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shtirl [24]
3 years ago
10

What would happen if Phosphorus disappeared?

Chemistry
1 answer:
ololo11 [35]3 years ago
3 0
<span>The cycles would eventually come to an end. If the sun were to burn out, the Earth would lose the energy and heat from the sun and soon lose its own heat and energy and fall apart into space. The magnetism would soon become demagnetized, plates would stop moving, trees would die, life etc. no more O2, CO2, cycle and all others would fade.</span>
You might be interested in
Which chemical equation represents a decomposition reaction?
MrMuchimi

Decomposition reaction D. 2H20⇒ 2H2 + O2

<h3>Further explanation </h3>

Every chemical reaction involves a compound consisting of reactants and products  

Reactants are compounds that react and form new compounds called products

There are several forms of reactions that can occur, including single replacement, double replacement, synthesis, decomposition, etc.

A. 2C2H6 + 702 ⇒ 4CO2 + 6H20

Combustion : reaction of Hydrocarbon and Oxygen

B. AgNO3 + LiCl → AgCl + LINO3

Double replacement : there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product

C. Ca + MgS → CaS + Mg

Single replacement :one element replaces the other elements of a compound to produce new elements and compounds

D. 2H20⇒ 2H2 + O2

Decomposition : One compound breaks down into 2 components

8 0
2 years ago
Which of the following reactions is a synthesis reaction? ​
tamaranim1 [39]
The first reaction is a synthesis reaction (A+B -> AB)
3 0
3 years ago
Read 2 more answers
What is the volume of a 5M solution with that has 16.5 moles?
kenny6666 [7]
Molarity can be used to calculate the volume of solvent or the amount of solute. The relationship between two solutions with the same amount of moles of solute can be represented by the formula c1V1 = c2V2, where c is concentration and V is volume.
7 0
3 years ago
Imagine you had HCl with a concentration of exactly 0.10 mol/dm3. If 0.023 dm3 of a sodium hydroxide solution, NaOH (aq), could
polet [3.4K]

Answer:

Explanation:

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 25.0 ÷ 1,000 = 0.0250 dm3

Rearrange:

Concentration in mol/dm3 =

Amount of solutein mol = concentration in mol/dm3 × volume in dm3

Amount of sodium hydroxide = 0.100 × 0.0250

= 0.00250 mol

Step 2: Find the amount of hydrochloric acid in moles

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.00250 mol of NaOH reacts with 0.00250 mol of HCl

Step 3: Calculate the concentration of hydrochloric acid in mol/dm3

Volume of hydrochloric acid = 20.00 ÷ 1000 = 0.0200 dm3

Concentration in mol/dm3 =

Concentration in mol/dm3 =

= 0.125 mol/dm3

Step 4: Calculate the concentration of hydrochloric acid in g/dm3

Relative formula mass of HCl = 1 + 35.5 = 36.5

Mass = relative formula mass × amount

Mass of HCl = 36.5 × 0.125

= 4.56 g

So concentration = 4.56 g/dm3

8 0
2 years ago
The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
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