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Rzqust [24]
3 years ago
7

What is the mass of heavy water, D2O(l), produced when 7.60 g of O2(g) reacts with excess D2(g)?

Chemistry
1 answer:
Rina8888 [55]3 years ago
3 0
<h3>Answer:</h3>

19.026 g

<h3>Explanation:</h3>

The reaction between D₂ and O₂ is given by the equation;

2D₂(g) + O₂(g) → 2H₂O(l)

We are given;

Mass of oxygen (O₂) that reacted as 7.60 g

We are required to calculate the mass of D₂O produced;

<h3>Step 1: Calculate the moles of O₂ used (limiting reactant)</h3>

To calculate the number of moles we divide mass by the molar mass.

Moles = mass ÷ Molar mass

Molar mass of O₂ is 16.0 g/mol

Therefore;

Moles = 7.60 g ÷ 16.0 g/mol

         = 0.475 moles

<h3>Step 2: Calculate the moles of D₂O produced </h3>

From the equation, 1 mole of oxygen reacts to produce 2 moles of D₂O

Thus, the mole ratio of O₂ : D₂O is 1 : 2

Therefore, moles of D₂O = Moles of O₂ × 2

                                          = 0.475 moles × 2

                                          = 0.95 moles

<h3>Step 3: Calculate the mass of heavy water produced</h3>

Mass = Number of moles × Molar mass

Molar mass of heavy water = 20.0276 g/mol

Therefore;

Mass of heavy water = 0.95 moles × 20.0276 g/mol

                                   = 19.026 g

Hence, the mass of heavy water produced is 19.026 g

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Potassium nitrate has a lattice energy of -163.8 kcal/mol and a heat of hydration of -155.5 kcal/mol.
Brut [27]

Answer:

<em>293.99 g </em>

OR

<em>0.293 Kg</em>

Explanation:

Given data:

Lattice energy of Potassium nitrate (KNO3) = -163.8 kcal/mol

Heat of hydration of KNO3 = -155.5 kcal/mol

Heat to absorb by KNO3 = 101kJ

To find:

Mass of KNO3 to dissolve in water = ?

Solution:

Heat of solution = Hydration energy - Lattice energy

                           = -155.5 -(-163.8)

                           = 8.3 kcal/mol

We already know,

1 kcal/mol = 4.184 kJ/mole

Therefore,

= 4.184 kJ/mol x 8.3 kcal/mol

= 34.73 kJ/mol

Now, 34.73 kJ of heat is absorbed when 1 mole of KNO3 is dissolved in water.

For 101 kJ of heat would be

= 101/34.73

= 2.908 moles of KNO3

Molar mass of KNO3 = 101.1 g/mole

Mass of KNO3 = Molar mass x moles

                         = 101.1 g/mole  x  2.908

                         = 293.99 g

                         = 0.293 kg

<em><u>293.99 g potassium nitrate has to dissolve in water to absorb 101 kJ of heat. </u></em>     

4 0
3 years ago
What is the usual charge on an ion formed from an atom that originally had seven valence electrons?
motikmotik
Negative because it'll have to gain an electron
4 0
3 years ago
Read 2 more answers
What is the mass of 0.100 mole of neon? (Watch sf’s)
dimaraw [331]

Answer:

The answer to your question is letter D. 2.02 g

Explanation:

Data

moles of Ne = 0.100

atomic mass of Neon = 20.18 g

Process

1.- Use proportions to find the answer

                   20.18 g of Ne ------------------  1 mol of Ne

                        x                 ------------------  0.1 moles

                        x = (0.1 x 20.18)/1

                        x = 2.018

2.- Consider the significant figures

      0.100 has three significant figures so the answer must be  2.02 g

4 0
3 years ago
Read 2 more answers
At 25 ∘C the reaction CaCrO4(s)←→Ca2+(aq)+CrO2−4(aq) has an equilibrium constant Kc=7.1×10−4. What is the equilibrium concentrat
Nitella [24]

Answer:

2.67 × 10⁻²

Explanation:

Equation for the reaction is expressed as:

CaCrO₄(s)    ⇄      Ca₂⁺(aq)         +        CrO₂⁻⁴(aq)

Given that:

Kc=7.1×10⁻⁴

Kc= [Ca^{2+}][CrO^{2-}_4]

Kc= [x][x]

Kc= [x²]

7.1×10⁻⁴ =  [x²]

x = \sqrt{7.1*10^{-4}}

x = 0.0267

x = 2.67*10^{-2}

6 0
3 years ago
So I saw this question: If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + Na
Nataly_w [17]

mass of PbI₂ = 27.6606 g

<h3>Further explanation</h3>

Given

Pb(NO₃)₂ + NaI → PbI₂ + NaNO₃

28.0 grams of Pb(NO₃)₂ react with 18.0 grams of NaI

Required

mass of PbI₂

Solution

Balanced equation

Pb(NO₃)₂ + 2NaI → PbI₂ + 2NaNO₃

The principle of a balanced reaction is the number of atoms in the reactants = the number of atoms in the product

mol Pb(NO₃)₂ :

= 28 : 331,2 g/mol

= 0.0845

mol NaI :

= 18 : 149,89 g/mol

= 0.12

Limiting reactant : mol : coefficient

Pb(NO₃)₂ : 0.0845 : 1 = 0.0845

NaI : 0.12 : 2 = 0.06

NaI limiting reactant (smaller ratio)

mol PbI₂ based on NaI

= 1/2 x 0.12 = 0.06

Mass PbI₂ :

= 0.06 x 461,01 g/mol

= 27.6606 g

4 0
3 years ago
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