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Maru [420]
3 years ago
14

Two comets are 300 m apart. One

Chemistry
1 answer:
ioda3 years ago
5 0

Answer:

a

Explanation:

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23°C = ______ K<br><br> -250<br> 296<br> 123<br> 23
strojnjashka [21]

Answer:

296

Explanation:

6 0
3 years ago
Read 2 more answers
Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =
BaLLatris [955]

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, 9.33\times 10^{-5}mol/L

Explanation :

As we know that CaCO₃ dissociates to give Ca^{2+} ion and CO_3^{2-} ion.

The solubility equilibrium reaction will be:

CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}][CO_3^{2-}]

Let solubility of CaCO₃ be, 's'

K_{sp}=(s)\times (s)

K_{sp}=s^2

8.7\times 10^{-9}=s^2

s=9.33\times 10^{-5}mol/L

Therefore, the solubility of CaCO₃ is, 9.33\times 10^{-5}mol/L

4 0
3 years ago
The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl
nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0
3 years ago
The standard reduction potentials of lithium metal and chlorine gas are as follows:Reaction Reduction potential(V)Li+(aq)+e−→Li(
meriva

Answer:

A) E° = 4.40 V

B) ΔG° = -8.49 × 10⁵ J

Explanation:

Let's consider the following redox reaction.

2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)

We can write the corresponding half-reactions.

Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq)      E°red = 1.36 V

Anode (oxidation):  2 Li(s) → 2 Li⁺(aq) + 2 e⁻         E°red = -3.04

<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>

The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.

E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V

<em>B) Calculate the free energy ΔG° of the reaction.</em>

We can calculate Gibbs free energy (ΔG°) using the following expression.

ΔG° = -n.F.E°

where,

n are the moles of electrons transferred

F is Faraday's constant

ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J

8 0
3 years ago
Which alternative energy source would be expected to be the least dependent upon the weather?
Elza [17]
Of the three sources listed, geothermal energy 
is the least dependent on the weather.
(Once it's installed and running, that is.)
3 0
3 years ago
Read 2 more answers
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