Molarity is moles per liter. Therefore you must covert grams->moles by dividing 9.82 by the molar mass of lead IV nitrate. Then convert mL->L. Once you have those two numbers you can plug them into the formula Molarity = mol/L.
The molarity of the HCl is 1 M when 12.0 of .500 M NaOH neutralized 6.0 ml of HCl solution.
Explanation:
Data given:
molarity of the base NaOH, Mbase =0. 5 M
volume of the base NaOH, Vbase = 12 ml
volume of the acid, Vacid = 6 ml
molarity of the acid, Macid = ?
The titration formula for acid and base is given as:
Mbase Vbase = Macid Vacid
Macid =
Macid = 1 M
we can see that 1 M solution of HCl was used to neutralize the basic solution of NaOH. The volume of NaOH is 12 ml and volume of HCl used is 6ml.
The amount of energy in kilocalories released from 49 g of glucose given the data is -4.4 Kcal
How to determine the mole of glucose
Mass of glucose = 49 g
Molar mass of glucose = 180.2 g/mol
Mole of glucose = ?
Mole = mass / molar mass
Mole of glucose = 49 / 180.2
Mole of glucose = 0.272 mole
How to determine the energy released
C₆H₁₂O₆ →2C₂H₆O + 2CO₂ ΔH = -16 kcal/mol
From the balanced equation above,
1 mole of glucose released -16 kcal of energy
Therefore,
0.272 mole of glucose will release = 0.272 × -16 = -4.4 Kcal
Thus, -4.4 Kcal were released from the reaction
Learn more about stoichiometry:
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The answer is [OH⁻] = 1 x 10⁻⁴.
[OH⁻] = H₂O ÷ [H₃O⁺]
[OH⁻] = 1 x 10⁻¹⁴ / 1 x 10⁻¹⁰
[OH⁻] = 10⁻⁴
Answer:
<h3>The answer is 2.0 g/cm³</h3>
Explanation:
The density of a substance can be found by using the formula

From the question
mass = 4 g
volume = 2 cm³
We have

We have the final answer as
<h3>2.0 g/cm³</h3>
Hope this helps you