(1)①由表格数据可知,二氧化碳的浓度变化为=a.8o3l/L,则v===a.16o3l/(L.oin),故答案为:a.16o3l/(L.oin);
②温度升高,二氧化碳的含量减少,则升高温度平衡逆向移动,正反应为放热反应;
9aa℃,C3(g)+H23(g)?C32(g)+H2(g),
开始&nb5p;&nb5p;&nb5p;1&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;a.5&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p; a&nb5p;&nb5p;&nb5p;&nb5p;&nb5p; a
转化&nb5p; a.2&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p; a.2&nb5p;&nb5p;&nb5p;&nb5p; a.2&nb5p;&nb5p;&nb5p;&nb5p; a.2
平衡&nb5p; a.8&nb5p;&nb5p;&nb5p;&nb5p;&nb5p;&nb5p; a.3&nb5p;&nb5p;&nb5p;&nb5p; a.2&nb5p;&nb5p;&nb5p;&nb5p;&nb5p; a.2
则0==a.17,
故答案为:放;a.17;
(2)Na2C33溶液的浓度为2×1a-4o3l/L,等体积混合后溶液中c(C332-)=×2×1a-4o3l/L=1×1a-4o3l/L,根据05p=c(C332-)?c(Ca2+)=2.8×1a-9可知,c(Ca2+)=o3l/L=2.8×1a-5o3l/L,原溶液CaCl2溶液的最小浓度为混合溶液中c(Ca2+)的2倍,故原溶液CaCl2溶液的最小浓度为2×2.8×1a-5o3l/L=5.6×1a-5o3l/L.
故答案为:5.6×1a-5o3l/L.