It is given that the person weighs 62 kg = 62,000 g
Natural abundances in mass percent are:
O = 65%
C = 18%
H = 10%
N = 3.0%
Ca = 1.6%
P = 1.2%
Corresponding weights of the elements are:
O = 65/100 * 62000 g = 40.30 * 10^3 g
C = 18/100 * 62000 g = 11.16 * 10^3 g
H = 10/100 * 62000 g = 62.00 * 10^2 g
N = 3.0/100 * 62000 g = 18.60 * 10^2 g
Ca = 1.6/100 * 62000 g = 9.92 * 10^2 g
P = 1.2/100 * 62000 g = 7.44 * 10^2 g
Answer:
0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl
Explanation:
Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.
Then, the more concentrated acid or base will be more acidic or basic.
CH3COOH. Weak acid
NaOH. Strong base
H2SO4. Strong acid
NH3. Weak base.
HCl. Strong acid
The less acid (More basic):
<h3>0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl</h3>
Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid
Remembering that
d = m ÷ v
d = ?
m = 104 g
v = 21 cm³
Therefore:
d = 104 ÷ 21
<span>d = 4,95 g÷cm³</span>
Density of liquid= 
.
so, density of liquid= 
= 1.2 gm/cm³.
The % yield if 500 g of sulfur trioxide reacted with excess water to produce 575 g of sulfuric acid is calculated using the below formula
% yield = actual yield/ theoretical yield x100
actual yield =575 grams
to calculate theoretical yield
find the moles of SO3 used =mass/molar mass
= 500g/ 80 g/mol =6.25 moles
SO3+H2O=H2SO4
by use of mole ratio of SO3 : H2SO4 which is 1:1 the moles of H2SO4 is also= 6.25 moles
the theoretical yield of H2SO4 is therefore = moles /molar mass
= 6.25 x98= 612.5 grams
%yield is therefore= 575 g/612 g x100= 93.9 %
