1. m = nM = 5 * 44 = 220g
V = n * V[m] = 5 * 22.4 = 112L
n[CO(2)] = n * N[A] = 5 * 6.02 * 10^23 = 3.01 * 10^24
n[O] = 2n * N[A] = 10 * 6.02 * 10^23 = 6.02 * 10^24
2. n[H(2)SO(4)] = cV = 2 * 0.4 = 0.8mol
m[H(2)SO(4)] = nM = 0.8 * 98 = 78.4g
c[H+] = 2c[H(2)SO(4)] = 4mol/L
c[SO(4)[2-]] = c[H(2)SO(4)] = 2mol/L
3. 所含分子數(shù)相等.
當(dāng) n[N(2)] 和 n[CO(2)] 相等時(shí),
m[N(2)] = nM[N(2)] = 28n
m[CO(2)] = nM[CO(2)] = 44n
所以后者質(zhì)量大
4. V = n[B] / c[B] = 1mol / (16mol/L) = 0.0625L
5. 設(shè) A 的相對(duì)原子質(zhì)量為 x,則：
x : 35.5*3 = 1:3.94, x = 27
7. n[HCl] = m / M = ρV / M = 1.19 * 4 * 0.37 / 36.5 = 0.048mol
c[B] = n[B] / V = 0.048 / 0.2 = 0.24mol/L