<span>By using the mole ratio, we can determine that 2 moles of NH3 are made when 3 moles of hydrogen gas are present. The numbers in front of the chemicals tell us the relative amounts consumed and produced. Since there is a 3 in front of H2 and a 2 in front of NH3, this tells us that for every 3 moles of H2 gas used, 2 moles of NH3 are made.</span>
We subtract the enthalpies of the reactants from that of the products:

Since this is < 0, this is an exothermic reaction.
The volume of the soft drink solution in milliliters that contains 102.5 g of sucrose is 11.93mL.
<h3>How to calculate volume?</h3>
The volume of a solution can be calculated by dividing the mass by the density. That is;
Volume = mass/density
According to this question, a soft drink contains 12.1% sucrose (C12H22O11) by mass. This means that the mass of the sucrose is
12.1/100 × 102.5 = 12.40g of sucrose
Volume = 12.40g ÷ 1.04g/mL
Volume = 11.93mL
Therefore, the volume of the soft drink solution in milliliters that contains 102.5 g of sucrose is 11.93mL.
Learn more about volume at: brainly.com/question/1578538
Answer:
<em>That's </em><em>because</em><em> </em><em>in </em><em>water</em><em> </em><em>NaF </em><em>will </em><em>dissolve</em><em> </em><em>to </em><em>produce </em><em>Na</em><em>+</em><em>,</em><em>the </em><em>conjugate </em><em>base </em><em>of </em><em>a </em><em>strong</em><em> </em><em>acid </em><em>which</em><em> </em><em>will </em><em>not </em><em>react </em><em>with </em><em>water.</em><em>h</em><em>o</em><em>w</em><em>e</em><em>v</em><em>e</em><em>r</em><em> </em><em>F- </em><em>will </em><em>behave </em><em>like </em><em>a </em><em>bronsted </em><em>base,</em><em> </em><em>and </em><em>accept</em><em> </em><em>a </em><em>proton </em><em>from </em><em>water.</em><em>t</em><em>h</em><em>i</em><em>s</em><em> </em><em>is </em><em>called </em><em>hydrolysis</em><em> </em><em>reaction,</em><em> because</em><em> </em><em>a </em><em>molecule</em><em> </em><em>of </em><em>water </em><em>is </em><em>broken </em><em>up.</em>
<em>a </em><em>conjugate</em><em> base</em><em> </em><em>is </em><em>what </em><em>I </em><em>leftover </em><em>after </em><em>an </em><em>acid </em><em>loses </em><em>a </em><em>hydrogen</em><em> </em><em>ion.</em>
<em>I </em><em>hope</em><em> this</em><em> helps</em>
Answer:
159 mg caffeine is being extracted in 60 mL dichloromethane
Explanation:
Given that:
mass of caffeine in 100 mL of water = 600 mg
Volume of the water = 100 mL
Partition co-efficient (K) = 4.6
mass of caffeine extracted = ??? (unknown)
The portion of the DCM = 60 mL
Partial co-efficient (K) = 
where;
solubility of compound in the organic solvent and
= solubility in aqueous water.
So; we can represent our data as:
÷ 
Since one part of the portion is A and the other part is B
A+B = 60 mL
A+B = 0.60
A= 0.60 - B
4.6=
÷ 
4.6 = 
4.6 ×
=
4.6 B
= 0.6 - B
2.76 B = 0.6 - B
2.76 + B = 0.6
3.76 B = 0.6
B = 
B = 0.159 g
B = 159 mg
∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.