Answer:
Hi There! I'm new so I hope I get This Right! ^^
Explanation:
double blind test. the control group receives a placebo. Why is a placebo used in a double-blind test? so that the effects on people in two different groups can be compared.
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Answer:
The limiting reacting is O2
Explanation:
Step 1: data given
Number of moles O2 = 21 moles
Number of moles C6H6O = 4.0 moles
Step 2: The balanced equation
C6H6O + 7O2 → 6CO2 + 3H2O
Step 3: Calculate the limiting reactant
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
O2 is the limiting reactant. It will completely be consumed (21 moles).
C6H6O is in excess.
For 7 moles O2 we need 1 mol C6H6O
For 21 moles O2 we'll need 21/7 = 3 moles C6H6O
There will remain 4.0 - 3.0 = 1 mol C6H6O
Step 4: calculate products
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
For 21 moles O2 we'll have 6/7 * 21 = 18 moles CO2
For 21 moles O2 we'll have 3/7 * 21 = 9 moles H2O
The limiting reacting is O2
Answer:
Tetrahedral electron geometry and trigonal pyramidal molecular geometry.
Explanation:
The Lewis structure is shown in Figure 1.
The central N atom has three bonding pairs and one lone pair, for <em>four electron groups</em>.
VSEPR theory predicts a tetrahedral electron geometry with bond angles of 109.5°.
We do not count the lone pair in determining the molecular shape.
The molecular geometry is trigonal pyramidal (see Figure 2).
You could argue that any solution with water in it has an equilibrium in it of some sort. If a solution is over saturated there is an equilibrium between the dissociated and solid solute (for example NaCl(s)⇄Na⁺(aq)+Cl⁻(aq) when in water). Even if the solution is not over saturated, water always has the reaction 2H₂O(l)⇄H₃O⁺(aq)+OH⁻(aq) since water can act as both an acid and a base (this reaction is also always at equilibrium and the equilibrium constant is 1×10⁻¹⁴).
Since we usually ignore the autoionization of water (unless dealing with acid base chemistry), I think the answer your teacher is looking for is over saturated solutions.
I hope this helps. Let me know if anything is unclear or if you need a different answer.
