For the equation 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O, how many units of NO3 are represented on the products side?

An unknown organic compound composed of carbon, hydrogen, and oxygen was analyzed and found to be 50.84% C, 8.53% H, and the res

MariettaO [177]

In every 100g of that compund there is 50.84 g of C, 8.53 g H and (100-59.37) g = 40.63 g of O.

Step 1: Convert each element's mass in moles. To do that we need to divide each element's mass by their respective molar mass.

For Carbon.

$C = \frac{50.84}{12} = 4.24 mol$

For Hydrogen.

$H = \frac{8.53}{1} = 8.53 mol$

For Oxygen.

$O = \frac{40.63}{16} = 2.54 mol$

Step 2: Divide each of the numbers by the smallest number.

For Carbon.

$C = \frac{4.24}{2.54} = 1.7$

For Hydrogen.

$H = \frac{8.53}{2.54} = 3.36$

For Oxygen.

$O = \frac{2.54}{2.54} = 1$

Step 3: So the empirical formula will be.

$C_{1.7} H_{3.36} O_1$

But using decimal will be messy. So we multiply the numbers by 3. The right empirical formula will be.

$C_{5} H_{10} O_3$

7 0

2 years ago

A metal is found that has a mass of 27g and will displace 6mL of water. Calculate its density and identify the metal.

melisa1 [442]

The density of metal = 4.5 g/ml, and the metal = Titanium

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

mass of metal=27 g

volume = 6 ml

The density :

$\tt \rho=\dfrac{27}{6}\\\\\rho=4.5~g/ml$

The metal with a density of 4.5 g/ml is Titanium

3 0

2 years ago

A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

Alexxandr [17]

Answer: The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

Explanation:

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

4 0

2 years ago

A 2.00 kg piece of lead at 40.0°C is placed in a very large quantity of water at 10.0°C,and thermal equilibrium is eventually re

sveticcg [70]

Answer:

Δ S = 26.2 J/K

Explanation:

The change in entropy can be calculated from the formula -

Δ S = m Cp ln ( T₂ / T₁ )

Where ,

Δ S = change in entropy

m = mass = 2.00 kg

Cp =specific heat of lead is 130 J / (kg ∙ K) .

T₂ = final temperature 10.0°C + 273 = 283 K

T₁ = initial temperature , 40.0°C + 273 = 313 K

Applying the above formula ,

The change in entropy is calculated as ,

ΔS = m Cp ln ( T₂ / T₁ ) = (2.00 )( 130 ) ln( 283 K / 313 K )

ΔS = 26.2 J/K

6 0

2 years ago

Valeric acid, HC5H9O2 (Ka = 1.5 ✕ 10−5), is used in the manufacture of magnesium valerate, a nerve-calming agent. What is the hy

Studentka2010 [4]

Answer:

$[H^+]=0.00332M$

Explanation:

Hello,

In this case, considering the dissociation of valeric acid as:

$HC_5H_9O_2 \rightleftharpoons C_5H_9O_2 ^-+H^+$

Its corresponding law of mass action is:

$Ka=\frac{[H^+][C_5H_9O_2^-]}{[HC_5H_9O_2]}$

Now, by means of the change $x$ due to dissociation, it becomes:

$Ka=\frac{(x)(x)}{0.737-x}=1.5x10^{-5}$

Solving for $x$ we obtain:

$x=0.00332M$

Thus, since the concentration of hydronium equals $x$ , the answer is:

$[H^+]=x=0.00332M$

Best regards.

7 0

2 years ago