The number of C2H5OH in a 3 m solution that contain 4.00kg H2O is calculate as below
M = moles of the solute/Kg of water
that is 3M = moles of solute/ 4 Kg
multiply both side by 4
moles of the solute is therefore = 12 moles
by use of Avogadro law constant
1 mole =6.02 x10^23 molecules
what about 12 moles
=12 moles/1 moles x 6.02 x10^23 = 7.224 x10^24 molecules
Answer: A. Is decomposition
B. Is synthesis where Na combines with Cl to form NaCl
C. Is single displacement or replacement. Mg displaces Cu.
Explanation:
Answer: Given the equation for reaction is
2
A
l
+C
u
S
O
4
→
A
l
2
(
S
O
4
)
3
+
3
C
u
.
Explanation:
Answer:
It is basic and has a pH of 9.8.
Explanation:
pOH = 4.2
we will determine its pH.
pOH + pH = 14
pH = 14 - pOH
pH = 14 - 4.2
pH = 9.8
According to pH scale the the pH lower than 7 is consider acidic while pH of seven is neutral and pH greater than 7 is basic.
The given solution has pH 9.8 it means it is basic.
Answer:
3.2 L
Explanation:
Given data:
Mass of oxygen = 3.760 g
Pressure of gas = 88.4 Kpa (88.4×1000 = 88400 Nm⁻²)
Temperature = 19°C (19+273.15 = 292.15 K)
R = 8.314 Nm K⁻¹ mol⁻¹
Volume occupied = ?
Solution:
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 3.760 g/ 32 g/mol
Number of moles = 0.12 mol
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant
T = temperature in kelvin
V = nRT/P
V = 0.12 mol × 8.314 Nm K⁻¹ mol⁻¹ × 292.15 K /88400 Nm⁻²
V = 291.472 Nm /88400 Nm⁻²
V = 0.0032 m³
m³ to L:
V = 0.0032×1000 = 3.2 L
