State the basicity of this acid HCOOH

A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of

allochka39001 [22]

Answer:

The answer to your question is C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

44 g of CO₂ --------------- 12 g of C

52.8 g --------------- x

x = (52.8 x 12)/44

x = 633.6/44

x = 14.4 g of C

12 g of C ------------------ 1 mol

14.4 g of C --------------- x

x = (14.4 x 1)/(12)

x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

18 g of H₂O --------------- 2 g of H

21.6 g of H₂O ------------- x

x = (21.6 x 2) / 18

x = 2.4 g of H

1 g of H -------------------- 1 mol of H

2.4 g of H ----------------- x

x = (2.4 x 1)/1

x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

= 6.4 g

16 g of O ---------------- 1 mol

6.4 g of O -------------- x

x = (6.4 x 1)/16

x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

C₃H₆O

8 0

2 years ago

A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

elena55 [62]

Answer:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Explanation:

Mass of an alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Mass of water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

$m_{a}$ $C_{a}$ [ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

This is the specific heat of the alloy.

4 0

2 years ago

Consider a process in which the entropy of a system increases by 125 J K−1 and the entropy of the surroundings decreases by 125

expeople1 [14]

Answer : The process is not spontaneous.

Explanation :

As, we know that:

Change in entropy = Change in entropy of system + Change in entropy of surrounding

As we are given in question, the entropy of surroundings decrease by the same amount as the entropy of the system increases.

For the given reaction to be spontaneous, the total change in entropy should be positive.

Given :

Entropy change of system = +125J/K

Entropy change of surroundings = -125J/K

Total change in entropy = Entropy change of system + Entropy change of surroundings

Total change in entropy = 125 J/K + (-125 J/K)

Total change in entropy = 0

The process is at equilibrium because the entropy change is equal to zero. So, the process is not spontaneous.

4 0

2 years ago

Write the formulas for the following ionic compounds: (a) copper bromide (containing the Cu+ ion), (b) manganese oxide (containi

ivann1987 [24]

Answer:The formulas of ionic compounds are:

a) $CuBr$

b) $Mn_2O_3$

c) $Hg_2I_2$

d) $Mg_3(PO_4)_2$

Explanation:

Formulas for the an ionic compounds is determine by:

Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

(a) Copper bromide :Given that it contains $Cu^+$ ion.

$Cu^++Br^-\rightarrow CuBr$

(b) Manganese oxide : Given that it contains $Mn^{3+}$ ion.

$Mn^{3+}+O^{2-}\rightarrow Mn_2O_3$

(c)Mercury iodide :Given that it contains $Hg_2^{2+}$

$Hg_2^{2+}+I^-\rightarrow Hg_2I_2$

(d) Magnesium phosphate :Given that it contains $PO_4^{3-}$

$Mg^{2+}+PO_4^{3-}\rightarrow Mg_3(PO_4)_2$

4 0

2 years ago

Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).

6 0

2 years ago

State the basicity of this acid HCOOH​

State the basicity of this acid HCOOH