Answer:
<em>When molecular hydrogen (H2) and oxygen (O2) are combined and allowed to react together, energy is released and the molecules of hydrogen and oxygen can combine to form either water or hydrogen peroxide.</em>
3.5 M has 3.5 moles per litre
so we have one litre, so we need 3.5 moles
moles = mass/molarmass
3.5 * 23 = 80.5
The molarity of (HNO₃) that was used if 2.00 L must be used to prepare 4.5 L of a 0.25M HNO₃ solution is 0.563 M
<u><em>calculation</em></u>
This is calculated usind M₁V₁=M₂V₂ formula
where,
M₁( molarity ₁) = ?
V₁( volume ₁) = 2.00 L
M₁ (molarity ₂) = 0.25M
V₂( volume₂) = 4.5 L
make M₁ the subject of the formula by diving both side of the formula by V₁
M₁ is therefore = M₂V₂/V₁
M₁ =[ (0.25 M x 4.5 L) / 2.00 L ] =0.563 M
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.
Answer:
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Explanation:
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