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Serjik [45]
3 years ago
14

Copper(II) sulfide is oxidized b molecular oxygen to produce gaseous sulfur trioxide and solid copper (II) oxide. The gaseous pr

oduc hen reacts with luquid water to produce liquiid hydrogen sulfate as they only product. Wroite the two equations which represens these reactions.
Chemistry
1 answer:
Tasya [4]3 years ago
3 0

Explanation:

Copper(II) sulfide reacts with oxygen gas to give solid copper(II) oxide and sulfur trioxide gas.

The reaction is given as:

CuS+2O_2\rightarrow CuO(s)+SO_3(g)

When 1 mol copper(II) sulfide react with 2 moles of oxygen gas it gives 1 mol of solid copper(II) oxide and 1 mol of sulfur trioxide gas

The gas formed in above reaction that is sulfur trioxide reacts with water to give sulfuric acid or hydrogen sulfate.

The reaction is given as:

SO_3(g)+H_2O(l)\rightarrow H_2SO_4(aq)

1 mol of sulfur trioxide gas reacts with 1 mol of liquid water to produce 1 molo of liquid hydrogen sulfate or sulfuric acid

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How much water would I need to add to 500 mL of a 2.4 M KCl solution to make a 1.0 M solution?
GalinKa [24]

Answer:

700 mL of water

Explanation:

This is the perfect example of dilation calculations. Along with this concept we have a formula c = n ( solute ) / V ( solution ). Let us first solve for n by changing this equation to isolate the solute,

n ( KCL ) = 2.4 mol / L * 500 * 10 ^ - 3 L,

n ( KCL ) = 1.2 moles ( KCL )

Knowing the amount of moles of potassium chloride, we have to now identify how much is present in the target solution,

V = 1 .2 moles /  ( 1.0 moles / L )

V = 1.2 L = 1200 mL

_______________________________________________________

Vadded = 1200 - 500 = 700 mL

<u><em>Hope that helps!</em></u>

3 0
3 years ago
Which of the following pair has a larger size? a)K or K- b)Br or Br- c)O2- or F- d) Li or Na​
Anna11 [10]

Answer:

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Explanation:

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7 0
2 years ago
Ammonia (NH3) burns in air to form nitrogen dioxide and water. Calculate grams of nitrogen dioxide produced if 4.0 grams of ammo
kenny6666 [7]
The balanced equation for the above reaction is;
4NH₃ + 7O₂  --> 4NO₂  + 6H₂O
stoichiometry of NH₃ to NO₂ is 4:4
the number of NH₃ moles consumed are - 4.0 g / 17 g/mol = 0.24 mol
number of NH₃ moles reacted are equivalent to number of NO₂ moles formed 
therefore number of NO₂ moles formed - 0.24 mol x 46 g/mol = 11.04 g
mass of NO₂ formed is 11.04 g
8 0
2 years ago
What is an alkali metal with fewer than 10 protons in its nucleus?
Trava [24]

Answer:

            Lithium

Explanation:

                  Alkali metals are group of metals which are present in first group of periodic table. As we know atomic number is equal to number of protons contained by a particular element. Therefore, the alkali metals along with there number of protons are listed below;

Alkali Metal                                         Number of Protons

Lithium                                                               3

Sodium                                                             11

Potassium                                                         19

Rubidium                                                         37

Cesium                                                             55

Francium                                                          87

Hence, it is cleared from above table that Lithium is having fewer protons than 10.

6 0
3 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
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