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OLga [1]
3 years ago
9

What is the atomic number of a sodium atom that has 11 protons and 12 neutrons?

Chemistry
1 answer:
galben [10]3 years ago
3 0
Answer should be 23.
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Each degree on the Kelvin scale equals:
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A change of 1 Kelvin is exactly the same as a change of 1 degree Celsius.
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What is the formal charge on the hydrogen atom in hf?
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The formal charge of H = +1, because F in compounds can have only -1.

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Heyy guys, so basically i need help with stoichiometric calculation I will give you 100 points just to answer all of these answe
jeka94

Answer:

3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

5. The mass of silver nitrate required is approximately 14.53 grams

6. The mass of copper oxide that would be needed is approximately 31.86 grams

7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams

b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams

c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams

Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g

The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g

4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles

The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes

5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles

The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams

6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles

The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams

7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles

2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams

b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams

3 0
2 years ago
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A 0.1 mm sample of human blood has approximately 6000 red blood cells. An adult typically has 5.0 L of blood. How many red blood
olga55 [171]

 3.0 × 10¹¹ RBC's    (or)      3E11 RBC's


Solution:

Step 1: Convert mm³ into L;


As,


                                            1 mm³  =  1.0 × 10⁻⁶ Liters


So,


                                         0.1 mm³  =  X  Liters


Solving for X,


                       X  =  (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³


                       X  =  1.0 × 10⁻⁷ Liters


Step 2: Calculate No. of RBC's in 5 Liter Blood:


As given


                        1.0 × 10⁻⁷ Liters Blood contains  =  6000 RBC's


So,


                         5.0 Liters of Blood will contain  =  X  RBC's


Solving for X,


                      X  =  (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters


                      X  =  3.0 × 10¹¹ RBC's


Or,


                     X  =  3E11 RBC's



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NH3-The limiting reactant is the reactant that get completely used up in a reaction
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