Answer:
0.319 L
Explanation:
M(MgSO4) = 120 g/mol
46.1 g * 1 mol/120 g = 0.384 mol MgSO4
0.384 mol * 1 L/1.20 mol = 0.319 L
<h3>Take the weighted average of the individual isotopes.</h3><h3 /><h3>Explanation:</h3><h3>63</h3><h3>C</h3><h3>u</h3><h3> has </h3><h3>69.2</h3><h3>%</h3><h3> abundance.</h3><h3 /><h3>65</h3><h3>C</h3><h3>u</h3><h3> has </h3><h3>30.8</h3><h3>%</h3><h3> abundance.</h3><h3 /><h3>So, the weighted average is </h3><h3>62.93</h3><h3>×</h3><h3>69.2</h3><h3>%</h3><h3> </h3><h3>+</h3><h3> </h3><h3>64.93</h3><h3>×</h3><h3>30.8</h3><h3>%</h3><h3> </h3><h3>=</h3><h3> </h3><h3>63.55</h3><h3> </h3><h3>amu</h3><h3> .</h3><h3 /><h3>If we look at the Periodic Table, copper metal (a mixture of isotopes but </h3><h3>63</h3><h3>C</h3><h3>u</h3><h3> and </h3><h3>65</h3><h3>C</h3><h3>u</h3><h3> predominate) has an approximate atomic mass of </h3><h3>63.55</h3><h3> </h3><h3>g</h3><h3>⋅</h3><h3>m</h3><h3>o</h3><h3>l</h3><h3>−</h3><h3>1</h3><h3> , so we know we are right.</h3>
Energy is released as a bond is formed
Answer:
the oxidation number of Pt in K₂PtCl₆ <u>is 4.</u>
K₂PtCl₆=0
2+x-6=0
x=4
