What other reactions is taking place?

You wish to prepare a tape-casting slip containing 50 vol% Al2O3 and 50 vol% polyvinyl butyral (PVB) binder. If the density of A

belka [17]

Answer: The mass of PVB required to produce 1000 grams of tape is 213.4 grams

Explanation:

To calculate the mass of aluminium oxide, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$ .......(1)

For $Al_2O_3$

We are given:

50% (v/v) of $Al_2O_3$

This means that 50 mL of aluminium oxide is present in 100 mL of tape

Calculating the mass of aluminium oxide by using equation 1:

Density of aluminium oxide = $3.98 g/cm^3$

Volume of aluminium oxide = $50mL=50cm^3$ (Conversion factor: $1mL=1cm^3$ )

Putting values in equation 1, we get:

$3.98g/cm^3=\frac{\text{Mass of aluminium oxide}}{50cm^3}\\\\\text{Mass of aluminium oxide}=(3.98g/cm^3\times 50cm^3)=199g$

Mass of aluminium oxide = 199 g

For PVB:

We are given:

50% (v/v) of PVB

This means that 50 mL of PVB is present in 100 mL of tape

Calculating the mass of PVB by using equation 1:

Density of PVB = $1.08 g/cm^3$

Volume of PVB = $50mL=50cm^3$

Putting values in equation 1, we get:

$1.08g/cm^3=\frac{\text{Mass of PVB}}{50cm^3}\\\\\text{Mass of PVB}=(1.08g/cm^3\times 50cm^3)=54g$

Mass of PVB = 54 g

Mass of tape = Mass of aluminium oxide + mass of PVB

Mass of tape = [199 + 54] g = 253 g

To calculate the mass of PVB required to produce 1000 g of tape, we use unitary method:

When 253 grams of tape is made, the mass of PVB required is 54 g

So, when 1000 grams of tape is made, the mass of PVB required will be = $\frac{54}{253}\times 1000=213.4g$

Hence, the mass of PVB required to produce 1000 grams of tape is 213.4 grams

4 0

3 years ago

There are two main items that cause the seasons what are they. There more than one answer

V125BC [204]

What causes the seasons?
The seasons are caused by the tilt of the Earth's rotational axis away or toward the sun as it travels through its year-long path around the sun.

8 0

3 years ago

Predict the initial and isolated products for the reaction. The starting material is a 6 carbon chain where there is a triple bo

satela [25.4K]

Answer:

See explanation and image attached

Explanation:

This reaction is known as mercuric ion catalyzed hydration of alkynes.

The first step in the reaction is attack of the mercuric ion on the carbon-carbon triple bond, a bridged intermediate is formed. This bridged intermediate is attacked by water molecule to give an organomercury enol. This undergoes keto-enol tautomerism, proton transfer to the keto group yields an oxonium ion, loss of the mercuric ion now gives equilibrium keto and enol forms of the compound. The keto form is favoured over the enol form.

7 0

3 years ago

The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r

Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈ 2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0

4 years ago

What's the difference between a mole of propane and a gram of propane?

kobusy [5.1K]

B. A gram would have a lot more molecules of propane than a mole

5 0

3 years ago