For Ar :
1 mol ------------ 22.4 L ( at STP )
7.6 mol ---------- x L
x = 7.6 * 22.4
x = 170.24 L
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For C2H3:
1 mol ------------ 22.4 ( at STP)
0.44 mol --------- y L
y = 0.44 * 22.4
y = 9.856 L
hope this helps !.
Answer:
Percentage error = 1.88 %
Solution:
Data Given:
Mass of Sample = 20.46 g
Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL
Formula Used:
Density = Mass / Volume
Putting values,
Density = 20.46 g / 3.0 mL
Density = 6.82 g.mL⁻¹
Percentage Error:
Experimental Value = 6.82 g.mL⁻¹
Accepted Value = 6.95 g.mL⁻¹
= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %
Percentage Error = 100 % - 98.12 %
Percentage error = 1.88 %
<span>Colligative properties are dependent upon the number of molecules or ions present in solution. Therefore, 1 mole of Na2SO4 will produce 3 moles of ions and so it will have 3 times as much of an effect as 1 mole of sugar, which is not an electrolyte and can't dissociate to an appreciable extent.</span>
It would be C) atoms of the opposite charges. iconic bonds are formed by the attraction of two atoms of the opposite charge.
Hope this helps :)
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