Answer:
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O=
Answer: 9.04 g of H2O
Explanation:
First set up equation: C4H10 (g)+ O2(g) -> CO2(g) + H2O(g)
Next balance it: 2C4H10 (g)+ 13O2(g) -> 8CO2(g) + 10H2O (g)
Use equation to get moles and plug given
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O
<span>Niels Henrik David Bohr is the scientist credited with developing the orbital model of the atom. He was Danish physicist who made foundational contributions to understanding atomic structure and quantum theory. He received the Nobel Prize in Physics in 1922</span>
Answer:
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)
Explanation:
Which ONE of the following is an oxidation–reduction reaction?
A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.
B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.
C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.
D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. <u>C is reduced</u> and <u>H is oxidized</u>.
Answer:
64J of energy must have been released.
Explanation:
Step 1: Data given
One reactant contains 346 J of chemical energy, the other reactant contains 153 J of chemical energy.
The product contains 435 J of chemical energy.
Step 2:
Since the energy is conserved
Sum of energy of Reactants = Energy of Products
Sum of energy of Reactants = 346 J + 153 J = 499 J
The energy of the product = 435 J
435 < 499
This means energy must have been lost as heat.
Step 3: Calculate heat released
499 J - 435 J = 64 J
64J of energy must have been released.