A) 3AgNO3 + GaCl3 --> 3AgCl + Ga(NO3)3
b) 3Ag(+) + 3NO3(-) + Ga(3+) + 3Cl(-) --> 3AgCl(s) + Ga(3+) + 3NO3(-)
c) 3Ag(+) + 3Cl(-) --> 3AgCl(s)
d) Ga(3+), 3NO3(-)
1 L ------- 1000 cm³
1.45 L ----- ???
1.45 * 1000 = 1450 cm³ ( volume )
Density = 0.710 g/cm³
mass = in Kg
m = D * V
m = 0.710 * 1450
m = 1029.5 g
1 Kg ------- 1000 g
kg -------- 1029.5 g
mass = 1029.5 / 1000
mass = 1.0295 Kg
hope this helps!
Calculate the number of molecules present in 3.50 mol feo2. 2.11 x 1024 molecules 2.107 x 1023 molecules 3.50 x 1023 molecules <u>3.5*(6.02*10^23)=2.107*10^24 molecules</u>
<h3>What is
molecules?</h3>
A molecule is the smallest unit of a substance that keeps its content and properties. It is made up of two or more atoms that are joined together by chemical bonds. Chemistry is built on molecules. The element symbol and a subscript indicating the number of atoms are used to identify molecules.
The fundamental building block of an element is an atom. They are made up of an electron-surrounded nucleus. An atom is considered to have valence electrons if its electron shell is not completely complete. A chemical (covalent) connection is created and a lower energy state is entered when two or more atoms join forces to share outer shell valence electrons. In an exothermic reaction, energy is released as atoms bond.
To learn more about molecules from the given link:
brainly.com/question/24722507
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Hi,
The answer is B, Urbanisation.
Hope this helps.
r3t40
The answer should be 2) electrons.
The elements in the periodic table are arranged by increasing atomic number, which is also the number of protons in an element. For example, Carbon has 6 protons, it is the sixth element on the table. While Nitrogen, which has 7 protons, is after Carbon.
In atoms, the number of protons equals to the number of electrons too, if the number of electrons does not equal to the number of protons, the substance is no longer atom, but ion. So, the answer to this is 2) electrons, as the elements on the table is arranged according to increasing atomic numbers.
