Hello!
First you need to calculate q
<span>delta U is change in internal energy </span>
<span>delta U = q + w </span>
<span>q is heat and w work done </span>
<span>here work was done by the system means energy leaving the system so w is negative </span>
<span>delta U = q + w </span>
<span>q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ </span>
<span>q = m x c x delta T </span>
<span>7211 J = 80.0 g x c x (225-25) °C </span>
<span>c = 0.451 J /g °C
</span>
Hope this Helps! Have A Wonderful Day! :)
The enthalpy of the creation of Methanol is negative. This means heat is released when the reaction proceeds. You would want to use a low temperature.
Answer:
h2+O ---> H2O
reactants: H2 & O
products: H2O
Explanation:
The simple reaction that produces a water molecule from H2 and O would be the one written above, even though there are 2 hydrogen molecules, they will form an H2 molecule rather than 2 individual H molecules (almost never seen) the reactants would be your hydrogen and oxygen molecules individually before they bond to form a molecule of water (H2O) which is the product
N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
mol = mass ÷ molar mass
If mass of hydrazine (N₂H₄) = 5.29 g
then mol of hydrazine = 5.29 g ÷ ((14 ×2) + (1 × 4))
= 0.165 mol
mole ratio of hydrazine to Nitogen is 1 : 1
∴ if moles of hydrazine = 0.165 mol
then moles of nitrogen = 0.165 mol
Mass = mol × molar mass
Since mol of nitrogen (N₂) = 0.165
then mass of hydrazine = 0.165 × (14 × 2)
= 4.62 g
the answer is c, more heat is being released.
