Answer is: 6,16 kJ.
1) changing temperature of ice from -25°C to 0°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 2 J/g·°C · 25°C
Q₁ = 900 J.
m(H₂O) = 1mol · 18 g/mol = 18 g.
C - <span>specific heat of ice.
</span>2) changing temperature of water from 0°C to 70°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 4,18 J/g·°C · 70°C
Q₁ = 5266,8 J.
C - specific heat of water.
Q = Q₁ + Q₂ = 900 J + 5266,8 J
Q = 6166,8 J = 6,16 kJ.
Answer:
290 g of hydrogen peroxide would fill the 200.0 mL container
Explanation:
Given data:
Volume = 200.0 mL
Mass of hydrogen peroxide = ?
Solution:
Density of hydrogen peroxide is 1.45 g/cm³.
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Now we can determine the mass of hydrogen peroxide.
D = m/v
1.45 g/cm³ = m/ 200.0 mL
1 mL = 1 cm³
m = 1.45 g/cm³× 200.0 cm³
m = 290 g
So 290 g of hydrogen peroxide would fill the 200.0 mL container.
Answer:
8.59 g
2.25 g
Explanation:
According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-
Moles of Pb(OH)CL is
= 0.0385 mol
Mass of PbO needed is
After solving the above equation we will get
= 8.59 g
Mass of NaCL needed is
After solving the above equation we will get
= 2.25 g
Therefore we have applied the above formula.
