Answer:
Succinic acid
Explanation:
The most common possibility is succinic acid
As it has decimals after whole no till hundredth it contains OH and C in most of the cases .
Let's check for succinic acid
- C4H_6O_6
- 4(12)+4(16)+6
- 64+48+6
- 118u
Yes approximately equal
Molecular formula is.
(CH_2)_2(CO_2H)_2
Balanced equation:
Pb(SO₄)₂ + 2 Zn → 2 ZnSO₄ + Pb
From the equation we found that 2 moles of Zn form 2 moles of ZnSO₄
so 0.582 mole of Zn will produce 0.582 ZnSO₄
Molar mass of ZnSO₄ = 161.47 g/mole
Mass of ZnSO₄ formed = 161.47 x 0.582 = 94.0 grams
Answer:
Step 1: List the known quantities and plan the problem.
Known
number of C atoms = 4.72 × 10 24
1 mole = 6.02 × 10 23 atoms
Unknown
4.72 × 10 24 = ? mol C
One conversion factor will allow us to convert from the number of C atoms to moles of C atoms.
Step 2: Calculate.
4.72 times 10^{24} text{atoms C} times frac{1 text{mol C}}{6.02 times 10^{23} text{atoms C}}=7.84 text{mol C}
Step 3: Think about your result.
The given number of carbon atoms was greater than Avogadro’s number, so the number of moles of C atoms is greater than 1 mole. Since Avogadro’s number is a measured quantity with three significant figures, the result of the calculation is rounded to three significant figures.
Explanation:
the image is 4 step2 mwa
Answer:
Exploding
Explanation:
(hope it helps you)
Edit: sorry Im pretty sure its vibrating and because they vibrate which is why it has kinetic energy
Answer:
See explanation
Explanation:
Carbon monoxide is toxic because it can successfully competes with oxygen for hemoglobin (Hb) sites according to the following equilibrium:
Hb(O₂)₄(aq) + 4CO(g) ⇄ Hb(CO)₄(aq) + 4O₂(g)
Based on LeChatelier’s Principle, CO poisoning can be reversed by applying O₂(g) to victim. This addition of O₂(g) increases system concentration of O₂(g) in the equilibrium reaction thereby overloading the product side of the equilibrium forcing it to shift to the left, decomposing the carboxyhemoglobin (Hb(CO)₄) and releasing the CO(g) and combining with O₂(g) to form more Hb(O₂)₄.
