Each element<span> can usually be classified as a metal or a non-metal based on their ... They are usually </span>dull<span>and therefore show no metallic </span>luster<span> and they do not reflect ... </span>Dull<span>, Brittle solids; Little or no metallic </span>luster<span>; </span>High<span> ionization energies; </span>High<span> ...</span>
I don't know the answer to long I just want points so plz like and thank me jk the answer is 12H2o
S + O2 → SO2
<span>z / (32.0655 g S/mol) x (1 mol SO2 / 1 mol S) x (64.0638 g SO2/mol) = (1.9979 z) g SO2 </span>
<span>C + O2 → CO2 </span>
<span>(9.0-z) / (12.01078 g C/mol) x (1 mol CO2 / 1 mol C) x (44.00964 g CO2/mol) = (32.9776 - 3.66418 z) g CO2 </span>
<span>Add the two masses of SO2 and CO2 and set them equal to the amount given in the problem: </span>
<span>(1.9979 z) + (32.9776 - 3.66418 z) = 27.9 </span>
<span>Solve for z algebraically: </span>
<span>z = 3.0 g S</span>
Answer:
Molarity is halved when the volume of solvent is doubled.
Explanation:
Using the dilution equation (volume 1)(molarity 1)=(volume 2)(molarity 2), we can demonstrate the effects of doubling volume.
Suppose the starting volume is 1 L and the starting molarity is 1 M, and doubling the volume would make the final volume 2 L.
Plugging these numbers into the equation, we can figure out the final molarity.
(1 L)(1 M)=(2 L)(X M)
X M= (1 L x 1 M)/(2 L)
X M= 1/2 M
This shows that the molarity is halved when the volume of solvent is doubled.
