Good electrical conductivity and electronegativities less than 1.7 are the properties and characteristic of Group 2 elements at STP.
<h3>What are the properties of group 2 elements?</h3>
Group 2 elements are metals so they are good conductors of heat and electricity. It has electronegativity values less than 1.7 and very reactive. They form 2+ charge in cationic form and also formed ionic bonds with other negatively charged elements.
So we can conclude that good electrical conductivity and electronegativities less than 1.7 are the properties and characteristic of Group 2 elements at STP.
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Answer:
Neutrons.
Explanation:
Isotopes can be defined as the atom of an element that has the same number of protons but different number of neutrons. This ultimately implies that, the isotopes of an element have the same atomic number (number of protons) but different atomic mass (number of nucleons).
The isotope of an element is denoted by
Where; X is the symbol of the element.
A is the atomic mass or number of nucleons.
Z is the atomic number or number of protons.
<em>Therefore, the number of neutrons = A - Z</em>
<em>Isotopes of carbon differ with respect to the number of neutrons.</em>
<em>Basically, there are three (3) Isotopes of Carbon and these are;</em>
<em>1. Carbon-12: it has an atomic mass of 12 with 6 numbers of proton and neutron respectively. </em>
<em>2. Carbon-13: it has an atomic mass of 13 with 6 numbers of proton and 7 numbers of neutron. </em>
<em>3. Carbon-14: it has an atomic mass of 14 with 6 numbers of proton and 8 numbers of neutron. </em>
<u>Answer:</u> 6.57 L of solution can be made.
<u>Explanation:</u>
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
.....(1)
Given values:
Molarity of LiBr = 3.5 M
Moles of LiBr = 23 moles
Putting values in equation 1, we get:

Hence, 6.57 L of solution can be made.
Let MM(x) be the molar mass of x.
MM(Pb) : MM(PbO)
=207.21 : 223.20 = 451.4 g : x g
cross multiply and solve for x
x=223.2/207.21*451.4
= 486.23 g
Percentage yield = 365.0/486.23= 0.75067 = 75.07% (rounded to 4 sign. fig.)
In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.
Let the mass of liquid be 100 g thus, mass of carbon, hydrogen and oxygen will be 34.31 g, 5.28 g and 60.41 g respectively.
To calculate molecular formula of compound, convert mass into number of moles as follows:

Molar mass of carbon, hydrogen and iodine is 12 g/mol, 1 g/mol and 126.90 g/mol.
Taking the ratio:

Putting the values,

Thus, molecular formula of compound will be
.