{"id":8567,"date":"2024-08-05T21:40:34","date_gmt":"2024-08-06T00:40:34","guid":{"rendered":"https:\/\/xequematenem.com.br\/blog\/?p=8567"},"modified":"2024-09-09T19:29:11","modified_gmt":"2024-09-09T22:29:11","slug":"questao-110-enem-ppl-2018","status":"publish","type":"post","link":"https:\/\/xequematenem.com.br\/blog\/questao-110-enem-ppl-2018\/","title":{"rendered":"Quest\u00e3o 110 \u2013 ENEM PPL 2018"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">O elemento radioativo t\u00f3rio (Th) pode substituir os combust\u00edveis f\u00f3sseis e baterias. Pequenas quantidades desse elemento seriam suficientes para gerar grande quantidade de energia. <strong>A part\u00edcula liberada em seu decaimento poderia ser bloqueada utilizando-se uma caixa de a\u00e7o<\/strong> <strong>inoxid\u00e1vel.<\/strong> A equa\u00e7\u00e3o nuclear para o decaimento do <sup>230<\/sup><sub>90<\/sub>Th \u00e9:<\/p>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" width=\"370\" height=\"43\" data-src=\"https:\/\/xequematenem.com.br\/blog\/wp-content\/uploads\/2024\/08\/image-181.png\" alt=\"\" class=\"wp-image-8568 lazyload\" style=\"--smush-placeholder-width: 370px; --smush-placeholder-aspect-ratio: 370\/43;width:310px;height:auto\" title=\"\" data-srcset=\"https:\/\/xequematenem.com.br\/blog\/wp-content\/uploads\/2024\/08\/image-181.png 370w, https:\/\/xequematenem.com.br\/blog\/wp-content\/uploads\/2024\/08\/image-181-300x35.png 300w\" data-sizes=\"(max-width: 370px) 100vw, 370px\" src=\"data:image\/svg+xml;base64,PHN2ZyB3aWR0aD0iMSIgaGVpZ2h0PSIxIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==\" \/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Considerando a equa\u00e7\u00e3o de decaimento nuclear, <strong>a part\u00edcula que fica bloqueada na caixa de a\u00e7o inoxid\u00e1vel \u00e9 o(a)<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A) alfa.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">B) beta.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">C) pr\u00f3ton.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">D) n\u00eautron.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">E) p\u00f3sitron.<\/p>\n\n\n\n<p class=\"has-text-align-center has-large-font-size wp-block-paragraph\"><strong>Solu\u00e7\u00e3o<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Para saber exatamente qual part\u00edcula foi emitida, \u00e9 necess\u00e1rio realizar um balan\u00e7o de carga e massa. Isso significa que o somat\u00f3rio das massas das esp\u00e9cies que est\u00e3o na posi\u00e7\u00e3o de reagente precisa ser igual ao somat\u00f3rio de quem ocupa a posi\u00e7\u00e3o dos produtos. O mesmo \u00e9 v\u00e1lido para a carga. Sendo assim, temos que:<\/p>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" width=\"562\" height=\"59\" data-src=\"https:\/\/xequematenem.com.br\/blog\/wp-content\/uploads\/2024\/08\/image-182.png\" alt=\"\" class=\"wp-image-8569 lazyload\" style=\"--smush-placeholder-width: 562px; --smush-placeholder-aspect-ratio: 562\/59;width:410px;height:auto\" title=\"\" data-srcset=\"https:\/\/xequematenem.com.br\/blog\/wp-content\/uploads\/2024\/08\/image-182.png 562w, https:\/\/xequematenem.com.br\/blog\/wp-content\/uploads\/2024\/08\/image-182-300x31.png 300w\" data-sizes=\"(max-width: 562px) 100vw, 562px\" src=\"data:image\/svg+xml;base64,PHN2ZyB3aWR0aD0iMSIgaGVpZ2h0PSIxIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==\" \/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A energia n\u00e3o apresenta nem carga e nem massa, como a radia\u00e7\u00e3o gama, por exemplo.<\/p>\n\n\n\n<div class=\"wp-block-group is-vertical is-layout-flex wp-container-core-group-is-layout-2c90304e wp-block-group-is-layout-flex\">\n<p class=\"wp-block-paragraph\"><strong>Para o balan\u00e7o de massa:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">230 = 226 + x<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">x = 230 \u2013 226<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>x = 4<\/strong><\/p>\n<\/div>\n\n\n\n<div class=\"wp-block-group is-vertical is-layout-flex wp-container-core-group-is-layout-2c90304e wp-block-group-is-layout-flex\">\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Para o balan\u00e7o de carga:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">90 = 88 + y<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">y = 90 &#8211; 88<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>y = 2<\/strong><\/p>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Possuir massa 4 e carga 2 \u00e9 caracter\u00edstico de uma part\u00edcula alfa ( ).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Alternativa A<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>O elemento radioativo t\u00f3rio (Th) pode substituir os combust\u00edveis f\u00f3sseis e baterias. Pequenas quantidades desse elemento seriam suficientes para gerar grande quantidade de energia. A part\u00edcula liberada em seu decaimento poderia ser bloqueada utilizando-se uma caixa de a\u00e7o inoxid\u00e1vel. A equa\u00e7\u00e3o nuclear para o decaimento do 23090Th \u00e9: Considerando a equa\u00e7\u00e3o de decaimento nuclear, a [&hellip;]<\/p>\n","protected":false},"author":11,"featured_media":8568,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[87,97],"tags":[],"area-do-conhecimento":[116,120],"assunto":[],"class_list":["post-8567","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-questoes-ppl","category-2018-questoes-ppl","area-do-conhecimento-ciencias-da-natureza-e-suas-tecnologias","area-do-conhecimento-quimica"],"_links":{"self":[{"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/posts\/8567","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/users\/11"}],"replies":[{"embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/comments?post=8567"}],"version-history":[{"count":3,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/posts\/8567\/revisions"}],"predecessor-version":[{"id":8572,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/posts\/8567\/revisions\/8572"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/media\/8568"}],"wp:attachment":[{"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/media?parent=8567"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/categories?post=8567"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/tags?post=8567"},{"taxonomy":"area-do-conhecimento","embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/area-do-conhecimento?post=8567"},{"taxonomy":"assunto","embeddable":true,"href":"https:\/\/xequematenem.com.br\/blog\/wp-json\/wp\/v2\/assunto?post=8567"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}