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Elina [12.6K]
2 years ago
6

The following molecular scenes depict an important reaction in nitrogen chemistry (nitrogen is blue; oxygen is red): Write a bal

anced equation for this reaction. ____N2O5 ___NO2 + ___O2
Chemistry
1 answer:
alexandr1967 [171]2 years ago
3 0

Answer:

2 N2O5 4 NO2 + 1 O2

Explanation:

For the last one you dont really have to add the 1 but you can if its nedded,

since 3 divided by 3 is 1.

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The atomic mass of Cu is 63.5. Find its electrochemical equivalent​
FrozenT [24]

Answer:

The electrochemical equivalent of copper, Cu, is 3.29015544 × 10⁻⁷ g/C

Explanation:

The given parameters are;

The element for which the electrochemical equivalent is sought = Copper

The atomic mass of copper = 63.5

The electrochemical equivalent, 'Z', of an element or a substance is the mass, 'm', of the element or substance deposited by one coulomb of electricity, which is equivalent to a 1 ampere current flowing for a period of 1 second

Mathematically, we have;

m = Z·I·t = Z·Q

We have;

Cu²⁺ (aq) + 2·e⁻ → Cu

Therefore, one mole of Cu, is deposited by 2 moles of electrons

The charge carried one mole of electrons = 1 Faraday = 96500 C

∴ The charge carried two moles of electrons, Q = 2 × 96500 C = 193,000 C

Given that the mass of an atom of Cu = 63.5 a.m.u., the mass of one mole of Cu, m = 63.5 g

Z = \dfrac{m}{Q} = \dfrac{63.5 \ g}{193,000 \ C} = 3.29015544 \times 10^{-4} \, g \cdot C^{-1}

∴ Z = 3.29015544 × 10⁻⁴ g/C = 3.29015544 × 10⁻⁷ g/C

The electrochemical equivalent of copper, Cu, is Z = 3.29015544 × 10⁻⁷ g/C

7 0
3 years ago
Write a hypothesis:<br> What is the mole ratio of the cation and the anion in a precipitate?
tankabanditka [31]

Answer:

The mole ratio of the cation and the anion in a precipitate is a simple fraction. ( im sorry if this dosent help a lot.)

Explanation:

8 0
2 years ago
To protect itself from herbivores, plants ______
9966 [12]
C.) both secrete toxins and have thorns (It simply depends on what kind of plant it is)
3 0
3 years ago
Read 2 more answers
A cylinder of argon contains 50 L of Ar at 12.4 atm and 127°C . How many moles of argon are in the cylinder
joja [24]

Answer:

18.9 moles

Explanation:

We have the following data:

V = 50 L

P = 12.4 atm

T= 127°C + 273 = 400 K

R = 0.082 L.atm/K.mol (it is the gas constant)

We use the ideal gas equation to calculate the number of moles n of the gas:

PV = nRT

⇒ n = PV/RT = (12.4 atm x 50 L)/(0.082 L.atm/K.mol x 400 K) = 18.9 mol

6 0
2 years ago
The reacted side of a balanced chemical equation is shown below. C3H8 + 5O2 How many oxygen atoms should there be on the product
atroni [7]
10 atoms. If there are 10 in the reactants you need the same number in the products
5 0
2 years ago
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