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zhuklara [117]
3 years ago
12

Please help! I have no idea what a galactic symbol is. It says it on one of my science assignments and I'm confused. All I know

is that it may have to do with the periodic table. If anyone could help that'll be great. My assignment is due today and I don't want to fail. (I have no Idea if this is Chemistry or biology. All I know is that its science. So I put Chemistry as the category. I apologize if I'm incorrect.)
Chemistry
2 answers:
gayaneshka [121]3 years ago
5 0
How is it used in the question?

Ilya [14]3 years ago
4 0
Hey, its alright bro! I gotchu.

Im pretty sure the galactic symbol is the same reference to the symbol of the element on the periodic table.
For an example (Hydrogen) is H 
So if im correct im pretty sure its referring to the symbols for each element

Hopefully this helps
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A chemical test has determined the concentration of a solution of an unknown substance to be 2.41 M. a 100.0 mL volume of the so
Oksana_A [137]

Answer : The molar mass of unknown substance is, 39.7 g/mol

Explanation : Given,

Mass of unknown substance = 9.56 g

Volume of solution = 100.0 mL

Molarity = 2.41 M

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of unknown substance}\times 1000}{\text{Molar mass of unknown substance}\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

2.41M=\frac{9.56g\times 1000}{\text{Molar mass of unknown substance}\times 100.0mL}

\text{Molar mass of unknown substance}=39.7g/mol

Therefore, the molar mass of unknown substance is, 39.7 g/mol

5 0
3 years ago
A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
natali 33 [55]

Answer:

There are 0.93 g of glucose in 100 mL of the final solution

Explanation:

In the first solution, the concentration of glucose (in g/L) is:

15.5 g / 0.100 L = 155 g/L

Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.

  • 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)

The concentration of the second solution is:

155 \frac{g}{L} *\frac{0.030L}{0.500L}=9.3\frac{g}{L}

So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:

1 L --------- 9.3 g

0.1 L--------- Xg

Xg = 9.3 g * 0.1 L / 1 L = 0.93 g

8 0
3 years ago
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