17.85 mL of 3.0 M stock solution of nitric acid is needed to make a 150 mL 0.357 M nitric acid.
The dilution of a stock solution can be solved by using the formula below, which indicates that the concentration of the solution changes but the number of moles of the solute remains the same.
(concentration of stock solution)(volume of stock solution) = (concentration of diluted solution)(volume of diluted solution)
Given:
- Stock solution: 3.0 M nitric acid
- Diluted solution: 150 mL of 0.357 M nitric acid
(3.0 M)(volume of stock solution)=(0.357 M)(150 mL)
volume of stock solution = [(0.357 M)(150 mL)] / (3.0 M)
volume of stock solution = 17.85 mL
To learn more about the dilution process, please refer to brainly.com/question/27097060.
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Answer: 2.26x 10²⁴ atoms C
Explanation: To get the number of atoms,
First we convert mass of C to moles
45g C x 1mole C / 12 g C
= 3.75 moles C
Cancel g of C
Next convert moles of C to atoms using Avogadro's Number.
3.75 moles C x 6.022x10²³ atoms C / 1 mole C
Cancel out the units of moles C.
= 2.26x 10²⁴ atoms of C
Answer:
(c.) erythrose
Explanation:
Erythrose 4-phosphate is an intermediate in the pentose phosphate pathway and the Calvin cycle
<span>Cell references in a formula are called electrode potentials. Electrode potentials are standard voltages for half-cell potentials. All half-cell potentials use the Standard Hydrogen Electrode (SHE) as a universal reference for redox reactions. For hydrogen, the standard electrode potential is zero for all temperatures.</span>
H20-h&o
Gold-au
Nitric acid-h,n,o
Cobalt-co
Oxygen-o
MAKE SURE THEY ARE CAPITALIZED