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3 years ago

Aqueous acetic acid is neutralized by aqueous potassium hydroxide. True or False

Chemistry

2 answers:

Stels [109]3 years ago

8 0

Answer:

i think its true

Explanation:

liberstina [14]3 years ago

4 0

Answer:

i think its true but I’m not sure

Explanation:

I know that they can. Be mixed

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a compound has an empirical formula of CH2 what is the molecular formula if it's molar mass is 252.5 grams/mol

Goryan [66]

Empirical formula is the simplest ratio of components making up the compound. the molecular formula is the actual ratio of components making up the compound.

the empirical formula is CH₂. We can find the mass of CH₂ one empirical unit and have to then find the number of empirical units in the molecular formula.

Mass of one empirical unit - CH₂ - 12 g/mol x 1 + 1 g/mol x 2 = 12 = 14 g

Molar mass of the compound is - 252 .5 g/mol

number of empirical units = molar mass / mass of empirical unit

= $\frac{252.5 g/mol}{14 g}$

= 18 units

Therefore molecular formula is - 18 times the empirical formula

molecular formula - CH₂ x 18 = C₁₈H₃₆

molecular formula is C₁₈H₃₆

8 0

3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

I need help with number two too?

KIM [24]

1. 20s, 22s, and 30s... height is the highest.

2. 34s, 38s, and 41s..... height is at its lowest.

i could be wrong.... but i tried.

8 0

3 years ago

So i have this question for science.- What is the element with the atomic number 7. Thank you.

vredina [299]

Answer: the atomic number 7 is

Nitrogen

Explanation:

3 0

3 years ago

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How is an oxidation half-reaction written using the reduction potential chart? How is the oxidation potential voltage determined

insens350 [35]

Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v

7 0

3 years ago

Read 2 more answers