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Elodia [21]
3 years ago
15

What composes a element

Chemistry
1 answer:
8_murik_8 [283]3 years ago
4 0

Answer:

2 or more atoms

Explanation:

An element is composed of 2 or more atoms

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Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)
Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation  Cr(s) → Cr3+ + 3e-  

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the  half-reactions

2(Cr → Cr3+ + 3e-)    E° ox = 0.74 V

3(Ni2+ +2e- → Ni)     E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox  = -0.25 + 0.74  = 0.49

E = E ° − 0.0257 V /n * ln Q  = E ° − 0.0257 V /n  *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

b)

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution)   MnO4-  +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e-   (oxidation)   Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the  half-reactions

MnO4-  +8H+ +10-e- ⇔ Mn2+ +4H2O    E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e-        E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox  = 1.51 + 2.12  = 3.63 V

E = E ° − 0.0257 V /n * ln Q  = E ° − 0.0257 V /n  *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0
2 years ago
a mass of mercury occupies 0.350 l. what volume would an equal mass of ethanol occupy? the density of mercury is 13.546 g/ml and
Fed [463]

The density told you the mass covered by one unit of volume of that substance.

Density defines how perfectly the molecules of a substance are packed in a unit of volume.

In the given problem the substances are expressed in g/ml which means that a unit of volume will be 1 mL.

Given ,

Density of mercury = 13.546 g/ml

It means 1mL of mercury will have a mass of 13.546 g, Similarly 1 mL of ethanol will have a mass of 0.789 g.

0.350 L * 10^3 ml/1 L* 13.546/1 mL = 4,741.1 g

The volume of ethanol will have an equal mass is

4741.1 g* 1 mL/0.789g = 6,008.9 mL

6,008.9 mL* 1L/10^3mL = 6.0081 L

Hence the answer is 6.0081 L

To know more about density occupies here :

brainly.com/question/18518241?referrer=searchResults

#SPJ4

8 0
1 year ago
Use the given heats of formation to calculate the enthalpy change for this reaction. B2O3(g) + 3COCl2(g) →2BCl3(g) + 3CO2(g) ΔHo
UkoKoshka [18]

Answer:

the enthalpy change for this reaction is -57.7 kJ/mol

Explanation:

Given:

HB₂O₃ = -1272.8 kJ/mol

HCOCl₂ = -218.8 kJ/mol

HBCl₃ = -403.8 kJ/mol

HCO₂ = -393.5 kJ/mol

Those are all standard enthalpies

Question: Calculate the enthalpy change for this reaction, ΔHreaction = ?

The enthalpy of the reaction is calculated using the standard enthalpies of formation of both products and reagents. To understand better, the reaction is as follows

B₂O₃ + 3COCl₂ → 2BCl₃ + 3CO₂

Where the compounds on the left are the reactants and the compounds on the right are the products

ΔHreaction = ∑ΔHproducts - ∑ΔHreactants

delta(H)_{reaction} =((2*(-403.2)+(3*(-393.5))-((1*(-1272.8)+(3*(-218.8))=-57.7kJ/mol

Please be careful with the signs.

4 0
3 years ago
What are three steps scientist take to evaluate a scientific explanation
Varvara68 [4.7K]
Prediction, control and experiment
5 0
2 years ago
An experiment with 55 co takes 47.5 hours. at the end of the experiment, 1.90 ng of 55-co remains. if the half-life is 18.0 hour
Andru [333]

Answer:

\boxed{\text{10.7 ng}}

Explanation:

Let A₀ = the original amount of ⁵⁵Co .

The amount remaining after one half-life is ½A₀.

After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.

After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀.

The general formula for the amount remaining is:

A =A₀(½)ⁿ

where n is the number of half-lives

n = t/t_½

Data:

   A = 1.90 ng

    t = 45 h

t_½ = 18.0 h

Calculation:

(a) Calculate n

n = 45/18.0 = 2.5

(b) Calculate A

1.90 = A₀ × (½)^2.5

1.90 = A₀ × 0.178

A₀ = 1.90/0.178 = 10.7 ng

The original mass of ⁵⁵Co was \boxed{\text{10.7 ng}}.

7 0
2 years ago
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