Answer:
#2 you multiply the numbers
Explanation:
When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 =
=74.8 %.
Answer:
It would take 3.11 J to warm 3.11 grams of gold
Explanation:
Step 1: Data given
Mass of gold = 3.11 grams
Temperature rise = 7.7 °C
Specific heat capacity of gold = 0.130 J/g°C
Step 2: Calculate the amount of energy
Q = m*c*ΔT
⇒ Q = the energy required (in Joules) = TO BE DETERMINED
⇒ m = the mass of gold = 3.11 grams
⇒ c = the specific heat of gold = 0.130 J/g°C
⇒ ΔT = The temperature rise = 7.7 °C
Q = 3.11 g * 0.130 J/g°C * 7.7 °C
Q = 3.11 J
It would take 3.11 J to warm 3.11 grams of gold
Answer:
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