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2 years ago

3. What's the empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen?

Chemistry

1 answer:

Hunter-Best [27]2 years ago

8 0

Answer:

Empirical formula is Li₂CO₃.

Explanation:

Percentage of oxygen= 65.0%

Percentage of lithium = 18.7%

Percentage of carbon= 16.3%

Empirical formula = ?

Solution:

Number of gram atoms of C = 16.3/12 = 1.4

Number of gram atoms of Li = 18.7/6.94 = 2.7

Number of gram atoms of O = 65.0/ 16 = 4.1

Atomic ratio:

Li : C : O

2.7/1.4 : 1.4/1.4 : 4.1/1.4

2 : 1 : 3

Li : C : O = 2 : 1 : 3

Empirical formula is Li₂CO₃.

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Consider the formation of nitryl fluoride: 2NO2(g)+F2(g)⇌2NO2F(g) The reaction is first order in F2 and second order overall. Wh

Dahasolnce [82]

Answer:

Rate = $k[NO_{2}][F_{2}]$

Explanation:

Two reactants are present in this reaction which are $NO_{2}and F_{2}$
We know overall order of a reaction is summation of individual order with respect to reactants present in rate law equation.
Here, overall order of reaction is 2 including first order with respect to $F_{2}$
So, rate of reaction should also be first order with respect to another reactant i.e. first order with respect to $NO_{2}$
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2 years ago

Which will not appear in the equilibrium constant expression for the reaction below?

n200080 [17]

Answer:

[C] carbon solid

Explanation:

Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.

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2 years ago

Determine the PH at the point in the titration of 40.0ml of 0.200M HC4H7o2 with 0.100 M Sr(OH)2 after 100ml of the strong base h

Dmitriy789 [7]

Answer:

Check the explanation

Explanation:

Mols HC4H7O2 = (volume in L)*(molarity) = (40.0 mL)*(0.200 M)

= (40.0 mL)*(1 L)/(1000 mL)*(0.200 M)

= 8.00*10-3 mol.

Mols Sr(OH)2 corresponding to 10.0 mL of 0.100 M solution =

(volume in L)*(molarity)

= (10.0 mL)*(0.100 M)

= (10.0 mL)*(1 L)/(1000 mL)*(0.100 M)

= 1.00*10-3 mol.

Consider the ionization of Sr(OH)2 as below.

Sr(OH)2 (aq) ----------> Sr2+ (aq) + 2 OH- (aq)

As per the stoichiometric equation,

1 mol Sr(OH)2 = 2 mols OH-.

Therefore,

0.0010 mol Sr(OH)2 = [0.0010 mol Sr(OH)2]*(2 mols OH-)/[1 mole Sr(OH)2]

= 0.0020 mol

= 2.00*10-3 mol

Set up the ICE charts as below.

HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)

Before (mol) 8.00*10-3 2.00*10-3 - -

Change (mol) -2.00*10-3 -2.00*10-3 - +2.00*10-3

After (mol) 6.00*10-3 0 - 2.00*10-3

The change in a pure substance, e.g., H2O is not considered in an acid-base reaction.

Volume of the solution = (40.0 + 10.0) mL = 50.0 mL = (50.0 mL)*(1 L)/(1000 mL) = 0.05 L.

The initial concentrations are obtained by dividing the numbers of moles by the volume, 0.05 L.

Set up the ICE charts as below.

HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)

Initial (M) 0.160 0.0400 - -

Change (M) -0.0400 -0.0400 - +0.0400

Equilibrium (M) 0.120 0 - 0.0400

The acid-ionization constant is written as

Ka = [H3O+][C4H7O2-]/[HC4H7O2] = 1.5*10-5

Plug in the known values and get

Ka = [H3O+]*(0.0400)/(0.120) = 1.5*10-5

======> [H3O+] = (1.5*10-5)*(0.120)/(0.0400) (ignore units)

======> [H3O+] = 4.5*10-5

The proton concentration of the solution is 4.5*10-5 M.

pH = -log (4.5*10-5 M)

= 4.346

≈ 4.35 (ans).

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1 year ago

when you burn a log in the fireplace, the resulting ashes have a mass less than that of the original log. Account for the differ

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2 years ago

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Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.

Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$C(s)+2H_2(g)\rightarrow CH_4(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ$

Hence, the enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

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2 years ago