Answer:
option C is correct (250 g)
Explanation:
Given data:
Half life of carbon-14 = 5700 years
Total amount of sample = 1000 g
Sample left after 11,400 years = ?
Solution:
First of all we will calculate the number of half lives passes during 11,400 years.
Number of half lives = time elapsed/ half life
Number of half lives = 11,400 years/5700 years
Number of half lives = 2
Now we will calculate the amount left.
At time zero = 1000 g
At first half life = 1000 g/2 = 500 g
At second half life = 500 g/2 = 250 g
Thus, option C is correct.
Answer: The weight/weight % or percent by mass of the solute is 5.41 %.
Explanation:
Mass of the sodium sulfate,w = 9.74 g
Volume of the water = 165 mL
Density of the water = 1 g/mL
Mass of the water =
Mass of the solution, W:
Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g
The weight/weight % or percent by mass of the solute is 5.41 %.
My Friday is going good and I might get to go over to my BFF house!!!
Answer:
102g
Explanation:
To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.
In this case, the given equation is already balanced.
From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.
Mole= Mass ÷Mr
Mass= Mole ×Mr
<u>Method 1: using the </u><u>mass of glucose</u>
Mr of glucose
= 6(12) +12(1) +6(16)
= 180
Moles of glucose reacted
= 200 ÷180
= 
mol
Amount of ethanol formed: moles of glucose reacted= 2: 1
Amount of ethanol
= 
= 
mol
Mass of ethanol
= ![\frac{20}{9} \times[2(12)+6+16]](https://tex.z-dn.net/?f=%5Cfrac%7B20%7D%7B9%7D%20%5Ctimes%5B2%2812%29%2B6%2B16%5D)
= 
= 102 g (3 s.f.)
<u>Method 2: using </u><u>mass of carbon dioxide</u><u> produced</u>
Mole of carbon dioxide produced
= 97.7 ÷[12 +2(16)]
= 97.7 ÷44
= 
mol
Moles of ethanol: moles of carbon dioxide= 1: 1
Moles of ethanol formed= mol
Mass of ethanol formed
= ![\frac{977}{440} \times[2(12)+6+16]](https://tex.z-dn.net/?f=%5Cfrac%7B977%7D%7B440%7D%20%5Ctimes%5B2%2812%29%2B6%2B16%5D)
= 102 g (3 s.f.)
Thus, 102 g of ethanol are formed.
Additional:
For a similar question on mass and mole ratio, do check out the following!
Hi, here is a basic summary of what we did in a lab; there were 3 reactions: The procedure: Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. NaOH(s)-> Na+(aq) + OH-(aq) ΔH1=-34.121kJ Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of HCl to form water and an aqueous solution of sodium chloride. NaOH(s) + H+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH2=-83.602kJ Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of HCl to form water an an aqueous solution of sodium chloride. H+(aq) + OH-(aq) + Na+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH3= -50.2kJ The ΔH values were calculated by dividing the heat gained by the number of moles (each reaction had 0.05moles of NaOH) The problem: Net ionic equations for reaction 2 & 3: 2: NaOH(s) + H+(aq) -> H2O + Na+(aq) 3: H+(aq) + OH-(aq) -> H2O i) In reaction 1, ΔH1 represents the heat evolved as solid NaOH dissolves. Look at the net ionic equations for reactions 2 and 3 and make similar statements as to what ΔH2 and ΔH3 represent. ii) Compare ΔH2 with (ΔH1 + ΔH3). Explain in sentences the similarity between these two values by using your answer to #5 above. Attempt at answering: i) Firstly, ΔH2 represents the heat evolved as the hydrogen ion displaces the sodium ion, creating a single displacement reaction. ΔH3 represents the heat evolved as the hydrogen and hydroxide ion form water via a neutralization reaction. ii) ΔH2 is equal to (or supposed to be, this is a source of error while calculating) (ΔH1 + ΔH3). The similarity between these two values is that .. (this is where I get confused!)
Source https://www.physicsforums.com/threads/calorimetry-help-chemistry.399653/
