Đáp án:
\( {m_K} = {6,47595.10^{ - 23}}{\text{ gam}}\)
\( {m_P} = {5,14755.10^{ - 23}}{\text{ gam}}\)
\({m_{Mg}} = {3,9852.10^{ - 23}}{\text{ gam}}\)
\( {m_{Al}} = {4,48335.10^{ - 23}}{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({m_C} = 12{\text{ đvC = 1}}{\text{,9926}}{\text{.1}}{{\text{0}}^{ - 23}}{\text{ gam}}\)
\( \to 1{\text{ đvC = }}\frac{{{{1,9926.10}^{ - 23}}}}{{12}} = {1,6605.10^{ - 24}}{\text{ gam}}\)
\( \to {m_K} = {1,6605.10^{ - 24}}.39 = {6,47595.10^{ - 23}}{\text{ gam}}\)
\( \to {m_P} = {1,6605.10^{ - 24}}.31 = {5,14755.10^{ - 23}}{\text{ gam}}\)
\( \to {m_{Mg}} = {1,6605.10^{ - 24}}.24 = {3,9852.10^{ - 23}}{\text{ gam}}\)
\( \to {m_{Al}} = {1,6605.10^{ - 24}}.27 = {4,48335.10^{ - 23}}{\text{ gam}}\)