30x10⁷ / 6.02x10²³ => 2.159x10⁻¹⁷ moles
hope this helps!
Answer:
c =0.2 J/g.°C
Explanation:
Given data:
Specific heat of material = ?
Mass of sample = 12 g
Heat absorbed = 48 J
Initial temperature = 20°C
Final temperature = 40°C
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 40°C -20°C
ΔT = 20°C
48 J = 12 g×c×20°C
48 J =240 g.°C×c
c = 48 J/240 g.°C
c =0.2 J/g.°C
