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OLga [1]
3 years ago
10

What mass of silicon dioxide contains 9.72 X10e24 molecules of silicon dioxide?

Chemistry
1 answer:
dangina [55]3 years ago
6 0

Answer:

1033.36g

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 molecules. This means that 1 mole of SO2 also contains 6.02x10^23 molecules.

1 mole SO2 = 32 + (16x2) = 64g.

Now, If 64g of SO2 contains 6.02x10^23 molecules,

Therefore, Xg of SO2 will contain 9.72x10^24 i.e

Xg of SO2 = (64x9.72x10^24)/6.02x10^23

Xg of SO3 = 1033.36g.

Therefore, 1033.36g of SO2 contains 9.72x10^24 molecules

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In Fig C, which pair of electrons is the lone pair?
arlik [135]

Answer:

bottom right top left fig

5 0
2 years ago
Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:
nignag [31]

Answer:  1) Maximum mass of ammonia  198.57g  

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of  the excess Reactant, that means the H2 with 3,44g

Explanation:

  • In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:

N2(g) + 3H2(g) ⟶2NH3(g)

Both equal amount of atoms side to side.

  • Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the  molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)

Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶  38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3

  • As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element  that would be completey consumed, and the maximum mass of ammonia will be produced from it.
  • We proceed calculating the maximum mass of NH3 from the 163.3g of N2.

11.66  mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3

  • In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:

163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of  H2, thus, 38.77g - 35.33 g = 3.44g H2 is left

3 0
3 years ago
I need help with these questions, please!
lbvjy [14]

Answer:

1. Ionic bond

2. High melting point and high boiling point for ionic bonds while covalent bonds have low melting and boiling point.

3. The similarity is that ionic and covalent bonding lead to the creation of stable molecules.

4. 4Fe + 3O2 → 2Fe2O3

5. It uses the process of fission.

6. Fission involves the splitting of radioactive elements into smaller particles/compounds while Fusion involves combining of two or more atomic nuclei to form one or more different atomic nuclei and subatomic particles.

7. Nuclear power plants produce little to no greenhouse gas.

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Nuclear is less expensive.

6 0
4 years ago
2 points
erica [24]

9 grams of hydrogen gas (H2) will SC  Johnson need to react in order to make 1 bottle of Windex.

Explanation:

Balance equation for the formation of ammonia from H2 gas.

N2 + 3H2 ⇒ 2 NH_{3}

Given

mass of ammonia in 1 bottle of windex = 51 gram

atomic mass of ammonia 17.01 gram/mole

number of moles = \frac{mass}{atomic mass of 1 mole}

number of moles = \frac{51}{17.01}

                               = 3 moles of ammonia is formed.

in 1 bottle of windex there are 3 moles of ammonia 0r 51 grams of ammonia.

From the equation it can be found that:

3 moles of hydrogen reacted to form 2 moles of ammonia

so, x moles of hydrogen will react to form 3 moles of ammonia.

\frac{2}{3} = \frac{3}{x}

x = 4.5 moles of hydrogen will be required.

to convert moles into gram formula used:

mass = atomic mass x number of moles  (atomic mass of H2 is 2grams/mole)

        = 2 x 4.5

        = 9 grams of hydrogen.

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3 years ago
Compare the amount of heat required to vaporize a 200.-gram sample of H2O(ℓ) at its boiling point to the amount of heat required
stira [4]
The enthalpy of vaporization of H2O is higher than the enthalpy of fusion of H2O, therefore vaporizing the same mass of H2O would require more heat/energy than melting the same mass of H2O.
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