Describe the light spectrum of a star
1 answer:
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Answer:
2 mol H₂O
Explanation:
With the reaction,
- 2H₂(g) + O₂(g) → 2 H₂O(g)
1.55 moles of O₂ would react completely with ( 2*1.55 ) 3.1 moles of H₂. There are not as many moles of H₂, thus H₂ is the limiting reactant.
Now we <u>calculate the moles of H₂O produced</u>, <em>starting from the moles of limiting reactant</em>:
- 2.00 mol H₂ *
= 2 mol H₂O
Answer:
Explanation:
1. find the molar mass (amu) of each element and add them to get the whole molar mass.
2. divide the 1 element molar mass with the whole molar mass
3. multiple by 100 and that gives you the % composition.
<h2><u><em>56-57: NaCl</em></u></h2>1. Na(22.99amu) + Cl (35.453amu)=58.443
2(Na): = .393
2(Cl): = .607
3(Na): .393 * 100=39.3%
3(Cl): .607 * 100= 60.7%
<h2><u>58-60 </u>1. K: (39.098)(2)=78.196
_ C: (12.011)(1)= 12.011
_O: (15.99)(3) = 47.997
78.196+12.011+47.997= 138.204
2:K: = .566 <u>Step </u>3: (.566)(100)= 56.6%
2: C: = .087 <u>Step 3</u>: (.087)(100)= 8.7%
2: O: = .347 <u>Step 3</u>: (.347)(100) = 34.7%
1. Fe (55.845)(3)= 167.535
_ O (15.999)(4) = 63.996
167.535+63.996=231.531
2: Fe: = .724 Step 3: (.724)(100)= 72.4%
2: O : = .276 Step 3: (.276)(100) = 27.6%
1.
C(12.011*3)=36.033
H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064
O(15.999*3)=47.997
add them: 92.094
2: C: = .391 Step 3: (.391)(100) = 39.1%
2: H: = .088 step 3: (.088)(100) = 8.8%
2: O: = .521 step 3: (.521)(100) = 52.1%