Answer:
14 mL
Explanation:
To prepare a solution by a concentrated solution, we must use the equation:
C1xV1 = C2xV2, where <em>C</em> is the concentration, <em>V</em> is the volume, 1 is the initial solution and 2 the final solution.
The final solution must have 2 mL and a concentration of 350 pg/mL, and the initial solution has a concentration of 50 pg/mL.
Then:
50xV1 = 350x2
50xV1 = 700
V1 = 700/50
V1 = 14 mL
Answer:

Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

Best regards.
The information that the third quantum number of an electron gives is the direction the electron in spinning. That is option A.
<h3>What are quantum numbers?</h3>
Quantum number are those numbers that are used to specify the properties of the atomic orbitals and the electrons in those orbitals.
The types of quantum numbers include the following:
The third quantum number is shows the direction of the electron while spinning through specifying its angular momentum.
Learn more about orbitals here:
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We are given –
- Mass of
is 57.1 g and we are asked to find number of moles present in 57.1 g of 






____________________
Now,Let's calculate the number of moles present in 57.1 g of 






__________________________________
Answer:
3.68 grams.
Explanation:
First we <u>convert 9.5 g of NaCl into moles of NaCl</u>, using its<em> molar mass</em>:
9.5 g ÷ 58.44 g/mol = 0.16 mol NaCl
In<em> 0.16 moles of NaCl there are 0.16 moles of sodium </em>as well.
We now <u>convert 0.16 moles of sodium into grams</u>, using <em>sodium's molar mass</em>:
0.16 mol * 23 g/mol = 3.68 g