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VashaNatasha [74]
3 years ago
5

0.68 moles of calcium to calcium atoms​

Chemistry
1 answer:
Oksanka [162]3 years ago
3 0

0.68 \: moles \: calcium \times  \frac{6.02 \times  {10}^{23} \: calcium \: atoms }{1 \: mole \: calcium} \\

The (( mole calcium ))s are simplified so the answer is based on the number of calcium atoms.

0.68 \times 6.02 \times  {10}^{23} = 68 \times  {10}^{ - 2} \times 602 \times  {10}^{ - 2} \times  {10}^{23} \\

= 68 \times 602 \times  {10}^{ - 4} \times  {10}^{23} = 40936 \times  {10}^{19} \\

calcium \: atoms = 40936 \times  {10}^{19}  \\

_________________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

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Complete the equation for the conversion of sucrose into glucose <br> (1)C12H22O11 + H2O
Svetlanka [38]

Answer:

C₁₂H₂₂O₁₁ + H₂O     →    C₅H₁₂O₆ + C₆H₁₂O₆

Explanation:

Chemical equation:

C₁₂H₂₂O₁₁ + H₂O     →    C₅H₁₂O₆ + C₆H₁₂O₆

Source of sucrose:

Sucrose is present in roots of plants and also in fruits. It is storage form of energy. Some insects and bacteria use sucrose as main food. Best example is honeybee which collect sucrose and convert it into honey.

Monomers of sucrose and hydrolysis:

Sucrose consist of monomers glucose and fructose which are join together through glycosidic bond. Hydrolysis break the sucrose molecule into glucose and fructose. In hydrolysis glycosidic bond is break which convert the sucrose into glucose and fructose. Hydrolysis is slow process but this reaction is catalyze by enzyme. The enzyme invertase catalyze this reaction.

The given reaction also completely follow the law of conservation of mass. There are equal number of atoms of elements on both side of chemical equation thus mass remain conserved.

8 0
3 years ago
If a sample of magnesium with a mass of 35.0 grams reacts with 35.0 grams of oxygen how much magnesium oxide will be produced?​
slamgirl [31]

Answer:

58.0 g of MgO

Explanation:

in a perfect world, 70 g, however we don't live in a perfect world

The equation of reaction

2Mg + O₂ --> 2MgO

first find which element is limiting:

35 g x 1 mol/24.3 g of Mg x 2 mol of MgO/ 2 mole of Mg = 1.44 moles of MgO

35 g x 1 mol/32g of Mg x 2 mol of MgO/ 1 mole of O₂ = 2.1875 moles of MgO

This means Mg is the limiting factor, so you will be using this moles to find grams of MgO

1.44 mols of MgO x 40.3 g of MgO/ 1 mol = 58.0 g of MgO

8 0
3 years ago
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO
Irina-Kira [14]

Answer:

Amount of excess Carbon (ii) oxide left over = 23.75 g

Explanation:

Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂

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Molar mass of Carbon (ii) oxide = 28 g/mol

From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g)  of carbon (ii) oxide

450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide

Therefore the excess reactant is carbon (ii) oxide.

Amount of excess Carbon (ii) oxide left over = 260 - 236.25

Amount of excess Carbon (ii) oxide left over = 23.75 g

5 0
3 years ago
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