I believe the third one if i am not mistaken
Answer:
Hello There!!
Explanation:
The answer is the last option: cells use oxygen to release energy from food.
hope this helps,have a great day!!
~Pinky~
Answer:
1.75 moles of H₂O
Solution:
The Balance Chemical Equation is as follow,
N₂H₄ + O₂ → N₂ + 2 H₂O
Step 1: Calculate the Limiting Reagent,
According to Balance equation,
32.04 g (1 mol) N₂H₄ reacts with = 32 g (1 mol) of O₂
So,
28 g of N₂H₄ will react with = X g of O₂
Solving for X,
X = (28 g × 32 g) ÷ 32.04 g
X = 27.96 g of O₂
It means 29 g of N₂H₄ requires 47.96 g of O₂, while we are provided with 73 g of O₂ which is in excess. Therefore, N₂H₄ is the limiting reagent and will control the yield of products.
Step 2: Calculate moles of Water produced,
According to equation,
32.04 g (1 mol) of N₂H₄ produces = 2 moles of H₂O
So,
28 g of N₂H₄ will produce = X moles of H₂O
Solving for X,
X = (28 g × 2 mol) ÷ 32.04 g
X = 1.75 moles of H₂O
Answer:
D) 5.15
Explanation:
Step 1: Write the equation for the dissociation of HCN
HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)
Step 2: Calculate [H⁺] at equilibrium
The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.
α% = [H⁺]eq / [HCN]₀ × 100%
[H⁺]eq = α%/100% × [HCN]₀
[H⁺]eq = 0.0070%/100% × 0.10 M
[H⁺]eq = 7.0 × 10⁻⁶ M
Step 3: Calculate the pH
pH = -log [H⁺] = -log 7.0 × 10⁻⁶ = 5.15
"The boron-nitrogen interaction in the studied molecules shows some similarities with the N→B bond in the H3N-BH3 molecule, formally understood as covalent-dative. ... The results show that all the studied BN bonds are triple, since three two-center orbitals have been obtained."
"Formation of a dative bond or coordinate bond between ammonia and boron trifluoride. When the nitrogen donates a pair of electrons to share with the boron, the boron gains an octet. ... In addition, a pair of non-bonding electrons becomes bonding; they are delocalized over two atoms and become lower in energy."