Answer:
They are both colorless, odorless, and tasteless. They have the same number of valence electrons too. And unbalanced electrons in their valence shell.
Explanation:
Answer:
The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg
Explanation:
Heat gain by ice = Heat lost by water
Thus,
Heat of fusion + 
Where, negative sign signifies heat loss
Or,
Heat of fusion + 
Heat of fusion = 334 J/g
Heat of fusion of ice with mass x = 334x J/g
For ice:
Mass = x g
Initial temperature = 0 °C
Final temperature = 6 °C
Specific heat of ice = 1.996 J/g°C
For water:
Volume = 353 mL
Density of water = 1.0 g/mL
So, mass of water = 353 g
Initial temperature = 26 °C
Final temperature = 6 °C
Specific heat of water = 4.186 J/g°C
So,
345.976x = 29553.16
x = 85.4197 kg
Thus,
<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>
Protons= 12, Electrons= 12, neutrons= 12
In such type of problems where reactant and product ratios are calculated one (Launa) should take start from balanced chemical equation. A balanced chemical equations helps in solving such problems. For this purpose chemical equation of this combustion reaction is drawn.
Equation:
C₃H₁₀ ⁺ 5 O₂ ⁻⁻⁻⁻⁻⁻⁻⁻⁻> 3 CO₂ ⁺ 4 H₂O
Now, According to this balanced equation one molecule of propane reacts with 5 molecules of oxygen to produce 3 molecules of carbon dioxide and 4 molecules of water.
As She is given with 2 molecules of propane and 10 molecules of oxygen, which means that the equation is multiplied by 2. So, the product side must also be multiplied by 2 to balance the equatio (Law of Conservation of Mass).
Result:
( C₃H₁₀ ⁺ 5 O₂ ⁻⁻⁻⁻⁻⁻⁻⁻⁻> 3 CO₂ ⁺ 4 H₂O ) ₓ 2
2 C₃H₁₀ ⁺ 10 O₂ ⁻⁻⁻⁻⁻⁻⁻⁻⁻> 6 CO₂ ⁺ 8 H₂O
So, 6 molecules of carbon dioxide and 8 molecules of water are produced.
