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When the enthalpy value is given, we can calculate how much heat is use or produces in a given equation.
67.6 kCal ---> 67.6 kCal= 1 mol of reaction
1 mol of reaction= 1 mol of CO (based on the coefficient)
so 1 mole of CO gives us 67.6 kCal of heat.
calculation:
1 mol CO
67.6 kCal ---> 67.6 kCal= 1 mol of reaction
1 mol of reaction= 1 mol of CO (based on the coefficient)
so 1 mole of CO gives us 67.6 kCal of heat.
calculation:
1 mol CO
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%