Answer:
2Mg + O₂ ⟶ 2MgO
Explanation:
Step 1. Start with the most complicated-looking formula (O₂?).
Put a 1 in front of it.
Mg + 1O₂ ⟶ MgO
Step 2. Balance O.
We have fixed 2 O on the left. We need 2O on the right. Put a 2 in front of MgO.
Mg + 1O₂ ⟶ 2MgO
Step 3. Balance Mg.
We have fixed 2 Mg on the right-hand side. We need 2 Mg atoms on the left. Put a 2 in front of Mg.
2Mg + 1O₂ ⟶ 2MgO
Every formula now has a coefficient. The equation should be balanced. Let’s check.
<u>Atom</u> <u>On the left</u> <u>On the righ</u>t
Mg 2 2
O 2 2
All atoms are balanced.
The balanced equation is
2Mg + O₂ ⟶ 2MgO
Using Avogadros number, we can get that 1 mole of an atom
contain 6.022 x 10^23 atoms. Therefore we can use this conversion factor to get
the number of moles:
moles ZnCO3 = 6.11 x 10^22 atoms * (1 mole / 6.022 x 10^23
atoms) = 0.10146 moles
The molar mass of ZnCO3 is about 125.39 g/mol, therefore the
mass is:
mass ZnCO3 = 0.10146 moles * (125.39 g / mol)
<span>mass ZnCO3 = 12.72 g</span>
Answer:
150.0 mL.
Explanation:
- Volume percent is the ratio of the solute volume to the solution volume multiplied by 100.
∵ V% = (V of solute/V of solution) x 100.
<em>∴ V of solution = (V of solute/V%) x 100</em> = (75 mL/50) x 100 = <em>150.0 mL.</em>
Answer:
-476.95 Kj
Explanation:
N2H4(l) + N2O4(g) = 2N2O(g) + 2H20(g)
∆Hrxn = n∆Hf(products) - m∆Hf(reactants)
Where n and m = stoichiometric coefficients of the products and reactants respectively from the balanced chemical equation, ∆Hf = standard enthalpy of formation, ∆Hrxn= standard enthalpy of reaction.
Using the following standard enthalpies of formation ( you did not provide any ):
N2H4(l) = +50.63Kj/mol; N2O4(g) = +9.08Kj/mol; N2O(g) =+33.18Kj/mol; H2O(g) = -241.8Kj/mol
∆Hrxn = [ (2(∆Hf(N2O)) + (2(∆Hf(H2O))] – [(1(∆Hf(N2H4)) + (1(∆Hf(N2O4))]
∆Hrxn = [ 2(+33.18) + 2(-241.8)] – [ (+50.63) + (+9.08)]
∆Hrxn = [ (+66.36)+(-483.6)] – [ +50.63+9.08]
∆Hrxn = [ +66.36-483.6] – [+59.71]
∆Hrxn = -417.24-59.71
∆Hrxn = -476.95 Kj
NOTE: Remember to use the standard enthalpies of formation given to you by your instructor if they differ from the values used herein, and follow the same procedure.
Answer:
Just add a plenty of dots in the first one and very few dots in the second one
