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AleksAgata [21]
3 years ago
9

Calculate the mass of h2o and c6 h12 o6 required to make 250 gram of 40% solution of glucose

Chemistry
1 answer:
Sedaia [141]3 years ago
5 0
40% solution of glucose is where the solution contains, by weight, 40% glucose and 60% water. 

Therefore, if the total weight of the solution is 250 g, 
mass of the glucose (C6H12O6) = 250 g * 40% = 100 g
mass of water (H2O) = 250 g * 60% = 150 g

Mass of water can also be calculated by subtracting the weight of glucose from the total weight of the solution:
mass of water = 250g-100g = 150g.
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How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with
konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = a=31 cm^2

Thickness of the gold plating  = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3

Density of the gold = d=19.3 g/cm^3

m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g

Moles of gold = \frac{29.915 g}{197 g/mol}=0.152 mol

Au^{3+}+3e^-\rightarrow Au

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

\frac{3}{1}\times 0.152 mol=0.456 mol of electrons

Number of electrons = N =0.456\times \times 6.022\times 10^{23}

Charge on single electron = q=1.6\times 10^{-19} C

Total charge required = Q

Q=N\times q

Amount of current passes = I = 8 Ampere

Duration of time  = T

I=\frac{Q}{T}

T=\frac{N\times q}{I}

=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0
3 years ago
State whether the following statement is true.justify the answer with examples
Ilya [14]

Answer:

False

Explanation:

Atoms only achieve complete outer electron shells if they contain an outer shell with 7 electrons before gaining another electron or an outer shell with 1 electron before losing an electron. This is assuming that the octet-rule can be applied to said atom. In addition, the number of valence electrons varies from atom to atom which is why not ALL atoms achieve complete outer electron shells after gaining or losing just ONE electron.

7 0
4 years ago
What type of central-atom orbital hybridization corresponds to each electron-group arrangement:
galina1969 [7]

Tetrahedral arrangement is resulted upon mixing one s and three p atomic orbitals, resulting in 4 hybridized sp^3 orbitals → sp^3 hybridization.

<h3>What is orbital hybridization?</h3>

In the context of valence bond theory, orbital hybridization (or hybridisation) refers to the idea of combining atomic orbitals to create new hybrid orbitals (with energies, forms, etc., distinct from the component atomic orbitals) suited for the pairing of electrons to form chemical bonds.

For instance, the valence-shell s orbital joins with three valence-shell p orbitals to generate four equivalent sp3 mixes that are arranged in a tetrahedral configuration around the carbon atom to connect to four distinct atoms.

Hybrid orbitals are symmetrically arranged in space and are helpful in the explanation of molecular geometry and atomic bonding characteristics. Usually, atomic orbitals with similar energies are combined to form hybrid orbitals.

Learn more about Hybridization

brainly.com/question/22765530

#SPJ4

3 0
2 years ago
If 10.00 g of iron metal is burned in the presence of excess of O2 how many grams of Fe2O3 will form
sergiy2304 [10]

14.292 grams of Fe2O3 is formed when 10 gram of iron metal is burned.

Explanation:

The balanced equation for the reaction is to be known so that number of moles taking part can be known.

The balanced chemical equation is

4Fe + 3O_{2}⇒ 2 Fe{2}O{3}

From the given weight of iron to be used for the production of Fe{2}O{3}, number of moles of Fe taking part in the reaction can be known by the formula:

Number of moles= mass ÷ Atomic mass of one mole of the element.

(Atomic weight of Fe is 55.845 gm/mole)

  Putting the values in equation  

Number of moles =  10 gm  ÷ 55.845 gm/mole

                               =  0.179 moles

Applying the stoichiometry concept

4 moles of Fe gives 2 Moles of Fe2O3

0.179 moles will produce x moles of Fe2O3

 So,  2÷ 4 = x ÷ 0.179

     2/4 = x/ 0.179

    2 × 0.179 = 4x

     2 × 0.179 / 4 = x

  x = 0.0895 moles

So from 10 grams of iron metal 0.0895 moles of Fe2O3 is formed.

Now the formula used above will give the weight of Fe2O3

weight = atomic weight × number of moles

            =  159.69 grams ×  0.0895

             = 14.292 grams of Fe2O3 formed.

4 0
3 years ago
How do you get the nitrogen you need ?
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We can't use the nitrogen present in the atmosphere although 80% of the atmosphere consists of nitrogen only.We get this from the plants called as leguminous plants which can convert or fix atmospheric nitrogen into usable ones..
8 0
3 years ago
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