All categories

3 years ago

Give a theoretical reason why lithium and oxygen combine in the ratio 2:1

Chemistry

1 answer:

salantis [7]3 years ago

7 0

If a substance is made of pure element, Chances are they're gonna react with one another.

You might be interested in

How many carbon atoms are present in 0.46 moles of C2H6O?

solmaris [256]

Here, Number of atoms = 0.46 * 6.23 * 10^23 [ Avogadro's number ]
= 2.77 * 10^23

In short, Your Answer would be Option B

Hope this helps!

4 0

4 years ago

Are cell phones scientific

Rasek [7]

Yes hope this helps you

6 0

4 years ago

Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction. half-reaction i

Natasha2012 [34]

Answer: Oxidation reaction: $Al(s)\rightarrow Al^{3+}(aq)+3e^-$

Reduction reaction: $Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$

Overall redox reaction : $2Al(s)+3Sn^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Sn(s)$

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

$Al(s)\rightarrow Al^{3+}(aq)+3e^-$

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

$Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$

Overall redox reaction : $2Al(s)+3Sn^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Sn(s)$

5 0

3 years ago

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

Sauron [17]

Answer:

The concentration of COF₂ at equilibrium is 0.296 M.

Explanation:

To solve this equilibrium problem we use an ICE Table. In this table, we recognize 3 stages: Initial(I), Change(C) and Equilibrium(E). In each row we record the <em>concentrations</em> or <em>changes in concentration</em> in that stage. For this reaction:

2 COF₂(g) ⇌ CO₂(g) + CF₄(g)

I 2.00 0 0

C -2x +x +x

E 2.00 - 2x x x

Then, we replace these equilibrium concentrations in the Kc expression, and solve for "x".

$Kc=8.30=\frac{[CO_{2}] \times [CF_{4}] }{[COF_{2}]^{2} } =\frac{x^{2} }{(2.00-2x)^{2} } \\8.30=(\frac{x}{2.00-2x} )^{2} \\\sqrt{8.30} =\frac{x}{2.00-2x}\\5.76-5.76x=x\\x=0.852$

The concentration of COF₂ at equilibrium is 2.00 -2x = 2.00 - 2 × 0.852 = 0.296 M

6 0

3 years ago

How much work is required to lift a 0.500 kg block 18.5 m?

kipiarov [429]

Answer:

Work = 90.65 j

Explanation:

Given data:

Mass = 0.500 Kg

Distance = 18.5 m

Work done = ?

Solution:

Work = force . distance

Force = mg

Work = mg.distance

Work = mgh

Work = 0.500 Kg × 9.8 m/s²× 18.5 m

Work = 90.65 Kg .m²/s²

Kg .m²/s² = j

Work = 90.65 j

3 0

4 years ago