Any advice for a GCSE chemistry student sitting first chemistry paper?

Chemistry

1 answer:

julia-pushkina [17]3 years ago

8 0

Answer:

when you go to gtake exam.be quite and don't make noise..

Explanation:

don't be scared

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Which of the following types of models is generally used to predict long-term events? a. idea model b. physical model c. compute

NikAS [45]

Answer:

Computer model is generally used to predict long-term events

I hope it helps.

7 0

3 years ago

How many molecules of nitrogen dioxide are in 1.28 g of nitrogen dioxide?

Kryger [21]

In order to determine the number of molecules of nitrogen dioxide that are in 1.28 g of nitrogen dioxide, we first need to know these values:

Molar mass of nitrogen dioxide = 46.0055 g/mol
Avogadro's number = 6.022 x 10^23 molecules/mol

Then, we divide 1.28 g by the molar mass and multiply the resulting value to Avogadro's number:

1.28 g/ 46.0055 g/mol x 6.022 x 10^23 molecules/mol = 1.6755 x 10^22 molecules of nitrogen dioxide

5 0

3 years ago

Read 2 more answers

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?

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