Answer:
A. 0.143 M
B. 0.0523 M
Explanation:
A.
Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).
KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄
The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:
1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol
The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.
5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:
M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M
B.
Let's consider the neutralization of potassium hydroxide and perchloric acid.
KOH + HClO₄ → KClO₄ + H₂O
When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.
Q = m c T
c= 0.140 j/(g x °c)
m= 250.0g
T =52
hope you can solve it now
Answer:
atomic particles
The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged).
Answer:
There is 52.33 grams of water produced.
Explanation:
Step 1: Data given
Mass of propane burned = 32.00 grams
Molar mass of propane = 44.1 g/mol
Oxygen is in excess
Molar mass of water = 18.02 g/mol
Step 2: The balanced equation
C3H8 + 5O2 → 4H2O + 3CO2
Step 3: Calculate moles of propane
Moles of propane = mass propane / molar mass of propane
Moles of propane = 32.00 grams / 44.1 g/mol
Moles of propane = 0.726 moles
Step 4: Calculate moles of H2O
Propane is the limiting reactant.
For 1 mol of propane consumed, we need 5 moles of O2 to produce 4 moles of H2O and 3 moles of CO2
For 0.726 moles of propane we'll have 4*0.726 = 2.904 moles of H2O
Step 5: Calculate mass of H2O
Mass of H2O = moles of H2O * molar mass of H2O
Mass of H2O = 2.904 moles * 18.02 g/mol
Mass of H2O = 52.33 grams
There is 52.33 grams of water produced.
A model showing that gases are made from the matter of particles that are too small to see and are moving freely around in space can explain many observations.
