Strontium atom loses 2 electrons to become an ion with 2 electrons lesser than its atom. Your answer is C.
The amount of current required to produce 75. 8 g of iron metal from a solution of aqueous iron (iii)chloride in 6. 75 hours is 168.4A.
The amount of Current required to deposit a metal can be find out by using The Law of Equivalence. It states that the number of gram equivalents of each reactant and product is equal in a given reaction.
It can be found using the formula,
m = Z I t
where, m = mass of metal deposited = 75.8g
Z = Equivalent mass / 96500 = 18.6 / 96500 = 0.0001
I is the current passed
t is the time taken = 75hour = 75 × 60 = 4500s
On subsituting in above formula,
75.8 = E I t / F
⇒ 75.8 = 0.0001 × I × 4500
⇒ I = 168.4 Ampere (A)
Hence, amount of current required to deposit a metal is 168.4A.
Learn more about Law of Equivalence here, brainly.com/question/13104984
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The water cycle regardless if it is in a lake, our bodies, food, or underground.
Answer:
94.325 g
Explanation:
We'll begin by converting 350 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
350 mL = 350 mL × 1 L /1000 mL
350 mL = 0.35 L
Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:
Volume = 0.35 L
Molarity of KC₂H₃O₂ = 2.75 M
Mole of KC₂H₃O₂ =?
Molarity = mole /Volume
2.75 = Mole of KC₂H₃O₂ / 0.35
Cross multiply
Mole of KC₂H₃O₂ = 2.75 × 0.35
Mole of KC₂H₃O₂ = 0.9625 mole
Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:
Mole of KC₂H₃O₂ = 0.9625 mole
Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)
= 39 + 24 + 3 + 32
= 98 g/mol
Mass of KC₂H₃O₂ =?
Mass = mole × molar mass
Mass of KC₂H₃O₂ = 0.9625 × 98
Mass of KC₂H₃O₂ = 94.325 g
Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g
