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2 years ago

Molten sodium chloride is broken down in an electrolytic cell into the elements sodium and chlorine. Where do the

1 answer:

5 0

Answer:Chlorine gas bubbles out of the melt above the anode. Sodium ions migrate to the cathode, where electrons enter the melt and are reduced to sodium metal

Explanation:

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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

2 years ago

How many grams of carbon should be burned in an excess of oxygen at stp to obtain 2. 21 l of carbon dioxide?.

snow_tiger [21]

You must burn 1.17 g C to obtain 2.21 L CO2 at
STP.
The balanced chemical equation is
C+02+ CO2.
Step 1. Convert litres of CO, to moles of CO2.
STP is 0 °C and 1 bar. At STP the volume of 1 mol
of an ideal gas is 22.71 L.
Moles of CO2= 2.21 L CO2 × (1 mol CO2/22.71 L
CO2) = 0.097 31 mol CO2
Step 2. Use the molar ratio of C:CO2 to convert
moles of CO to moles of C
Moles of C= 0.097 31mol CO2 × (1 mol C/1 mol
CO2) = 0.097 31mol C
Step 3. Use the molar mass of C to calculate the
mass of C
Mass of C= 0.097 31mol C × (12.01 g C/1 mol C) =
1.17 g C
It looks as if you are using the old (pre-1982)
definition of STP. That definition gives a value of
1.18 g C.

4 0

2 years ago

How many molecules are in 2.570 moles of H2

docker41 [41]

A lot of molecules will be in 2.570 moles of H2

5 0

3 years ago

Read 2 more answers

During science, Jim's teacher gives this simple rule to use as a first attempt to increase solubility of the solute and the rate

gladu [14]

B. Agitate the solute in the solvent.

3 0