Answer:
Mg(s)+Zn2+>>>>>Mg2++Zn(s) B
Explanation:
Answer:
To supply the required ions it is necessary to inject 5,6mL of 6g/30mL solution and 131,1 mL of 0,9% solution.
Explanation:
1mEq of sodium are 59mg of NaCl and 1mEq of potassium are 75mg KCl
in intravenous infusion 15 mEq of K are:
15x75mg KCl = 1,125g of KCl
And 20 mEq of Na are:
20x59mg NaCl = 1,18g of NaCl
To supply the potassium ion it is necessary to inject:
1,125g of KCl×
=<em> 5,6mL of 6g/30mL solution</em>
And, to supply the sodium ion it is necessary to inject:
1,18g of NaCl×
= <em>131,1 mL of 0,9% solution</em>
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I hope it helps!
Which of these what?
This isn't clear, can you put the answer choice or something?
Answer:
They reduce the bond angle to be slightly lower than the tetrahedral bond angle, approximately 104.45 degrees.
Explanation:
The unshared pair of electrons or lone pair electrons in order to have the minimum repulsion possible with each other pushes the other bonding pairs closer together making the bond angle smaller or bent.
The bond angle is slightly lower than the tetrahedral bond angle of 108 degrees, leaving the water molecule with a bent molecular geometry.
Answer:
Theoretical yield of the reaction = 34 g
Excess reactant is hydrogen
Limiting reactant is nitrogen
Explanation:
Given there is 100 g of nitrogen and 100 g of hydrogen
Number of moles of nitrogen = 100 ÷ 28 = 3·57
Number of moles of hydrogen = 100 ÷ 2 = 50
Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation
N2 + 3H2 → 2NH3
From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed
Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen
⇒ We require 10·71 moles of hydrogen
But we have 50 moles of hydrogen
∴ Limiting reactant is nitrogen and excess reactant is hydrogen
From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia
Molecular weight of ammonia = 17 g
∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g
