<span>For equation A + 3B + 2C ---> 2D,
1 mole of A will produce 2 moles of D
3 moles of B will produce 2 moles of D, so 1 mole of B will produce 2/3 moles of D
2 moles of C will produce 2 moles of D, so 1 mole of C will produce 1 mole of D
If only 1 mole of B is present, only 2/3 moles of D can be produced. This is regardless of the number of moles of A and C. B is the limiting reactant and the maximum number of moles of D expected is 2/3.</span>
Q = M * C *ΔT
Q / <span>ΔT = M
</span>Δf - Δi = 98.4ºC - 62.2ºC = 36.2ºC
<span>
C = 1137 J / 140 * 36.2
C = 1137 / 5068
C = 0.224 J/gºC</span>
Hey there,
Answer:
4 valence electrons.
Hope this helps :D
<em>~Top☺</em>
The law of conservation of mass states that mass is neither created nor destroyed. Since we have 2 g/mol of A and 3 g/mol of B then AB should be equal to the sum of their molar mass that is
2 g/mol + 3 g/mol = 5 g/mol AB
for the case of A2B3
A2 = 2 * 2 = 4 g/mol
B3 = 3 * 3 = 9 g/mol
therefore A2B3 = 13 g/mol
The question is incomplete. The complete question is:
The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
570 years
Explanation:
The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.