More precisely, we need to specify its position<span> relative to a convenient reference frame. .... Also you s</span>hould know<span> that some people use the subscript "0" to refer to the ... mx, </span>start<span> subscript, 0, end subscript, equals, 1, </span>point<span>, 5, space, m and her </span>final<span> ... </span>between<span> two </span>points<span>, or we </span>can<span> talk about the distance traveled by an </span>object<span>.</span>
Answer:
3.089 L
Explanation:
From the given information, provided that the no of moles and the temperature remains constant;
= 15.6 psi
= ???
= 25.43 psi
= 1.895 L
Using Boyle's law:
Answer is: <span>the mass of the glucose is 81,07 grams.
</span>c(C₆H₁₂O₆) = 0,3 M = 0,3 mol/L.
V(C₆H₁₂O₆) = 1,500 L.
n(C₆H₁₂O₆) = c(C₆H₁₂O₆) · V(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 0,3 mol/L · 1,5 L.
n(C₆H₁₂O₆) = 0,45 mol.
m(C₆H₁₂O₆) = n(C₆H₁₂O₆) · M(C₆H₁₂O₆).
m(C₆H₁₂O₆) = 0,45 mol · 180,156 g/mol.
m(C₆H₁₂O₆) = 81,07 g.
First, we write the reaction for CH3OH combustion
CH3OH+3/2O2--->CO2+2H2O
for 1 mole of methanol, we get 1 mole of CO2, therefore for 5,25 moles of methanol we will get 5,25 moles of CO2
