Which of the following reactions would have a decrease in entropy?

What is the percent composition of NaHCO3?

zhuklara [117]

Answer:

Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)

Explanation:

Percent Composition is defined as the %age by mass of each element present in a compound. Therefore, it is a relative amount of each element present in a compound.

Calculating Percent Composition of NaHCO₃:

1: Calculating Molar Masses of all elements present in NaHCO₃:

a) Na = 22.99 g/mol

b) H = 1.01 g/mol

c) C = 12.01 g/mol

d) O₃ = 16.0 × 3 = 48 g/mol

2: Calculating Molecular Mass of NaHCO₃:

Na = 22.99 g/mol

H = 1.01 g/mol

C = 12.01 g/mol

O₃ = 48 g/mol

----------------------------------

Total 84.01 g/mol

3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:

For Na:

= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 27.36 %

For H:

= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 1.20 %

For C:

= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 14.29 % ≈ 14.30 %

For O:

= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 57.13 % ≈ 57.14 %

8 0

3 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

HELPPPP Name the three states of matter. Describe the arrangement and movement of atoms in a substance for each state.

olga nikolaevna [1]

Solid, Liquid, and Gas

Solid is the state in which matter maintains a fixed volume and shape; liquid is the state in which matter adapts to the shape of its container but varies only slightly in volume; and gas is the state in which matter expands to occupy the volume and shape of its container.

3 0

3 years ago

Read 2 more answers

What type of bond is made between amino acids in the ribosome?

sleet_krkn [62]

Answer:

Peptide bonds form between the amino group of the amino acid attached to the A-site tRNA and the carboxyl group of the amino acid attached to the P-site tRNA. The formation of each peptide bond is catalyzed by peptidyl transferase, an RNA-based enzyme that is integrated into the 50S ribosomal subunit.

8 0

3 years ago

Part 1: Name the type of chemical reaction that occurs when calcium hydroxide (Ca(OH)2) reacts with nitric acid (HNO3).

liberstina [14]

Answer:

For 1: Neutralization reaction

For 2: Zinc is more reactive than lead and less reactive than calcium.

Explanation:

For (1):

When a base reacts with an acid to form a salt and water molecule, it is known as a neutralization reaction. The general equation follows:

$HX+BOH\rightarrow BX+H_2O$

The chemical equation for the reaction of calcium hydroxide and nitric acid follows:

$Ca(OH)_2(aq)+2HNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+2H_2O(l)$

For (2):

A single displacement reaction is defined as the reaction in which a more reactive metal displaces a less reactive metal from its salt solution. The general chemical equation follows:

$A+BX\rightarrow AX+B$