<span>Answer:
Enthalpy is delta-H-
We need to look at the molecule and determine which bonds are broken adn which bonds are formed.
Bonds that are broken are H-H (from the H2 molecule) and the C=O from acetone.
their energies add up like this: 436 kJ + 745 kJ = 1181 kJ
looking at the bonds formed, these are C-O, O-H, and C-H. these add up to 1229 kJ
solving for delta H by taking the sum of the broken bonds and subtracting the sum of the formed bonds, like so:
1181 - 1229 = -48 kJ</span>
Answer:
2.9 grams.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:
no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.
no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.
<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.074 mol of Mg produce → 0.074 mol of MgO.
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>
Answer:
By atomic number?
Explanation:
fingers crossed its right :/
Answer:
d. 127 g/mol.
Explanation:
Hello!
In this case, since we have the amount of molecules of this this compound, we are able to compute the moles out there by using the Avogadro's number:
Which correspond to the moles of X2. Then, by using the mass we are able to compute the molar mass of X2:
It means that the atomic mass of X halves the molar mass of X2, which is then d. 127 g/mol.
Best regards!
Answer:
398 mL
Explanation:
Using the equation for molarity,
C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L
V₂ = V₁ + V' where V' = volume of water added.
So, From C₁V₁ = C₂V₂
V₁ = C₂V₂/C₁
= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L
= 0.875 mol/8.61 mol/L
= 0.102 L
So, V₂ = V₁ + V'
0.5 L = 0.102 L + V'
V' = 0.5 L - 0.102 L
= 0.398 L
= 398 mL
So, we need to add 398 mL of water to the nitric solution.
