Answer:
d
. H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
Explanation:
a
. 2HgO → 2Hg + O₂,
It is a decomposition reaction, that HgO is decomposed to give Hg and O.
b
. Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl,
It is a double replacement reaction where two salts replaces their cations and anions with each others producing 2 new salts.
c
. Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag,
It is a single replacement reaction, that Zn replaces Ag from its salt.
d
. H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
It is acid-base reaction "neutralization reaction" where H₂SO₄ (acid) reacts with KOH (base) producing K₂SO₄ (salt) and water.
<em>d
. H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O</em>
<em></em>
Answer:
520 kg
Explanation:
Let's consider the combustion of isooctane.
C₈H₁₈(l) + 12.5 O₂(g) → 8 CO₂(g) + 9 H₂O(l)
We can establish the following relations:
- 1 mL of C₈H₁₈ has a mass of 0.690 g (ρ = 0.690 g/mL).
- The molar mass of C₈H₁₈ is 114.22 g/mol.
- The molar ratio of C₈H₁₈ to O₂ is 1:12.5.
- The mole fraction of O₂ in air is 0.21.
- The molar mass of air is 28.96 g/mol.
50 L of isooctane require the following mass of air.
Answer:
chemical change
Explanation:
burning and cooking are both chemical changes.
Answer:
Al4C3 + 12H2O = 3CH4 + 4Al(OH)3
Explanation:
Not sure if any explanation is needed but always start with the most complex compound. In this case it is Al(OH)3. You can see that there is 4 Aluminiums on the other side so I would start by putting a 4 next to the Al(OH)3. This now gives me 12 Hydrogens and 12 Oxygens on the right side. I put a 3 next to the CH4 to balance the Carbons on the left side. This leaves me with 12 Oxygens and 24 Hydrogens on the right side. This ends up being perfect because I can put a 12 next to the H2O.
Here's how to do it:
<span>Balanced equation first: </span>
<span>Mg + HCl = H2 + MgCl2 unbalanced </span>
<span>Mg + 2 HCl = H2 = MgCl balanced </span>
<span>Therefore 1 mole Mg reacts with 2 moles Hcl. </span>
<span>50g Mg = ? moles (a bit over 2; you work it out) </span>
<span>75 g HCl = ? moles (also a bit over 2; you work it out) </span>
<span>BUT, you need twice the moles HCl; therefore it is the Mg that is in excess. (you can now work out how many moles are in excess, and therefore how much mg is left over). </span>
<span>So, 2 moles HCl produce 1 mole H2(g) </span>
<span>therefore, the amount of H2 produced is half the number of moles of HCl </span>
<span>At STP, there are X litres per mole of gas (look it up - about 22 from memory) </span>
<span>Therefore, knowing the moles of H2, you can calculate the volume</span>
