求解答过程:实验室有一包暗红色粉
\u5b9e\u9a8c\u5ba4\u4e00\u5305\u6df7\u6709\u5c11\u91cf\u94c1\u7c89\u7684\u94dc\u7c89\uff0c\u4f60\u80fd\u7528\u54ea\u4e9b\u65b9\u6cd5\u9664\u53bb\u94dc\u7c89\u4e2d\u94c1\u7c89\uff08\u7b80\u8ff0\u64cd\u4f5c\u8fc7\u7a0b\uff09\u65b9\u6cd5\u4e00\uff08\u78c1\u94c1\u5438\u5f15\uff09\uff1a\u628a\u91d1\u5c5e\u7c89\u672b\u5e73\u94fa\uff0c\u7528\u78c1\u94c1\u8fdb\u884c\u5145\u5206\u5438\u5f15\uff0c\u94c1\u7c89\u88ab\u5438\u9644\u5728\u78c1\u94c1\u4e0a\uff0c\u7559\u4e0b\u7684\u7c89\u672b\u5373\u4e3a\u7eaf\u51c0\u7684\u94dc\u7c89\uff1b\u65b9\u6cd5\u4e8c\uff08\u52a0\u7a00\u76d0\u9178\uff09\uff1a\u628a\u91d1\u5c5e\u7c89\u672b\u653e\u5165\u70e7\u676f\u4e2d\uff0c\u5012\u5165\u7a0d\u8fc7\u91cf\u7684\u7a00\u76d0\u9178\uff0c\u6405\u62cc\uff0c\u5f85\u4e0d\u518d\u653e\u51fa\u6c14\u4f53\u8fc7\u6ee4\uff0c\u5e72\u71e5\u6ee4\u6e23\uff0c\u5373\u5f97\u5230\u7eaf\u51c0\u7684\u94dc\u7c89\uff1b\u65b9\u6cd5\u4e09\uff08\u52a0\u786b\u9178\u94dc\u6eb6\u6db2\uff09\uff1a\u628a\u91d1\u5c5e\u7c89\u672b\u653e\u5165\u70e7\u676f\u4e2d\uff0c\u5012\u5165\u7a0d\u8fc7\u91cf\u7684\u786b\u9178\u94dc\u6eb6\u6db2\uff0c\u6405\u62cc\uff0c\u5f85\u9ed1\u8272\u7c89\u672b\u5b8c\u5168\u6d88\u5931\uff0c\u8fc7\u6ee4\uff0c\u5e72\u71e5\uff0c\u5373\u5f97\u5230\u7eaf\u51c0\u7684\u94dc\u7c89\uff0e
\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 |
绛旓細y= p/(x-0.4)鎶妜=0.65 ,y=0.8浠e叆 p=y*(x-0.4)=0.8*0.25=0.2 鎵浠 y=0.2/(x-0.4)鍖栫畝寰楋細5xy-2y-1=0 杩欏氨鏄粬浠殑鍏崇郴寮 (2) 璁捐皟鍒皒鏃舵弧瓒虫潯浠 锛坸-0.3)*y=(0.8-0.3)*(1+20%)=0.6 涓よ竟鍚屼箻10 10xy-3y-6=0 缁撳悎 5xy-2y-1=0 姹傚緱y=4 x...
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+...
绛旓細1銆佺洏瀛愯〃闈㈢殑鍙嶅皠涓嶆槸鍏ㄩ暅闈㈠弽灏勩傛棤璁烘庢牱鍏夋粦鐨勭洏瀛愮殑琛ㄩ潰閮芥槸瀛樺湪婕弽灏勩備綘鍐嶇湅鐪嬭嚜宸卞涓殑鐩樺瓙锛岄獙璇佷竴涓嬨傝嫻鏋滅殑琛ㄩ潰涔熸槸鍙戠敓鐨勬极鍙嶅皠銆傛墍浠ユ槸鑳界湅鍒板畠浠殑棰滆壊鐨勩2銆佹恫浣撳湪浠讳綍鎯呭喌涓嬮兘鏈夋恫鍖栫幇璞″彂鐢燂紝娑插寲鐨勬柟寮忔湁浜岀锛屼竴绉嶆槸钂稿彂锛屼竴绉嶆槸娌歌吘銆傚湪涓鑸殑鎯呭喌涓嬶紝娑蹭綋鐨勮〃闈㈤兘鏈夎捀鍙戯紝鍙湁...
