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Most wine is prepared by the fermentation of the glucose in grape juice by yeast: C6H12O6(aq) --> 2C2H5OH(aq) + 2CO2(g) How many grams of glucose should there be in grape juice to produce 725 mLs of wine that is 11.0% ethyl alcohol, C2H5OH (d=0.789 g/cm3), by volume?

Answers

Answer 1

Answer:

Answer:

123.41 g

Explanation:

Given that the ethyl alcohol produced is 11.0 % by volume.

It means that 1000 mL contains 110 mL of ethyl alcohol

Given that the volume is:- 725 mL

So,

Volume of ethyl alcohol = $(110)/(1000)* 725\ mL$ = 79.75 mL

Given that:- Density = 0.789 g/cm³ = 0.789 g/mL

So, Mass = Density*Volume = $0.789* 79.75\ g$ = 62.92 g

Calculation of the moles of ethyl alcohol as:-

Molar mass of ethyl alcohol = 46.07 g/mol

The formula for the calculation of moles is shown below:

$moles = (Mass\ taken)/(Molar\ mass)$

Thus,

$Moles= (62.92\ g)/(46.07\ g/mol)$

$Moles=1.37\ mol$

According to the reaction:-

$C_6H_(12)O_6_((aq))\rightarrow 2C_2H_5OH_((aq)) +2CO_2_((g))$

2 moles of ethyl alcohol is produced when 1 mole of glucose reacts

Also,

1.37 moles of ethyl alcohol is produced when $(1)/(2)* 1.37$ mole of glucose reacts

Moles of glucose = 0.685 Moles

Molar mass of glucose = 180.156 g/mol

Mass = Moles*Molar mass = $0.685* 180.156\ g$ = 123.41 g

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How many electrons are in Fe3+ ?

Answers

23 electrons hope this helps

Answer:

There are 23 electrons in Fe3+

Measurements show that unknown compound has the following composition: element mass 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. Write the empirical chemical formula of this compound?(A) Ca2PO4
(B) Ca3PO6
(C) Ca4P2O4
(D) Ca3P2O8 (or Ca3(PO4)2)
(E) CaPO4

Answers

Answer:

D) empirical formula is: C₃P₂O₈

Explanation:

Given:

Mass % Calcium (Ca) = 38.7%

Mass % Phosphorus (P) = 19.9%

Mass % oxygen (O) = 41.2 %

This implies that for a 100 g sample of the unknown compound:

Mass Ca = 38.7 g

Mass P = 19.9 g

Mass O = 41.2 g

Step 1: Calculate the moles of Ca, P, O

Atomic mass Ca = 40.08 g/mol

Atomic mass P = 30.97 g/mol

Atomic mass O = 16.00 g/mol

$Moles\ Ca = (38.7g)/(40.08g/mol) =0.966\ mol\n\nMoles\ P = (19.9g)/(30.97g/mol) =0.643\ mol\n\nMoles\ O = (41.2g)/(16.00g/mol) =2.58\ mol$

Step 2: Calculate the molar ratio

$C = (0.966)/(0.643) =1.50\n\nP = (0.643)/(0.643) = 1.00\n\nO = (2.58)/(0.643) =4.00$

Step 3: Calculate the closest whole number ratio

C: P: O = 1.50 : 1.00 : 4.00

C : P : O = 3:2:8

Therefore, the empirical formula is: C₃P₂O₈

Final answer:

The mass percentage composition of a compound can be used to determine its empirical formula. For a compound with 38.7% calcium (Ca), 19.9% phosphorus (P), and 41.2% oxygen (O), the empirical formula is Ca3(PO4)2.

Explanation:

To solve this problem, we're going to use the atomic mass percentages to determine the empirical formula of the compound.

We do this by assuming we have a 100g sample of the compound. Therefore:

The mass of calcium (Ca) is 38.7g.

The mass of phosphorus (P) is 19.9g.

The mass of oxygen (O) is 41.2g.

Next, we calculate how many moles we have of each element:

Ca: 38.7g / 40.08g/mol (the atomic mass of calcium) = 0.965 moles
P: 19.9g / 30.97g/mol (the atomic mass of phosphorous) = 0.643 moles
O: 41.2g / 16.00g/mol (the atomic mass of oxygen) = 2.575 moles

Then, we divide each of these numbers by the smallest number of moles, which is 0.643 (P):

Ca: 0.965/0.643 = 1.5 (~1)
P: 0.643/0.643 = 1
O: 2.575/0.643 = 4

Learn more about Empirical Formula here:

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A construction worker is transporting pieces of plastic in two stacks with touching pieces. In Stack 1, both pieces of plastic are the same size. In Stack 2, the bottom piece is the same size as those in Stack 1, and the top piece is larger. The diagram above shows the energy of the molecules in the pieces of plastic before they touched.After a while, which of the two top plastic pieces will be cooler, and why?

Answers

Answer:

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Answers

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64.92 grams of Hg(NO3)2 to make 5.00 liters of solution.

Answers

Answer:

$\large\boxed{\large\boxed{0.0400mol/liter}}$

Explanation:

Knowing that you have 64.92 grams of Hg(NO₃)₂ to make 5.00 liters of solution, you can calcualte the molarity of the solution.

Molarity is a measure of concentration, defined as the number of moles of solute per liter of soluiton. Mathematically:

$Molarity=\frac{\text{number of moles of solute}}{\text{liters of solution}}$

Then, first you must calculate the number of moles of solute. The formula is:

$\text{number of moles}=\frac{\text{mass in grams}}{\text{molar mass}}$

You can either calculate the molar mass of the compound using the chemical formula or search it in the internet.

The molar mass of Hg(NO₃)₂ is found to be 324.7 g/mol.

Now you have everything to calculate the molarity of the solution:

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Calculate the molality of a solution containing 22.75 g of glycerol (C3H8O3) in 79.6 g of ethanol (C2H5OH).

Answers

Answer:

3.11 mol/kg

Explanation:

Molality M = number of moles of solute, n/mass of solvent, m

To calculate the number of moles of glycerol (C₃H₈O₃) in 22.75 g of glycerol, we find its molar (molecular) mass, M',

So, M' = 3 × atomic mass of carbon + 8 × atomic mass hydrogen + 3 × atomic mass of oxygen

= 3 × 12 g/mol + 8 × 1 g/mol + 3 × 16 g/mol = 36 g/mol + 8 g/mol + 48 g/mol = 92 g/mol.

So, number of moles of glycerol, n = m'/M' where m' = mass of glycerol = 22.75 g and M' = molecular mass of glycerol = 92 g/mol

So, n = m'/M'

n = 22.75 g/92 g/mol

n = 0.247 mol

So, the molality of the solution M = n/m

Since m = mass of ethanol = 79.6 g = 0.0796 kg, substituting the value of n into the equation, we have

M = 0.247 mol/0.0796 kg

M = 3.11 mol/kg

So, the molality of the solution is 3.11 mol/kg.

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