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grandymaker [24]
3 years ago
12

The most common fuel for nuclear fission reactors is:

Chemistry
2 answers:
Ostrovityanka [42]3 years ago
7 0
C. Uranium-235

Thus, your answer. 

(I took this on a quiz I had, known as a Quick Check to INCA students.)

XD
Arada [10]3 years ago
5 0
C) uranium-235 is the correct answer
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Suppose a platinum atom in the oxidation state formed a complex with two chloride anions and two ammonia molecules. Write the ch
Korolek [52]

Answer: Pt(Cl)2(NH3)2

Explanation:

In the formation of the complex, the oxidation number of platinum is plus two (+2) and two chloride ions cancel it out by their oxidation number of -1 each. Hence the complex has an overall charge of zero. It is thus neutral with no charge attached to its formula.

7 0
3 years ago
How much potassium chloride, KCI, is produced during the decomposition of 100.0 grams of potassium chlorate KCIO3? Please show s
Ne4ueva [31]
Answer: the reaction will produce 15.3 g of
KCl.

explanation:
1. write the balanced equation.
2KClO
3
→
2KCl
+
3O
2

2. calculate the moles of
KClO
3
.
Moles of KClO
3
=
25.0
g KClO
3
×
1 mol KClO
3
122.55
g KClO
3
=
0.2046 mol KClO
3

3. calculate the moles of
KCl
.
Moles of KCl
=
0.2046
mol KClO
3
×
2 mol KCl
2
mol KClO
3
=
0.2046 mol KCl

4. calculate the mass of
KCl
.
Mass of KCl
=
0.2046
mol KCl
×
74.55 g KCl
1
mol KCl
=
15.3 g KCl
4 0
3 years ago
Analyze: What pattern do you see? Make a rule: Based on your data, how are elements arranged into chemical families?
julia-pushkina [17]

Answer: in order of increasing atomic number.

Explanation: Elements in the same group have similar chemical properties. This is because they have the same number of outer electrons and the same valency.

7 0
3 years ago
Read 2 more answers
This means that the steel bar lost 14900 J of thermal energy. What is the change in temperature of the steel bar? Recall that th
stiv31 [10]

Answer:

ΔT=-747,13°C

Explanation:

Sensible heat is<em> the amount of thermal energy that is required to change the temperature of an object</em>, the equation for calculating the heat change is  given by:

Q=msΔT

where:

  • Q, heat that has been absorbed or realeased by the substance [J]
  • m, mass of the substance [g]
  • s, specific heat capacity [J/g°C] (
  • ΔT, changes in the substance temperature [°C]

To solve the problem, we clear ΔT of the equation and then replace our data:

Q=msΔT

ΔT=Q/ms

ΔT=\frac{-14900 J}{40,7g*0,49\frac{J}{gC} }=-747,13°C

<em>(Note that Q=-14900 J because there is a </em><u><em>LOST</em></u><em> of thermal energy)</em>

Thus, the change in temperature of the steel bar is -747,13°C, meaning that the temperature of the bar decreases.

5 0
4 years ago
Read 2 more answers
Suppose a laboratory wants to identify an unknown pure substance. The valence electrons of the substance's atoms feel an effecti
zalisa [80]

Answer:

  • The answer is the third option in the list:<em> It would have smaller atomic radii than Si and higher ionization energies than Si.</em>

Explanation:

The<em> effective nuclear charge</em> is that portion of the total nuclear charge that a given electron in an atom feels.

Since, the inner electrons repel the outer electrons, t<em>he effective nuclear charg</em>e of a determined electron is the sum of the positive charge (number of protons or atomic number) that it feels from the nucleus less the number of electrons that are in the shells that are are closer to the nucleus than the own shell of such (determined) electron.

Mathematically, <em>the effective nuclear charge (Zeff)</em> is equal to the atomic number (Z) minus the amount (S) that other electrons in the atom shield the given (determined) atom from the nucleus.

  • Zeff = Z - S.

Since, the valence electrons are the electrons in the outermost shell of the atom, you can find certain trend for the value Zeff.

Let's look at the group to which Si belongs, which is the group 14. This table summarizes the relevant data:

Element   Z   Group   # valence electrons     S                      Zeff = Z - S

C              6      14                      4                     6 - 4 = 2             6 -  2 = +4

Si             14     14                      4                     14 - 4 = 10         14 - 10 = +4

Ge           32     14                     4                     32 - 4 = 28       32 -28 = +4

Sn           50     14                     4                     50 - 4 = 46       50 - 46 = +4

Pb           82     14                     4                     82 - 4 = 78        82 - 78 = +4  

With that, you have shown that the valence electrons of the unknown substance's atoms feel an effective nuclear charge of +4 and you have a short list of 4 elements which can be the unknown element: C, Ge, Sn or Pb.

The second known characteristic of the unknown substance's atoms is that it has a <em>higher electronegativity than silicon (Si)</em><em>.</em>

So, you must use the known trend of the electronegativity in a group of the periodic table: the electronegativity decreases as you go down in a group. So, three of the elements (Ge, Sn, and Pb) have lower electronegativity than Si, which has left us with only one possibility: the element C. The valence electrons of carbon (C) atoms feel an effective nuclear charge of +4 and it carbon has a higher electronegativity than silicon.

Other two periodic trends attending the group number are the <em>atomic radii and the ionization energy</em>.

The atomic radii generally increases as you go from top to bottom in a group. This is because you are adding electrons to new higher main energy levels. So, you can conclude that the originally unknwon substance (carbon) has a smaller atomic radii, than Si.

The ionization energies generally decreases as you go from top to bottom in a group. This os due to the shielding effect: as seen, the effective nuclear charge of the atom's valence electrons remains constant, while the distance of the electrons from the nucleus increases (the valence electrons are farther away from the nucleus), which means the upper the element in a given group, the larger the ionization energy of the atoms.

With this, our conclusions about the unnkown substance are:

  • Since it has a higher electronegativity value than silicon (Si), it is right up of Si, and there is on only element possible element than can be (C).

  • Since, it is upper than silicon (Si), it would have smaller atomic radii.

  • Due to the shielding effect, it would have larger ionization energies.

  • The answer is the third option in the list: It would have smaller atomic radii than Si and higher ionization energies than Si.

6 0
3 years ago
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