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2 years ago

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

1 answer:

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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What is the standard potential, e∘celle∘cell, for this galvanic cell? use the given standard reduction potentials in your calcul

Olin [163]

The standard potential for the given galvanic cell is 0.477 V

<h3>What is electrode potential?</h3>

The electrode potential is the electromotive force of a galvanic cell built using a standard reference electrode and another electrode whose potential is to be found.

There are two types of electrode potential

Oxidation potential - The potential associated with oxidation reaction is known as oxidation potential

Reduction potential - The potential associated with reduction reaction is known as reduction potential

At the anode, oxidation occurs

$Sn(s)\rightarrow Sn^{2+}(aq)+2e^-$

At the cathode, reduction occurs

$Cu^{2+}(aq)+2e^-\rightarrow Cu(s)$

$E^o_{cell} =E^o_{cathode} -E^o_{anode}$

= 0.337 - (-0.140)

= 0.477 V

Thus, The standard potential for the given galvanic cell is 0.477 V

Learn more about electrode potential:

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Disclaimer: The question was given incomplete on the portal. Here is the complete question

Question: What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.

$Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$ , E°red=−0.140 V

$Cu^{2+}(aq)+2e^-\rightarrow Cu(s)$ , E°red=+0.337 V

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Caz(PO4)2 + H2504 - CaSO4+H3PO4

m_a_m_a [10]

Answer:

Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + 2H3PO4

The coefficients are 1, 3, 3, 2

Explanation:

Ca3(PO4)2 + H2SO4 —> CaSO4 + H3PO4

From the above equation,

There are 3 atoms of Ca on the left and 1 atom of Ca on the right. To balance Ca, put 3 in front of CaSO4 as shown below

Ca3(PO4)2 + H2SO4 —> 3CaSO4 + H3PO4

Now, we have 3 atoms of SO4 on the right and 1atom on the left. To balance SO4, put 3 in front of H2SO4 as shown below:

Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + H3PO4

Looking closely, there are 6 atoms of H on the left and 3 on the right. Therefore, it is balanced by by putting 2 in front of H3PO4 as shown below:

Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + 2H3PO4

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