The Bohr model proposed that electrons could just have characterized vitality levels thus when rotting back to a lower vitality level discharge a specific measure of vitality. Since the measure of vitality could be changed over to a specific recurrence then particular emanation lines were found in the electromagnetic range. Alternate speculations couldn't clarify the discharge lines.
Answer:
Pb(NO3)2 + KI = KNO3 + PbI2
Volume perhaps ?
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Answer:
38.16 mL
Solution:
Data Given;
Density = 1.31 g/mL
Mass = 50 g
Volume = ?
Formula Used;
Density = Mass ÷ Volume
Solving for Volume,
Volume = Mass ÷ Density
Putting values,
Volume = 50 g ÷ 1.31 g.mL⁻¹
Volume = 38.16 mL
Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
