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Agata [3.3K]
3 years ago
14

The formula for lithium oxide is A) Li2O B) LiO2 C) LiO D) Li2O3

Chemistry
2 answers:
Greeley [361]3 years ago
4 0
Li2O is the formula for <span> lithium oxide</span>
AnnyKZ [126]3 years ago
4 0

you are correct it is A)

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Identify and describe 2 changes of state that release energy
soldier1979 [14.2K]

Answer: All changes of state involve the transfer or energy

Explanation: i got my information from this site

https://www.esrl.noaa.gov/gmd/education/info_activities/pdfs/CTA_the_water_cycle.pdf

4 0
3 years ago
Type the formula of the following compound: Aluminum nitride
evablogger [386]

The formula for that compound is AlN

6 0
3 years ago
A solution that contains a large amount of solute would be described as what
antiseptic1488 [7]

Answer:

A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute.

8 0
3 years ago
3. In an experiment it was found that 40.0cm of 0.2M sodium hydroxide solution just neutralized 0.2g
hjlf

The relative molecular mass of acid A : 50 g/mol

<h3>Further explanation</h3>

Given

40.0 cm³(40 ml) of 0.2M sodium hydroxide

0.2g  of a dibasic acid

Required

the relative molecular mass of acid A

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence(number of H⁺/OH⁻)

NaOH ⇒ n = 1

Dibasic acid =  diprotic acid (such as H₂SO₄)⇒ n = 2

mol = M x V

Input the value in the formula :(1 = NaOH, 2=dibasic acid)

0.2 x 40 x 1 = M₂ x V₂ x 2

M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A

The relative molecular mass of acid A (M) :

\tt M_A=\dfrac{mass }{mol}=\dfrac{0.2~g}{4.10^{-3}}=50~g/mol

5 0
3 years ago
A mixture of water and graphite is heated to 600 k in a 1 l container. when the system comes to equilibrium it contains 0.15 mol
Lunna [17]
(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???

when the reaction equation is:

C(s) + H2O(g)  ↔ H2(g) + CO(g) 

∴ Kc = [H2] [CO] / [H2O]

and we have Kc = 0.0393 (given missing in the question)

when the O2 is added so, the reaction will be:

2H2(g) + O2(g) → 2H2O(g)

that means that 0.15 mol H2 gives 0.15 mol of H2O

∴ by using ICE table:

            [H2O]          [H2]        [CO]

initial 0.57 + 0.15      0               0.15

change  -X                +X              +X

Equ   (0.72-X)             X            (0.15+X)

by substitution:

0.0393 = X (0.15+X) / (0.72-X)  by solving for X

∴ X = 0.098 

∴[H2] = X = 0.098 M

∴[CO] = 0.15 + X
 
           = 0.15 + 0.098 = 0.248 M

∴[H2O] = 0.72 - X

             = 0.72 - 0.098

             = 0.622 M


8 0
3 years ago
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