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3 years ago

Identify the spectator ions in the following molecular equation.

1 answer:

3 0

We can rewrite the equation KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq) into its net ionic equation into
K+ + Br- + Ag + + NO3- = AgBr + K + NO3-only aqueous solutions can dissociate. Spectator ions are present in both sides, hence these are K+ and NO3-. THe rules of assigning oxidation numbers is to identify the number of valence electrons of elements and may be arbitrary depending on the charge of the molecule.

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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

Draw the Lewis structure for NF3. What orbitals on N and F overlap to form bonds between these elements

jonny [76]

Is there a picture ????????????

4 0

2 years ago

In order to gain one pound of body weight, the average person must consume 3500 more calories

Fed [463]

Answer:

17500 calories of chocolate bars are needed to eat to gain 5 pounds.

Explanation:

We can use ratios to calculate the answer using the information given in the question.

1 pound : 3500 grams

5 pounds : x grams

As it is given that the individual is burning no calories, we do not have to factor in any additional numbers.

Method A:

To go from 1 in the first ratio to 5 in the second ratio, they multipled 1 by 5. Hence, to go from 3500 in the first ratio to x in the second ratio, we must multiply by 5.

x = 3500 × 5

x = 17500

Method B:

To solve for the answer x, we can convert the ratios into fractions.

1 / 5 = 3500 / x

3500 / x = 1 / 5

To make x the subject, multiply the denominator of the left fraction with the numerator of the right fraction and place it on the left side. Then multiply the numerator of the left fraction with the denominator of the right fraction and place it on the right side.

x = 5 × 3500

x = 17500

4 0

3 years ago

Standard temperature and pressure (STP), in the context of gases, refers to a) 298.15 K and 1 atm b) 273.15 K and 1 atm c) 298.1

lina2011 [118]

Answer : The correct option is, (B) 273.15 K and 1 atm

Explanation :

STP : STP stands for standard temperature and pressure.

STP conditions :

The temperature is, $0^oC$ or $273.15K$ or $32^oF$

The pressure is, $1atm$ or $101.325kPa$

The volume is 22.4 L for 1 mole of a substance.

Hence, the correct option is, (B) 273.15 K and 1 atm

7 0

3 years ago

For the chemical reaction

polet [3.4K]

Answer:

806.3g

Explanation:

Given parameters:

Number of moles of silver nitrate = 4.85mol

Unknown:

Mass of silver chromate = ?

Solution:

2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃

To solve this problem, we work from the known to the unknown;

The known specie here is AgNO₃ ;

From the balanced chemical equation;

2 moles of AgNO₃ will produce 1 mole of Ag₂CrO₄

4.85 moles of AgNO₃ will produce $\frac{4.85}{2}$ = 2.43moles of Ag₂CrO₄

Mass of silver chromate produced;

mass = number of moles x molar mass

Molar mass of Ag₂CrO₄

Atomic mass of Ag = 107.9g/mol

Cr = 52g/mol

O = 16g/mol

Input the parameters and solve;

Molar mass = 2(107.9) + 52 + 4(16) = 331.8g/mol

So,

Mass of Ag₂CrO₄ = 2.43 x 331.8 = 806.3g

8 0

3 years ago