Which statement is in accordance with the steady state model of the universe?

Chemistry

1 answer:

Sloan [31]3 years ago

7 0

E. The nuclear fuel in the stars will eventually burn out, making the universe a cold and dark place

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What unit is mass generally measured in

Veronika [31]

Answer:

kilograms

Explanation:

hope this helps, pls mark brainliest :D

6 0

3 years ago

Read 2 more answers

3. What is the formula for the compound formed from the magnesium ion and nitrate?

bija089 [108]

Answer:

D.) Mg(NO₃)₂

Explanation:

Since magnesium (Mg) is a metal and nitrate (NO₃) is a polyatomic anion, they would combine to form an ionic compound. Magnesium would form the cation Mg²⁺ because it generally has 2 valence electrons. Nitrate always has a -1 charge.

In order for the overall compound to be neutral (have a charge of 0), there must be one Mg²⁺ and two NO₃⁻ ions in the compound (+2 + (-1) + (-1) = 0).

Therefore, the formula for the compound formed is Mg(NO₃)₂.

8 0

2 years ago

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?

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