Answer:
ork out which of the displacement (S), initial velocity (U), acceleration (A) and time (T) you have to solve for final velocity (V).
If you have U, A and T, use V = U + AT.
If you have S, U and T, use V = 2(S/T) - U.
If you have S, U and A, use V = SQRT(U2 + 2AS)
2.4 x 10²² atoms
<h3>Further explanation</h3>
Atomic mass is the average atomic mass of all its isotopes
In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu
So the atomic mass obtained is the mass of the atom relative to the 12th carbon atom
mass single Uranium atom=4.7 x 10⁻²² g
then for 111 mg=0.111 g
You must verify that the number of atoms of each type is equal on both sides of the chemical equation: same number of C, same number of H and same number of O on both sides.
<span>A. C4H6 + 5.5O2 ---> 4CO2 + 3H2O
element reactant side product side
C 4 4
H 6 3*2 = 6
O 5.5 * 2 = 11 4*2 + 3 = 11
Then, this equation is balanced.
</span>Do the same with the other equations if you want to verify that they are not balanced.
Answer: option A.
Answer:
The answer to your question is: letter D.
Explanation:
Noble gases are located in group VIIIA of the periodic table, this means that they have 8 eight electrons in their outermost shell.
Due to this characteristic, they are stable and do not react with other elements.
a. 1s22s22p4 The outermost shell of this electron configuration has 6 electrons, then this element has 6 electrons not 8. This configuration is of an element of the group VIA.
b. [Ne]2s22p2 The outermost shell of this element has 4 electrons, so this is not the configuration of a noble gas.
c. [Ar] 3s1 This element only has one electron in its outermost shell, so this is the electron configuration of an alkaline metal.
d. 1s22s22p6 This element has 8 electrons in its outermost shell, so this is the electron configuration of a noble gas.
