Explanation:
The first one is series circuit as it is joined one after another while the second one is parallel circuit joined in a parallel way. Ok the line just like this --------- is the wire , the sign like this | | is the battery and the circles with cross are the bulbs
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Answer:
561 g P₂O₃
Explanation:
To find the mass of P₂O₃, you need to (1) convert moles H₃PO₃ to moles P₂O₃ (via mole-to-mole ratio from equation coefficients) and then (2) convert moles P₂O₃ to grams P₂O₃ (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the amount of sig figs in the given value.
Atomic Mass (P): 30.974 g/mol
Atomic Mass (O): 15.998 g/mol
Molar Mass (P₂O₃): 2(30.974 g/mol) + 3(15.998 g/mol)
Molar Mass (P₂O₃): 109.942 g/mol
1 P₂O₃ + 3 H₂O -----> 2 H₃PO₃
10.2 moles H₃PO₃ 1 mole P₂O₃ 109.942 g
---------------------------- x -------------------------- x ------------------- = 561 g P₂O₃
2 moles H₃PO₃ 1 mole
Answer:
12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.
Explanation:
For first solution of sulfuric acid :
C₁ = 40% , V₁ = ?
For second solution of sulfuric acid :
C₂ = 10% , V₂ = ?
For the resultant solution of sulfuric acid:
C₃ = 28% , V₃ = 20L
Also,
<u>V₁ + V₂ = V₃ = 20L</u> ......................................(1)
Using
<u>C₁V₁ + C₂V₂ = C₃V₃</u>
<u>40×V₁ + 10×V₂ = 28×20</u>
So,
40V₁ + 10V₂ = 560........................................(2)
Solving 1 and 2 as:
V₂ = 20 - V₁
Applying in 2
40V₁ + 10(20 - V₁) = 560
40V₁ + 200 - 10V₁ = 560
30V₁ = 360
<u>V₁ = 12 L</u>
So,
<u>V₂ = 20 - V₁ = 8L</u>
<u><em>12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.</em></u>
Aluminum is more reactive than iron
but it forms Al-oxide, that form a thin layer on the surface of Al, and protect Al from reaction with water.
Answer: Very unreactive aluminum oxide forms a thin layer on aluminum.
