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dedylja [7]
3 years ago
12

4. How many atoms are in 32.6 g of Oxygen?

Chemistry
2 answers:
Natali5045456 [20]3 years ago
7 0

Hey there!

Oxygen has a molar mass of 16. That means 16g of oxygen is 1 mole.

32.6 ÷ 16 = 2.0375 moles

We have 2.0375 moles.

There are 6.022 x 10²³ atoms in one mole.

2.0375 x 6.022 x 10²³

1.3 x 10²⁴

There are 1.3 x 10²⁴ atoms in 32.6 grams of oxygen.

Hope this helps!

Hitman42 [59]3 years ago
4 0

Answer:

To determine the number of atoms in 32 grams of oxygen, one needs to know that there are 6.02x10^23 atoms per mole of any element. For elemental oxygen, the atomic mass is 16, so.. 6.02x10^23 atoms/mole x 1 mole/16 grams x 32 grams = 1.2x10^24 atoms.

Explanation:

Hope this helps! Have a good day!

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Answer:

2KCl + F₂   →     2KF + Cl₂

Explanation:

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According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

This law was given by French chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

2KCl + F₂   →     2KF + Cl₂

In this equation mass of reactant and product is equal. There are 2 potassium 2 chlorine and fluorine atoms on both side of equation it means mass remain conserved.

All other options are incorrect because mass is not conserved.

Mg₂ + LiBr ---> LiMg + Br

In this equation mass of magnesium is more on reactant side.

Na +O₂ ---> Na₂O

In this equation there is more oxygen and less sodium on reactant side while there is more sodium and less oxygen on product side.

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5 0
3 years ago
True or False: In every gram of sucrose (table sugar) there is 0.513 g of oxygen
Ber [7]

Answer:

True

Explanation:

Step 1: Find molecular formula of sucrose

C₁₂H₂₂O₁₁

Step 2: Convert moles of oxygen present to grams

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11 mol O = 176 g O

Step 3: Find molar mass of sucrose

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H - 1.01 g/mol

O - 16.00 g/mol

12.01(12) + 22(1.01) + 11(16.00) = 342.34 g/mol C₁₂H₂₂O₁₁

Step 4: Set up dimensional analysis

1 gC_{12}H_{22}O_{11}(\frac{176 gO}{342.34gC_{12}H_{22}O_{11}} )

Step 5: Multiply/Divide and cancel out units

Grams of C₁₂H₂₂O₁₁ and grams of C₁₂H₂₂O₁₁ cancel out.

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4 0
3 years ago
A sheet of polyethylene 1.5-mm thick is being used as an oxygen barrier at a temperature of 600K. If the flux is 2.48 x 10-5 kg/
Eddi Din [679]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The concentration of  high-pressure side is  C_1 =   0.722 \ kg/m^3

Explanation:

From the question we are told that

    The thickness of the polyethylene is  d  =  1.5 \ mm = 0.0015 \ m

     The  temperature is  T  =  600 \   K

      The flux is  JA  =  2.48 *10^{-5} \  kg/m^2\cdot s

      The concentration on the low-pressure side is  C_2 =  0.5 \ kg/m^3

       The initial diffusivity  is  D_o  =  6.2 *10^{-4} \ m^2 /s

       The activation energy for  diffusion is   Q_d  =  41 \ kJ /mol  =  41*10^3 J /mol

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        D   = D_o * e^{- \frac{Q_d}{R * T  } }  

substituting values  

         D   = 6.2*10^{-4} * e^{- \frac{41 *10^3}{8.314 * 600  } }  

          D   = 1.671 *10^{-7} \ m^2 /s  

Generally the flux is mathematically represented as

          JA  =  D  *  \frac{C_1 -C_2}{d}

Where C_1 is the concentration of oxygen at the higher side

       So  

             C_1 =    d  * \frac{JA}{D}  + C_2

substituting values  

             C_1 =    0.0015   * \frac{2.48*10^{-5}}{1.671*10^{-7}}  + 0.5

              C_1 =   0.722 \ kg/m^3

 

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