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NNADVOKAT [17]
3 years ago
15

Please help, chemistry question

Chemistry
2 answers:
natulia [17]3 years ago
8 0
I think that should be evaporation :)
pashok25 [27]3 years ago
7 0

"NOT" the answer:

Try not getting it wrong idk what the answer is but i hope you do good

"NO" Explanation:

XD

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What is 1500 in scientific notation
Svet_ta [14]
To form the value of 1500 in scientific notation, it is <span><span>
1.    </span>By simply moving the period which separates the whole numbers from the decimal numbers between 1 and 5. </span> <span><span>
2.    </span>Thus, it then becomes 1.5 </span> <span><span>
3.    </span>Next, is you have to count how many moves the period made from its point of origin hence, for this value is 3</span>
<span><span>4.    </span>Therefore, the scientific notation for the number is 1.5 x 10^3</span>



7 0
3 years ago
What is a proper adjective?
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This is what a proper adjective is.

7 0
3 years ago
Calculate the concentration of the following solution in mol/dm3 0.1 moles of NaCl in 200 cm3
taurus [48]
1 cm ^{3} = 0.001 dm ^{3}. Therefore 200 cm ^{3} = 0.2 dm ^{3}. Molarity = \frac{number of moles of NaCl}{volume of the solution} =  \frac{0.1}{0.2} = 0.5 mol/dm^{3}. Hope this helps.
8 0
3 years ago
A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the
baherus [9]

Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
3 years ago
How many moles of methane are in 20.32 x 10^16 molecules
Marizza181 [45]

Taking into account the definition of avogadro's number, 3.37×10⁻⁷ moles of methane are 20.32×10¹⁶ molecules.

First of all, you have to know that Avogadro's number indicates the number of particles of a substance (usually atoms or molecules) that are in a mole.

Its value is 6.023×10²³ particles per mole and it applies to any substance.

Then you can apply the following rule of three: if 6.023×10²³ molecules are contained in 1 mole of methane, then 20.32×10¹⁶ molecules are contained in how many moles of methane?

amount of moles of methane= (20.32×10¹⁶ molecules × 1 mole)÷ 6.023×10²³ atoms

Solving:

<u><em>amount of moles of methane= 3.37×10⁻⁷ moles</em></u>

Finally, 3.37×10⁻⁷ moles of methane are 20.32×10¹⁶ molecules.

Learn more about Avogadro's Number:

  • <u>brainly.com/question/11907018?referrer=searchResults </u>
  • <u>brainly.com/question/1445383?referrer=searchResults </u>
  • <u>brainly.com/question/1528951?referrer=searchResults</u>
7 0
2 years ago
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