Answer:
-608KJ/mol
Explanation:
3 C2H2(g) -> C6H6(g)
ΔHrxn = ΔHproduct - ΔHreactant
ΔHrxn= ΔHC6H6 - 3ΔHC2H2
ΔHrxn = 83 - 3(230)
ΔHrxn = -608
Answer:
[NH₃] = 14.7 mol/L
Explanation:
28 wt% is a type of concentration that indicates that 28 g of ammonia is contained in 100 g of solution.
Let's determine the amount of ammonia:
28 g . 1 mol / 17.03g = 1.64 moles of NH₃
You need to consider that, when you have density's data it is always referred to solution:
Mass of solution is 100 g, let's find out the volume
0.90 g/mL = 100 g /V
V = 100 g / 0.90mL/g → 111.1 mL
We convert the volume to L → 111.1 mL . 1 L/1000mL = 0.1111 L
mol/L = 1.64 mol/0.1111L → 14.7 M
mol/L = M → molarity a sort of concentration that indicates the moles of solute in 1L of solution
Answer:
What was the acceleration of the cart with Low fan speed? ⇒ 18.0 cm/s2
What was the acceleration of the cart with Medium fan speed? ⇒ 24.0 cm/s2
What was the acceleration of the cart with High fan speed? 32.0cm/s2
Explanation:
Answer:
We need 0.095 moles of ethanol
Explanation:
Step 1: Data given
Number of moles water = 0.095 moles
Step 2: The balanced equation
CH3CH2OH + O2 ⇒ H2O + CH3COOH
Step 3: Calculate moles of ethanol
For 1 mol ethanol we need 1 mol oxygen to produce 1 mol water and 1 mol acetic acid
For 0.095 moles water, we need 0.095 moles ethanol and 0.095 moles oxygen
We need 0.095 moles of ethanol
