求解答过程：实验室有一包暗红色粉
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\uff081\uff09\u94dc\u4e0e\u6c27\u6c14\u5728\u52a0\u70ed\u6761\u4ef6\u4e0b\u53cd\u5e94\u751f\u6210\u6c27\u5316\u94dc\uff0e\u8be5\u53cd\u5e94\u7684\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1a2Cu+O 2 \u25b3 . 2CuO\uff0e\u78b1\u5f0f\u78b3\u9178\u94dc\u5728\u52a0\u70ed\u6761\u4ef6\u4e0b\u53cd\u5e94\u751f\u6210\u6c27\u5316\u94dc\u3001\u6c34\u548c\u4e8c\u6c27\u5316\u78b3\uff0e\u8be5\u53cd\u5e94\u7684\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1aCu 2 \uff08OH\uff09 2 CO 3 \u25b3 . 2CuO+H 2 O+CO 2 \u2191\uff0e\u6c27\u5316\u94dc\u4e0e\u7a00\u786b\u9178\u53cd\u5e94\u751f\u6210\u786b\u9178\u94dc\u548c\u6c34\uff0e\u8be5\u53cd\u5e94\u7684\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1aCuO+H 2 SO 4 =CuSO 4 +H 2 O\uff0e\uff082\uff09\u8bbe\u7eff\u8272\u7c89\u672b\u4e2d\u94dc\u7684\u8d28\u91cf\u4e3ax\uff0c\u52a0\u70ed\u540e\u751f\u6210\u6c27\u5316\u94dc\u7684\u8d28\u91cf\u4e3ay\uff0e2Cu+O 2 \u25b3 . 2CuO128 160x y 128 160 = x y \uff0cy= 5x 4 \u8bbe\u7eff\u8272\u7c89\u672b\u4e2d\u94dc\u7eff\u5728\u52a0\u70ed\u540e\u53cd\u5e94\u751f\u6210\u6c27\u5316\u94dc\u7684\u8d28\u91cf\u4e3az\uff0eCu 2 \uff08OH\uff09 2 CO 3 \u25b3 . 2CuO+H 2 O+CO 2 \u2191222 16080.0g-x z 222 160 = 80.0g-x z \uff0cz= 80(80.0g-x) 111 \u7531\u9898\u610f\u5f97 5x 4 + 80(80.0g-x) 111 =80.0gx= 1984 47 g\u94dc\u7eff\u7684\u8d28\u91cf\u4e3a80.0g- 1984 47 g= 1776 47 g\u7eff\u8272\u7c89\u672b\u4e2d\u5355\u8d28\u94dc\u548c\u94dc\u7eff\u7684\u8d28\u91cf\u6bd4\u4e3a 1984 47 g\uff1a 1776 47 g=124\uff1a111\uff083\uff09\u8bbe\u751f\u6210\u786b\u9178\u94dc\u7684\u8d28\u91cf\u4e3aw\uff0eCuO+H 2 SO 4 =CuSO 4 +H 2 O80 16080.0g w 80 160 = 80.0g w \uff0cw=160g\u6240\u5f97\u6eb6\u6db2\u4e2d\u786b\u9178\u94dc\u7684\u8d28\u91cf\u5206\u6570\u4e3a 160g 80.0g+320g \u00d7100%=40%\u6545\u7b54\u6848\u4e3a\uff1a\uff081\uff092Cu+O 2 \u25b3 . 2CuO\uff0cCu 2 \uff08OH\uff09 2 CO 3 \u25b3 . 2CuO+H 2 O+CO 2 \u2191\uff0cCuO+H 2 SO 4 =CuSO 4 +H 2 O\uff0e\uff082\uff09 124 111 \uff0e\uff083\uff0940%\uff0e
| Ⅰ.KMnO 4 溶液、稀硫酸；取少量样品溶于试管中，配成溶液后，再滴入几滴稀硫酸和几滴KMnO 4 溶液，只要溶液不褪色或不变浅，即说明样品中无FeO Ⅱ．（1）BDAC；（2）将反应产生的CO 2 气体尽可能彻底的赶入装置A中，使之完全被Ba(OH) 2 溶液吸收；（3）ad；（4）61.5%；（5）偏低 |
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绛旓細B
绛旓細(1) f(x)涓哄鍑芥暟=> f(0)=0 => a-2/2^0+1=0 => a=1 (2) 濡傛灉宸茬粡瀛﹁繃瀵兼暟锛屽彲浠ョ洿鎺ユ眰涓闃跺锛歞f/dx= -2*ln2*2^(-x)>0锛屾晠f(x)鍗曡皟閫掑銆傚鏋滄病瀛﹁繃瀵兼暟锛屽氨鐩存帴鐢ㄥ畾涔夎瘉鏄庯細璁緓1>x2锛宖(x1)-f(x2) = -2/2^x1-(-2/2^x2) = 2*(2^x1-2^x2)/2^(x1+...
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绛旓細姹傚浘褰㈣〃绀虹殑鏁:1/10 15 8瑙ｅ喅闂:1.2梅4/7脳40%=1.4(鍚)2.1梅(1/10-1/15)=30(灏忔椂)3.18脳80%梅(1+20%)=12(鍏)4.绛旀涓嶅敮涓,鍙煡鎬昏垂鐢4000鍏冦佷紮椋1000鍏冦佷氦閫800鍏冪瓑,寤鸿:鍚堢悊寮鏀,鐣ユ湁鑺備綑銆侾40-41鍏堝寲绠,鍐嶆眰姣斿:1.1:2000=1/20002.9:10=9/103.3:1=3鍒嗘瀽鍗曚綅鈥1鈥濆拰鍏蜂綋閲忕殑瀵瑰簲...
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