Answer:
2 grams of salt must be added by chef to make the recipe noticeably saltier.
Explanation:
Weber's Law :
Where =
I = Initial stimulus intensity
ΔI = Difference threshold
k = Weber constant
According to Weber’s constant for saltiness , k= 
Amount of salt required for the recipe , I= 10 g
2 grams of salt must be added by chef to make the recipe noticeably saltier.
Step 1
<em>The reaction involved:</em>
CO + 2 H2 → CH3OH (completed and balanced)
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Step 2
<em>Data provided:</em>
19.7 g H2 (the limiting reactant)
Excess reactant = CO
144.5 g CH3OH = actual yield
----
<em>Data needed:</em>
The molar masses of:
H2) 2.00 g/mol
CH3OH) 32.0 g/mol
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Step 3
The theoretical yield:
By stoichiometry,
CO + 2 H2 → CH3OH (The molar rate between H2 and CH3OH = 2:1)
2 x 2.00 g H2 --------- 32.0 g CH3OH
19.7 g H2 --------- X
X = 19.7 g H2 x 32.0 g CH3OH/2 x 2.00 g H2
X = 157.6 g CH3OH (The theoretical yield)
-----------
Step 4
The % yield is defined as follows:
Answer: d. 93% (it is the nearest value in comparison to my result)
Below are I think the data for this problem:
Given the following data:
<span>Ca (s) + 2 C (graphite) → CaC2 (s) ∆H = -62.8 kJ </span>
<span>Ca (s) + ½ O¬2 (g) → CaO (s) ∆H = -635.5 kJ </span>
<span>CaO (s) + H2O (l) → Ca(OH)2 (aq) ∆H = -653.1 kJ </span>
<span>C2H2 (g) + 5/2 O¬2 (g) → 2 CO2 (g) + H2O (l) ∆H = -1300 kJ </span>
<span>C (graphite) + O¬2 (g) → CO2 (g) ∆H = -393.51 kJ
</span>
Below is the answer:
CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq) + C2H2 (g)
<span>So what you do is: </span>
<span>Times the first equation by -1 Second by 1 Third By 1 Fourth by -1 and Fifth by 2 </span>
<span>So This gives us: </span>
<span>1.CaC2--> Ca+2C </span>
<span>2.Ca+1/2O2-->CaO </span>
<span>3.CaO+H2O-->Ca(OH)2 </span>
<span>4.2CO2+H2O-->C2H2+5/2O2 </span>
<span>5.2C+202-->2CO2 </span>
<span>Now you cancel out like terms on either sides of the equation and you end up with </span>
<span>CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq) + C2H2 (g) Just what you wanted </span>
<span>So to calculate ∆H: </span>
<span>62.8-635.5-653.1+1300-787.02= -712.82</span>
<span>An orbital can only have two copies of its electron. Primarily there is only one orbital however it can be filled twice which therefore suggests in total there would be 2 copies. As an example, 3 orbitals would be able to possess 6 electronic copies</span>
I want to say 59 atomic mass units<span> is beta decay, but I'm just guessing from previous Chem. class.</span>
