Group 1: +1
Group 2: +2
Group 13: +3
Group 14: +4
Group 15: -3
Group 16: +II
Group 17: -1
C. The number of moles of H in 0.109 mole of N₂H₄ is 0.436 mole
D. The number of moles of H in 34 moles of C₁₀H₂₂ is 748 moles
<h3>C. How to determine the number of mole of H in 0.109 mole of N₂H₄</h3>
1 mole of N₂H₄ contains 4 moles of H
Therefore,
0.109 mole of N₂H₄ will contain = 0.109 × 4 = 0.436 mole of H
<h3>D. How to determine the number of mole of H in 34 mole of C₁₀H₂₂</h3>
1 mole of C₁₀H₂₂ contains 22 moles of H
Therefore,
34 mole of C₁₀H₂₂ will contain = 34 × 22 = 748 mole of H
Learn more about mole:
brainly.com/question/13314627
#SPJ1
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!
The product of prime polynomials is equivalent to 36x3 – 15x2 – 6x is letter B which is 3x(3x – 2)(4x 1). Below is the solution.
3x(3x - 2) (4x + 1)
= 9x2 - 6x (4x + 1)
= 36x3 + 9x2 + - 24x2 - 6x
= 36x3 - 15x2 - 6x
A
The number of electrons/protons is the one that determines the atomic number of an element.
