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Naddika [18.5K]
4 years ago
11

2. 2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2

Chemistry
1 answer:
Slav-nsk [51]4 years ago
3 0

Answer:

84.4g of AgCl

Explanation:

Based on the reaction:

2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂

<em>2 moles of AgNO₃ and 1 mole of CaCl₂ priduce 2 moles of AgCl and 1 mole of Ca(NO₃)₂</em>

<em />

100g of each reactant are:

AgNO₃: 100g × (1mol / 169.87g) = 0.589 moles

CaCl₂: 100g × (1mol / 110.98g) = 0.901 moles

For a complete reaction of 0.901 moles of CaCl₂ are necessaries 0.901×2 = <em>1.802 moles of AgNO₃. </em>As there are just 0.589moles, <em>AgNO₃ is limitng reactant</em>

<em></em>

0.589 moles of AgNO₃ produce:

0.589 moles × ( 2 moles AgCl / 2 moles AgNO₃) =

<em>0.589 moles of AgC</em>l. In mass:

0.589 moles of AgCl × (143.32g / mol) =<em> 84.4g of AgCl</em>

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99<br> 96<br> 040Zr<br> +<br> 42He<br> ----&gt;<br> + "42Mo
Ahat [919]

he mass defect of the helium nucleus ⁴He₂ is 0.030377 u

Further explanation

Mass defect means the difference between the mass of particles forming an atom with an atomic mass.

Δm = mass defect ( u )

mp = mass of proton ( u )

me = mass of electron ( u )

mn = mass of neutron ( u )

M = atomic mass ( u )

A = mass number

Z = atomic number

Let us now tackle the problem !

Given :

Unknown :

Δm = ?

Solution :

Learn more

Rutherford’s major achievements : brainly.com/question/1552732

Unit of radius of an atom : brainly.com/question/1968819

Fusion : brainly.com/question/11395223

Answer details

Grade: College

Subject: Physics

Chapter: Nuclear Physics

Keywords: Mass , Defect , Nucleon , Number , Atomic , Proton , Electron , Neutron

4 0
3 years ago
An acidified solution was electrolyzed using copper electrodes. A constant current of 1.18 A caused the anode to lose 0.584 g af
Alexxx [7]

Answer:

\boxed{\text{(a) 209 mL; (b) } 6.09 \times 10^{23}}

Explanation:

(a) Gas produced at cathode.

(i). Identity

The only species known to be present are Cu, H⁺, and H₂O.

Only the H⁺ and H₂O can be reduced.

The corresponding reduction half reactions are:

(1) 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻;     E° = -0.8277 V

(2) 2H⁺ +2e⁻ ⇌ H₂;                     E° =  0.0000 V

Two important points to remember when using a table of standard reduction potentials:

  • The higher up a species is on the right-hand side, the more readily it will lose electrons (be oxidized).
  • The lower down a species is on the left-hand side, the more readily it will accept electrons (be reduced}.

H⁺ is below H₂O, so H⁺ is reduced to H₂.

The cathode reaction is 2H⁺ +2e⁻ ⇌ H₂, and the gas produced at the cathode is hydrogen.

(ii) Volume

a. Anode reaction

The only species that can be oxidized are Cu and H₂O.

The corresponding half reactions  are:

(3) Cu²⁺ + 2e⁻ ⇌ Cu;                E° =  0.3419 V

(4) O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O     E° =   1.229   V

Cu is above H₂O, so Cu is more easily oxidized.

The anode reaction is Cu ⇌ Cu²⁺ + 2e⁻.

b. Overall reaction:

Cu           ⇌ Cu²⁺ + 2e⁻

<u>2H⁺ +2e⁻ ⇌ H₂            </u>        

Cu + 2H⁺ ⇌ Cu²⁺ + H₂

c. Moles of Cu lost

n_{\text{Cu}} = \text{0.584 g } \times \dfrac{\text{1 mol}}{\text{63.55 g}} = 9.190 \times 10^{-3}\text{ mol Cu}

d. Moles of H₂ formed

n_{\text{H}_{2}}} = 9.190 \times 10^{-3}\text{ mol Cu} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol Cu}} =9.190 \times 10^{-3}\text{ mol H}_{2}

e. Volume of H₂ formed

Volume of 1 mol at STP (0 °C and  1 bar) = 22.71 mL

V = 9.190 \times 10^{-3}\text{ mol}\times \dfrac{\text{22.71 L}}{\text{1 mol}}  = \text{0.209 L} = \boxed{\textbf{209 mL}}

(b) Avogadro's number

(i) Moles of electrons transferred

\text{Moles of electrons} = 9.190 \times 10^{-3}\text{ mol Cu}\times \dfrac{\text{2 mol electrons}}{\text{1 mol Cu}}\\\\\\= \text{0.018 38 mol electrons}

(ii) Number of coulombs

Q  = It  

Q = \text{1.18 C/s} \times 1.52 \times 10^{3} \text{ s} = 1794 C

(iii). Number of electrons

n = \text{ 1794 C} \times \dfrac{\text{1 electron}}{1.6022 \times 10^{-19} \text{ C}} = 1.119 \times 10^{22} \text{ electrons}

(iv) Avogadro's number

N_{\text{A}} = \dfrac{1.119 \times 10^{22} \text{ electrons}}{\text{0.018 38 mol}} = \boxed{6.09 \times 10^{23} \textbf{ electrons/mol}}

6 0
3 years ago
knowing that the reaction of potassium in water is extremely exothermic, what would be the problem with adding a large piece of
anygoal [31]

Answer:

Explanation:

If the reaction is really exothermic (and it is) then the water would spatter all over the place. It would boil off if the container could hold it. It would also react according to the following reaction.

You are talking about a reaction like

2K + 2HOH = 2KOH + H2

8 0
3 years ago
Please help (show work if needed)
vlada-n [284]

Basically, this figure represents scattering of alpha particles by the atoms of a gold foil.

  • Here, A represents that a few alpha particles are deflected through large angle. It shows that the nucleus of the atom is positively charged which repels the positively charged alpha particles and deflect them from their original path.
  • B represents that most alpha particles pass straight through the gold foil without any deflection from their original path. The reason is that there is a lot of empty space in the atom.
  • C represents that some alpha particles are deflected through small angles. The reason is same as that of A.

Hope you could understand.

If you have any query, feel free to ask.

3 0
3 years ago
How many joules are required to raise the temperature of a 6.50 g sample of silver from 34°c to 189°C? The specific heat of gold
nata0808 [166]

Answer:

236.76 J

Explanation:

Parameters given:

Mass of sample, m = 6.50 g

Initial temperature, T_1 = 34^o C

Final temperature, T_2 = 189^oC

Specific heat capacity of gold, c = 0.235 J/g^oC

Heat of a substance is related to specific heat capacity by the formula:

H = mc(T_2 - T_1)

H = 6.5 * 0.235 * (189 - 34)\\\\\\H = 6.5 * 0.235 * 155\\\\\\H = 236.76 J

236.76 J of heat is required to raise the temperature of the sample.

5 0
4 years ago
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