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Papessa [141]
3 years ago
11

Which of the following does NOT represent a balanced reaction? ​

Chemistry
1 answer:
Gwar [14]3 years ago
8 0

Answer:

<h3>letter D i think</h3>

Explanation:

correct me if im wrong

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KClKCl has a lattice energy of −701 kJ/mol.−701 kJ/mol. Consider a generic salt, ABAB , where A2+A2+ has the same radius as K+,K
elena-14-01-66 [18.8K]

Explanation:

It is given that lattice energy is -701 kJ/mol.

Whereas it is known that realtion between lattice energy and radius is as follows.

               Lattice energy \propto \frac{Z_{+}Z_{-}}{r}

where,          Z_{+} = +2,    and Z_{-} = -2

Therefore, lattice energy of AB = 4 \times \text{lattice energy of KCl}

                                                    = 4 \times -701 kJ/mol

                                                    = -2804 kJ/mol

Thus, we can conclude that lattice energy of the salt ABAB is -2804 kJ/mol.

6 0
3 years ago
The lab procedure you just studied involves
Svetllana [295]

Answer: c , v, c, c, v

Explanation:

8 0
3 years ago
이
Nataliya [291]

The daughter isotope  : Radon-222 (Rn-222).

<h3>Further explanation</h3>

Given

Radium (Ra-226) undergoes an  alpha decay

Required

The daughter nuclide

Solution

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,  

  • alpha α particles ₂He⁴
  • beta β ₋₁e⁰ particles
  • gamma particles ₀γ⁰
  • positron particles ₁e⁰
  • neutron ₀n¹

The decay reaction uses the principle: the sum of the atomic number and mass number before and after decay are the same

Radium (Ra-226) : ₈₈²²⁶Ra

Alpha particles : ₂⁴He

So Radon-226 emits alpha α particles ₂He⁴  , so the atomic number decreases by 2, mass number decreases by 4

The reaction :

₈₈²²⁶Ra ⇒ ₂⁴He + ₈₆²²²Rn

4 0
3 years ago
There is an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+.
leonid [27]

Answer:

See explanation

Explanation:

Anode;

Sn(s) ------> Sn^2+(aq) + 2e

Cathode;

Mn^2+(aq) + 2e ------> Mn(s)

The minimum voltage required to drive the reaction is the cell voltage. The cell voltage is obtained from;

E°cell= E°cathode - E°anode

E°cell= -1.19 - (-0.14)

E°cell= -1.05 V

4 0
3 years ago
A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24
Setler79 [48]

<u>Answer:</u>

<u>For Part A:</u> The partial pressure of Helium is 218 mmHg.

<u>For Part B:</u> The mass of helium gas is 0.504 g.

<u>Explanation:</u>

  • <u>For Part A:</u>

We are given:

p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg

To calculate the partial pressure of helium, we use the formula:

P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}

Putting values in above equation, we get:

745=245+119+163+p_{He}\\p_{He}=218mmHg

Hence, the partial pressure of Helium is 218 mmHg.

  • <u>For Part B:</u>

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

PV=\frac{m}{M}RT

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g

Hence, the mass of helium gas is 0.504 g.

6 0
3 years ago
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