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2 years ago

3Cu + 8HNO = 3Cu(NO3)2 +2NO + 4H2O

1 answer:

7 0

Taking into account the reaction stoichiometry, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 Cu+ 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Cu: 3 moles
HNO₃: 8 moles
Cu(NO₃)₂: 3 mole
NO: 2 moles
H₂O: 4 moles

The molar mass of the compounds is:

Cu: 63.55 g/mole
HNO₃: 63 g/mole
Cu(NO₃)₂: 187.55 g/mole
NO: 30 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Cu: 3 moles ×63.55 g/mole= 190.65 grams
HNO₃: 8 moles ×63 g/mole= 504 grams
Cu(NO₃)₂: 3 moles ×187.55 g/mole= 562.65 grams
NO: 2 moles ×30 g/mole= 60 grams
H₂O: 4 moles ×18 g/mole= 72 grams

<h3>Mass of Cu(NO₃)₂ produced</h3>

The following rule of three can be applied: if by reaction stoichiometry 504 grams of HNO₃ form 562.65 grams of Cu(NO₃)₂, 4.69 grams of HNO₃ form how much mass of Cu(NO₃)₂?

$mass of Cu(NO_{3} )_{2} =\frac{4.69 grams of HNO_{3} x562.65 grams of Cu(NO_{3} )_{2} }{504 grams of HNO_{3} }$

<u><em>mass of Cu(NO₃)₂= 5.2634 grams</em></u>

Then, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

Learn more about the reaction stoichiometry:

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2 years ago

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Which of the following statements correctly describe the trends in ionic size?

8_murik_8 [283]

the spin quantum number has values+1212 of -1212 describes the trends in ionic size.

spin quantum is denoted by (s) as it represents the electronic configuration of the atom and the ionic size also depends on the electronic configuration of the sub-shells of the atom like l, m, n, s.

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8 0

8 months ago

One crystalline form of silica (SiO2) has a cubic unit cell, and from X-ray diffraction data it is known that the cell edge leng

tigry1 [53]

Answer:

There are 8Si atoms and 16 O atoms per unit cell

Explanation:

From the question we are told that:

Edge length $l=0.700nm=>0.7*10^9nm$

Density $\rho=2.32g/cm^3$

Generally the equation for Volume is mathematically given by

$V=l^3$

$V=(0.7*10^9)m^3$

$V=3,43*10^-{22}cm$

Where

Molar mass of (SiO2) for one formula unit

$M=28+32$

$M=60g/mol$

Therefore

Density of Si per unit length is

$\rho_{si}=\frac{9.96*10^{23}}{3.43*10^22}$

$\rho=0.29$

Molar mass of (SiO2) for one formula unit

$M=28+32$

$M=60g/mol$

Therefore

There are 8Si atoms and 16 O atoms per unit cell

6 0

3 years ago

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle

Sedaia [141]

Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of $4.71\times 10^{-15}J$ . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{\sqrt{2mE_k}}$

where,

= De-Broglie's wavelength = ?

h = Planck's constant = $6.624\times 10^{-34}Js$

m = mass of beta particle = $9.1094\times 10^{-31} kg$

$E_k$ = kinetic energy of the particle = $4.71\times 10^{-15}J$

Putting values in above equation, we get:

$\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}$

$\lambda = 6.762\times 10^{-12} m$

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

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3 years ago

What is the volume of a balloon of gas at 842 mm Hg and -23° C, if its volume is 915 mL at a pressure of 1,170 mm Hg and a tempe

garik1379 [7]

Answer:
V₂ = 1070 mL or 1.07 L

Solution:

Data Given;
P₁ = 1170 mmHg

V₁ = 915 mL

T₁ = 24 °C + 273 K = 297 K

P₂ = 842 mmHg

V₂ = ?

T₂ = - 23 °C + 273 K = 250 K

According to Ideal gas equation,

P₁ V₁ / T₁ = P₂ V₂ / T₂

Solving for V₂,

V₂ = P₁ V₁ T₂ / P₂ T₁

Putting Values,

V₂ = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)

V₂ = 1070 mL or 1.07 L

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3 years ago