Which of the following is an example of representative money?

40. What is the mass of 3 moles of H 2 O

Leya [2.2K]

Answer:

54g

Explanation:

Given parameters:

Number of moles of H₂O = 3 moles

Unknown:

mass of water = ?

Solution:

To solve this problem, we use the expression below:

mass = number of moles x molar mass

Molar mass of H₂O = 2(1) + 16 = 18g/mol

Mass of water = 3 x 18 = 54g

3 0

3 years ago

Analysis of a gaseous chlorofluorocarbon, CClxFy, shows that it contains 11.79% C and 69.57% Cl. In another experiment, you find

uranmaximum [27]

Answer:

The molecular formula = $C_2Cl_{4}F_2$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 11.79

Molar mass of C = 12.0107 g/mol

% moles of C = 11.79 / 12.0107 = 0.9816

% of Cl = 69.57

Molar mass of Cl = 35.453 g/mol

% moles of Cl = 69.57 / 35.453 = 1.9623

Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,

% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64

Molar mass of F = 18.998 g/mol

% moles of F = 18.64 / 18.998 = 0.9812

Taking the simplest ratio for C, Cl and F as:

0.9816 : 1.9623 : 0.9812

= 1 : 2 : 1

The empirical formula is = $CCl_2F$

Also, Given that:

Pressure = 21.3 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

Volume = 458 mL = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K

⇒n = 0.00052445 moles

Given that :

Amount = 0.107 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.00052445= \frac{0.107\ g}{Molar\ mass}$

$Molar\ mass= 204.0233\ g/mol$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol

Molar mass = 204.0233 g/mol

So,

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

The molecular formula = $C_2Cl_{4}F_2$

6 0

3 years ago

Partner Namets) For each of the following reactions carried out: Write the balanced chemical equation, the full ionic equation,

torisob [31]

Answer : The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

Reactants are lead nitrate and potassium iodide.

Products are lead iodide and potassium nitrate.

The spectator ions are, $K^+,NO_3^-$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.

The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

The ionic equation in separated aqueous solution will be,

$2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)$

In this equation, $K^+\text{ and }NO_3^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)$

4 0

4 years ago

Propanoic acid, ch3ch2cooh, has a pka =4.9. draw the structure of the conjugate base of propanoic acid and give the ph above whi

xenn [34]

Conjugate base of Propanoic acid ( $CH_{3}CH_{2}COOH$ is propanoate where -COOH group gets converted to -CO $O^{1-}$ . The structure of conjugate base of Propanoic acid is shown in the diagram.

The $p^{H}$ above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.