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babunello [35]
2 years ago
10

Standard reduction potentials are based on which element?

Chemistry
2 answers:
KonstantinChe [14]2 years ago
6 0
They're based on hydrogen.
Nadusha1986 [10]2 years ago
3 0

Answer:

Hydrogen

Explanation:

Standard reduction potential is a measure of the tendency of an atom to attract electrons and get reduced. Higher the reduction potential i.e. more positive the value greater will be the tendency to get reduced. These values are measured at 298 K and 1 atm and expressed in terms of volts.

The standard reduction potential of Hydrogen is taken as zero and the values of all other elements are deduced based on hydrogen. These are tabulated in the thermochemical reduction potential table. The reduction equation for hydrogen is:

H2(g) + 2e-  → 2H+           where E⁰ = 0.00 Volts

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How much energy is required to heat 40g of water from -7 degrees Celsius to 108 degrees Celsius (Lf= 335,000J/Kg for ice)
solniwko [45]
The   energy  required  to heat  40g  of water  from -7 c  to 108 c is
1541000  joules

     calculation

Q(heat)=  M( mass)  x c(specific heat capacity) xdelta t( change in temperature)

M=  40g=  40/1000= 0.04 Kg
C=  335,000 j/kg/c
delta T   (    108 --7= 115  c)

Q  is therefore   =  0.04 g x  335000 j/kg/c  x 115 c  = 1541,000  joules


6 0
3 years ago
The image compares the arrangement of electrons in two different neutral atoms.
jekas [21]

Answer:

  • <u>Option b. Atom P has an estimated Zeff of 7 and is therefore to the right of Atom Q, which has a Zeff of 6.</u>

Explanation:

Please, find attached the figures of both atom Q and atom P corresponding to this question.

The <u>features of atom Q are</u>:

  • Each <em>black sphere</em> represents an electron
  • In total this atom has 8 electrons: 2 in the inner shell and 6 in the outermost shell.
  • Since it is assumed that the atom is neutral, it has 8 protons: one positive charge of a proton balances one negative charge of an electron. Thus, the atomic number of this atom is 8.
  • Since only two shells are ocuppied, you can assert that the atom belongs to the period 2 (which is confirmed looking into a periodic table with the atomic number 8).
  • <em>Zeff </em>is the effective nuclear charge of the atom. It accounts for the  net positive charge the valence electrons experience. And may, in a very roughly way, be estimated as the number of protons less the number of electrons in the inner shells. Thus, for this atom, an estimated  Z eff = 8 - 2 = 6.

The <u>features of atom P</u> are:

  • Again, each black sphere represents an electron
  • In total this atom has 9 electrons: 2 in the inner shell and 7 in the outermost shell.
  • Since it is assumed that the atom is neutral, it has 9 protons.
  • The atomic number of this atom is 9.
  • Using the same reasoning used for atom Q, this atom is also in the period 2.
  • Estimated Z eff = 9 - 2 = 7.

Then, since atom P has a greater Z eff than atom Q (an estimated Zeff of 7 for atom P against an estimated Z eff of 6 for atom Q),  and both atoms are in the same period, you can affirm that <em>atom P</em> has a greater atomic number and<em> is therefore to the right of atom Q</em>.

8 0
3 years ago
Please help me find what type of chemical reaction takes place from numbers 1-5.
Marrrta [24]
Sorry I’m not sure sksnsnsns. Sksnsnsnsnss sksnsnsnsnsnssnsnsns
7 0
3 years ago
Read 2 more answers
2. (2 pts) How would you prepare 1.5 liters of 2 M KCI (MW=74.55 g/mol)
ra1l [238]

Answer:

Dissolve 226 g of KCl in enough water to make 1.5 L of solution

Explanation:

1. Calculate the moles of KCl needed

n = \text{1.5 L} \times \dfrac{\text{2 mol}}{\text{1 L}}= \text{3.0 mol}

2. Calculate the mass of KCl

m = \text{3.0 mol} \times \dfrac{\text{74.55 g}}{\text{1 mol}}= \text{224 g}

3. Prepare the solution

  • Measure out 224 g of KCl.
  • Dissolve the KCl in a few hundred millilitres of distilled water.
  • Add enough water to make 1.5 L of solution. Mix thoroughly to get a uniform solution.
8 0
2 years ago
Will mark brainliest!!
olga2289 [7]

Answer:

I think its B im not sure

but i hope this helps

6 0
3 years ago
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