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...
绛旓細璁惧娌℃湁杩炴帴濂姐佹崲鍒板悗闈㈢殑USB鎺ュ彛璇曡瘯銆
绛旓細2锛屽彉閰嶇數鎵鐨勯珮鍘熺粷缂樼洃瑙嗙數鍘嬭〃涓鐩告寚绀洪檷涓洪浂锛屽叾浠栦袱椤圭數鍘嬪崌楂樼殑鍘熷洜鏄怌銆慉銆傜數鍘嬩簰鎰熷櫒鐔旀柇鍣鏈変竴鐩哥啍鏂 B銆傚彉鍘嬭〃鎹熷潖銆侰銆傜郴缁熷崟鐩告帴鍦般 D銆傜郴缁熶袱鐩告帴鍦般4锛屽父鐢ㄧ殑鐢垫祦浜掓劅鍣ㄤ簩娆¢瀹氱數娴佷负銆5銆慉銆侫銆1 B銆2 C銆3 D銆4 6锛岀敱GL鍨嬫劅搴斿紡杩囨祦缁х數鍣ㄦ瀯鎴愰珮鍘嬬數...
绛旓細涓銆佷腑瀛﹀寲瀛﹀疄楠屾搷浣滀腑鐨勪竷鍘熷垯 鎺屾彙涓嬪垪涓冧釜鏈夊叧鎿嶄綔椤哄簭鐨勫師鍒,灏卞彲浠ユ纭瑙g瓟"瀹為獙绋嬪簭鍒ゆ柇棰"銆 1."浠庝笅寰涓"鍘熷垯銆備互Cl2瀹為獙瀹鍒舵硶涓轰緥,瑁呴厤鍙戠敓瑁呯疆椤哄簭鏄:鏀惧ソ閾佹灦鍙扳啋鎽嗗ソ閰掔簿鐏啋鏍规嵁閰掔簿鐏綅缃浐瀹氬ソ閾佸湀鈫掔煶妫夌綉鈫掑浐瀹氬ソ鍦嗗簳鐑х摱銆 2."浠庡乏鍒板彸"鍘熷垯銆傝閰嶅鏉傝缃簲閬靛惊浠庡乏鍒板彸椤哄簭銆傚涓婅缃...
绛旓細A銆佷竴绉嶉傚簲鎬 B銆侀傚簲鐨勭浉瀵规 C銆侀暱鏈熺殑閬椾紶涓庡彉寮傜殑缁撴灉 D銆佸彈闈炵敓鐗╁洜绱犲埗绾︾殑缁撴灉 4銆佽妫楠岀豢鑹叉鐗╁湪鍛煎惛杩囩▼涓斁鍑轰簩姘у寲纰筹紝蹇呰鐨瀹為獙鏉′欢鏄紙 锛堿.瑕佺敤涓鏍彾瀛愬鐨勬鐗 B.鎶婃鐗╂饭娌″湪姘翠腑 C.瑕佸湪榛戞殫鏉′欢涓嬪疄楠 D.鐢ㄤ竴鏍辜灏忔鐗 5銆佸櫒瀹樼Щ妞嶇殑鎴愬姛鏄汉绫诲尰瀛︾殑浼熷ぇ...
绛旓細鐢叉瘮涔欐瘡灏忔椂澶氳12鍗冪背锛4灏忔椂鍚庣敳鍒拌揪瑗跨珯. 姝ゆ椂 鐢茶溅 姣斾箼杞 澶氳浜 12 4=48鍗冪背銆傛鍚 鐢茶溅杩斿洖 琛屼簡36鍗冪背閬囧埌涔欒溅锛岃鏄庡湪杩欑浉鍚岀殑鏃堕棿閲岋紝涔欒溅鍙蛋浜12鍗冪背銆傛牴鎹矾绋嬮櫎浠ラ熷害 绛変簬鏃堕棿 锛屽彲浠ュ垪鍑烘柟绋嬨 瑙 锛氳鐢茶溅鐨勯熷害鏄瘡灏忔椂 x 鍗冪背锛屽垯鐢茶溅鐨勯熷害鏄 x+12鍗冪背銆12/...