What must the melting point of the mold be compared to the temperature at which class get soft and why

1 answer:

7 0

The answer is that the mold melting point must be higher then molten glass, otherwise the mold would melt when molten glass is poured into it .

I hope this helps :)

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A gas has v= 4.0L, T =27 degree C and P=2.0 atm. What is the V if T=327 degree C and P=8.0 atm?

wariber [46]

As I am reading the problem,i can tell the question gives you two temperatures, two pressures, one volume and asking for the other. this should be an indication that you need to use the following gas formula

P1V1/T1= P2V2/T2

P1= 2.0 atm
V2= 4.0 L
T1= 27= 300 K
P2= 8.0 atm
V2= ?
T2= 327= 600 k

let's plug in the values into the formula

(2.0 x 4.0)/ 300= (8.0 x V2)/ 600 K

V2= 2.0 Liters

answer is D

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Consider the following reaction: 2 Bi(s) + 3 Cl2(g) → 2 BiCl3(s)

Mandarinka [93]

Taking into account the reaction stoichiometry, 1.119 grams of chlorine gas are required to produce 3.32 grams of bismuth chloride is produced.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Bi + 3 Cl₂ → 2 BiCl₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Bi: 2 moles
Cl₂: 3 moles
BiCl₃: 2 moles

The molar mass of the compounds is:

Bi: 209 g/mole
Cl₂: 70.90 g/mole
BiCl₃: 315.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Bi: 2 moles ×209 g/mole=418 g

Cl₂: 3 moles ×70.90 g/mole= 212.7 g

BiCl₃: 2 moles ×315.45 g/mole= 630.9 g

<h3>Mass of Cl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 630.9 grams of BiCl₃ are formed by 212.7 grams of Cl₂, 3.32 grams of BiCl₃ are formed by how much mass of Cl₂?

$mass of Cl_{2} =\frac{3.32 grams of BiCl_{3}x212.7 grams of Cl_{2} }{630.9 grams of BiCl_{3}}$