Answer:
0.5188 M or 0.5188 mol/L
Explanation:
Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.
***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".
The formula for molarity is:
n = moles (unit mol)
V = volume (unit L)
<u>Find the molar mass (M) of potassium hydroxide.</u>
<u>Calculate the moles of potassium hydroxide.</u>
Carry one insignificant figure (shown in brackets).
<u>Convert the volume of water to litres.</u>
Here, carrying an insignificant figure doesn't change the value.
<u>Calculate the concentration.</u>
<= Keep an insignificant figure for rounding
<= Rounded up
<= You use the unit "M" instead of "mol/L"
The concentration of this standard solution is 0.5188 M.
Answer is: beryllium-10.
The diagram shows that atom has 4 protons, 5 neutrons and 2 valence electrons.
Atomic number is the number of protons, which is characteristic of a chemical element, beryllium (Be) is an element with atomic number 4.
Two valence electrons means that atom is from 2. group of periodic table, only beryllium is from that group; sodium (1. group), boron (13. group) and carbon (14. group).
Beryllium-10 has 6 neutrons, so it is isotope (different number of neutrons or mass number).
Answer:
1: Balsatic
2:Rhyolitic
3:Andesitic
Explanation:
I did an investigation of the volcanoes
Direct Proportion and The Straight Line Graph.Straight<span> line graphs that go through the origin, like the one immediately below, show that the quantities on the graph are in direct proportion.
thanks
cbuck763
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3CH4 refers to three molecules of a
compound and C3H12 refers to only one
molecule of a compound.
3CH4 and C3H12 are actually different
molecules though they contain the same
number and types of atoms. Recall that both
of them are hydrocarbons hence they
contain only hydrogen and carbon.
3CH4 refers to three molecules of methane
while C3H12 refers to one molecule of
diphenyl methane. Hence 3CH4 refers to
three molecules of a compound and C3H12
refers to only one molecule of a compound.