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Ksenya-84 [330]
3 years ago
11

And ironic bonds what happens to electrons? No

Chemistry
1 answer:
iren [92.7K]3 years ago
5 0

Answer:

metals donate electrons to nonsmetals

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What is the percent copmosition of chromium in CrF3
zepelin [54]
47.707% is the answer
5 0
3 years ago
Calculate the concentration of clo2− at equilibrium if the initial concentration of hclo2 is 2. 25×10^−2 m.
jek_recluse [69]

The concentration of ClO₂⁻ at equilibrium if the initial concentration of  HClO₂ is 0.0654.

<h3>What is concentration?</h3>

The concentration of any substance is the quantity of that substance in per square of the space or container.

The reaction is

HClO₂ + H₂O <=> H₃O⁺ + ClO₂⁻

The pH is 0.454 M

Ka = [H₃O⁺][ClO₂⁻ ] / [HClO₂]

2. 25 × 10⁻² m = [x][x] / 0.454-x]

2 + 0.011 - 0.004994 = 0

solve the quadratic equation

x = 0.0654 = [H3O+] = [ClO2-]

pH = -log (H3O+)

pH = -log(0.0654)

pH = 1.2

equilibrium concentrations of

[HClO2] = 0.454 -x = 0.454 -0.0654 = 0.3886 M

[ClO2- ] = x = 0.0654

Thus, the equilibrium concentrations  is 0.0654.

To learn more about concentration, refer to the link:

brainly.com/question/16645766

#SPJ4

6 0
1 year ago
what is matter that can not be divided into two or more diffrent types of matter by any physical or chemical called/
VikaD [51]

Answer an atom

Explanation:

6 0
3 years ago
Propane gas (C3H8) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide gas (CO2) and water vapor (H2O).
Sindrei [870]
Oxen beacon is great for your self halllway
7 0
3 years ago
An analytical chemist is titrating 118.3 mL of a 0.3500 M solution of butanoic acid (HC3H7CO2) with a 0.400 M solution of KOH. T
TEA [102]

Answer:

pH = 12.33

Explanation:

Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.

The titration reaction is

HA + KOH ---------------------------- A⁻ + H₂O + K⁺

number of moles of HA :   118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA

number of  moles of OH  : 115.4 mL/1000ml/L x 0.400 mol/L  = 0.046 mol A⁻

therefore the weak acid will be completely consumed and what we have is  the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.

n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH

pOH = - log (KOH)

M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M

pOH = - log (0.0021) = 1.66

pH = 14 - 1.96 = 12.33

Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.

4 0
3 years ago
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