Answer:
37064 J
Explanation:
Data Given:
mass of Steam (m) = 16.4 g
heat released (Q) = ?
Solution:
This question is related to the latent heat of condensation.
Latent heat of condensation is the amount of heat released when water vapors condenses to liquid.
Formula used
Q = m x Lc . . . . . (1)
where
Lc = specific latent heat of condensation
Latent heat of vaporization of water is exactly equal to heat of condensation with - charge
So, Latent heat of vaporization of water have a constant value
Latent heat of vaporization of water = 2260 J/g
So
Latent heat of condensation of water will be = - 2260 J/g
Put values in eq. 1
Q = (16.4 g) x (- 2260 J/g)
Q = - 37064 J
So, 37064 J of heat will be released negative sign indicate release of energy
A protective layer composed of overlapping cells, like fish scales or roof tiles, but facing downwards. The outer cuticle holds your hair in your hair follicle by means of a Velcro-like bond. It also minimizes the movement of water (moisture) in and out of the underlying cortex.
Please please thank you lord lord thank
Step 1: Write Imbalance Equation
CH₃CHO + O₂ → CO₂ + H₂O
Step 2: Balance Carbon Atoms:
There are 2 carbon atoms at reactant side and one at product side. So multiply CO₂ with 2 to balance them. i.e.
CH₃CHO + O₂ → 2 CO₂ + H₂O
Step 3: Balance Hydrogen Atoms:
There are 4 hydrogen atoms at reactant side and 2 Hydrogen atoms at product side. So, multiply H₂O by 2 to balance Hydrogen on both sides. i.e.
CH₃CHO + O₂ → 2 CO₂ + 2 H₂O
Step 4: Balance Oxygen Atoms:
There are 3 Oxygen atoms at reactant side and 6 Oxygen atoms at product side. In order to balance them multiply O₂ on reactant side by 2.5 (5/2). i.e
CH₃CHO + 5/2 O₂ → 2 CO₂ + 2 H₂O
Step 6: Eliminate Fraction:
Multiply overall equation by 2 to eliminate fraction. i.e.
2 CH₃CHO + 5 O₂ → 4 CO₂ + 4 H₂O
2 Na₂CO₃ + 2 CuSO₄ + H₂O → CuCO₃.Cu(OH)₂ + 2 Na₂SO₄ + CO₂
Malachite molar mass = 221.1 g / mol
So 2 moles CuSO₄ produce 1 mole of malachite
so 1.5 mole CuSO₄ produce (0.75) mole malachite
Mass of malachite = 0.75 mole * 221.1 g/ mol = 165.83 g
