They have free electron(s) on their outermost energy levels making them good conductors.
They have metallic bonds in their chemical structure.
They readily lose the electrons on their outermost energy levels, to bond with non-metals in ionic bonds to form chemical compounds called "salts"
Answer:
67.4 % of C₉H₈O₄
Explanation:
To make titrations problems we know, that in the endpoint:
mmoles of acid = mmoles of base
mmoles = M . volume so:
mmoles of acid = 20.52 mL . 0.1121 M
mmoles of acid = mg of acid / PM (mg /mmoles)
Let's determine the PM of aspirin:
12.017 g/m . 9 + 1.00078 g/m . 8 + 15.9994 g/m . 4 = 180.1568 mg/mmol
mass (mg) = (20.52 mL . 0.1121 M) . 180.1568 mg/mmol
mass (mg) = 414.4 mg
We convert the mass to g → 414.4 mg . 1g / 1000mg = 0.4144 g
We determine the % → (0.4144 g / 0.615 g) . 100 = 67.4 %
I attended the answer. I hope this helps you.
Answer:
There will 3.95 grams of Na2 and H2O that should be added to form a concentric required solution.
Answer:
A) = 4.7 × 10⁻⁴atm
Explanation:
Given that,
Kp = 1.5*10³ at 400°C
partial pressure pN2 = 0.10 atm
partial pressure pH2 = 0.15 atm
To determine:
Partial pressure pNH3 at equilibrium
The decomposition reaction is:-
2NH3(g) ↔N2(g) + 3H2(g)
Kp = [pH2]³[pN2]/[pNH3]²
pNH3 =√ [(pH2)³(pN2)/Kp]
pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm
![K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cfrac%7B%5BpH_2%5D%20%5E3%5BpN_2%5D%7D%7B%5BpNH_3%5D%5E2%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%28pH_2%29%5E3%28pN_2%29%7D%7BpNH_3%7D%20%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%280.15%29%5E3%280.10%29%7D%7B1.5%20%5Ctimes%2010%5E3%7D%20%7D%20%5C%5C%3D4.74%20%5Ctimes%2010%5E-%5E4atm)
= 4.7 × 10⁻⁴atm
