Theoritical discussion of calorimetry? ​

Answers

  • Réponse publiée par: girly61

    the answer is Mol. concentration/time

  • Réponse publiée par: jemuelpogi

    a. Atoms are can be created or destroyed in a chemical reaction

    It is false, atoms are indivisible and indestructible. It although today breakthroughs have proven it possible to destroy atoms but it is ony made possible by nuclear reactions and NOT by chemical reactions.

  • Réponse publiée par: kimashleybartolome

    Hello!

    An antacid tablet containing 0.50 g of NaHCO3 is dissolved in 250 mL of water. What is the molar concentration of NaHCO3 in the solution ?

    We have the following data:

    M (Molar Concentration or Molarity) = ? (in mol/L)

    m (mass) = 0.50 g

    V (volume) = 250 mL → 0.25 L

    MM (Molar Mass of NaHCO3)

    Na = 1*(23u) = 23 u

    H = 1*(1u) = 1 u

    C = 1*(12u) = 12 u

    O = 3*(16u) = 48 u

    -------------------------------

    MM (Molar Mass of NaHCO3) = 23 + 1 + 12 + 48 = 84 g/mol

    We apply the data found to the formula of molar Concentration or Molarity, let's see:

    M = \dfrac{m}{MM*V}

    M = \dfrac{0.50}{84*0.25}

    M = \dfrac{0.50}{21}

    M = 0.023809... \to \boxed{\boxed{M \approx 0.024\:mol/L}}\:\:\:\:\:\:\bf\purple{\checkmark}

    The Molar Concentration is approximately 0.024 mol/L

    ________________________

    \bf\purple{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

  • Réponse publiée par: cbohol56

    Hello!

    How many moles of CdBr2 are in a 23.75 gram sample of CdBr2 ?

    We have the following data:

    Cd = 1*(112.4 u) = 112.4 u

    Br = 2*(79.9 u) = 159.8 u

    ------------------------------------

    Molar Mass of CdBr2 = 112.4 + 159.8 = 272.2 g/mol

    Now, let's make the rule three, grams relative to the mole (g/mol), let's see:

    272.2 g ---------------- 1 mol

    23.75 g -------------  y mol

    \dfrac{272.2}{23.75} = \dfrac{1}{y}

    Product of extremes equals product of means

    272.2*y = 23.75*1

    272.2\:y = 23.75

    y = \dfrac{23.75}{272.2}

    \boxed{\boxed{y \approx 0.087\:mol}}\end{array}}\qquad\checkmark

    ____________________________________

    I Hope this helps, greetings ... Dexteright02! =)

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Theoritical discussion of calorimetry? ​...