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sergejj [24]
3 years ago
6

A student prepares an iron standard solution in a 50.00 mL volumetric flask. The iron stock concentration is 0.0001974 M. A stud

ent dispenses 29.80 mL of stock iron solution into the volumetric flask, adds the reagents, and dilutes to volume with H2O. The concentration of the iron standard is:
Chemistry
1 answer:
Serhud [2]3 years ago
4 0

Answer:

0.00011765 M

Explanation:

When a solution is prepared by dilution, the volumes and concentrations are related by:

C1*V1 = C2*V2

Where C1 is the concentration of the solution 1, V1 is the volume of the solution 1, C2 is the concentration of solution 2, and V2 is the volume of solution 2.

The stock solution is the solution 1, and the standard solution, the solution 2, so:

0.0001974*29.80 = C2*50.00

C2 = 0.00011765 M

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To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, how many moles of nitrogen gas (n2) are required?
evablogger [386]

To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, 3.0 moles of nitrogen gas (n2) are required.

<h3>What are moles?</h3>

The mole is a SI unit of measurement that is used to calculate the quantity of any substance.

<h3 />

The given reaction is \rm  N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

By the stoichiometry rule of ratio hydrogen: nitrogen

3 : 1

The reacted moles of nitrogen is equals to H/3 moles of reacted hydrogen

So, moles of nitrogen  

\rm Moles\; of\; nitrogen = \dfrac{9.0 }{3} =3.0\;mol

Thus, 3.0 moles of nitrogen gas (n2) are required.

Learn more about moles

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3 0
2 years ago
1. The electron configuration below represents the ground state of an atom.
Natalka [10]

Answer:

{S}^{16}

group 16 period 2 of the periodic table

note: that is not the electronic configuration, that is the Bohr model.

4 0
2 years ago
a student determined that 8.2 milligrams of oxygen is dissolved in a 1000 gram sample of water at 15 degrees Celsius and 1 atm w
melamori03 [73]

Answer: We do not know. We have not been given the solubility of oxygen in water at a given temperature nor have we been given the Henry's laws constant. We also do not know whether you mean 1 atmosphere of air, or 1 atmosphere of oxygen.

8 0
3 years ago
Boiling point of a solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water. kB= 0.512 c/m.
Nookie1986 [14]

100.133 degree celsius is the boiling point of the solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water.

Explanation:

Balanced eaquation for the reaction

CaCl2 + 2H20 ⇒ Ca(OH)2 + HCl

given:

mass of CaCl2 = 15.2 grams

mass of the solution = 57 grams

Kb (molal elevation constant) = 0.512 c/m

i = vont hoff factor is 1 as 1 mole of the substance is given as product.

Molality is calculated as:

molality = \frac{grams of solute}{grams of solution}

              = \frac{15.2}{57}

               = 0.26 M

Boiling point is calculated as:

ΔT = i x Kb x M

     = 1 x 0.512 x 0.26

      = 0.133 degrees

The boiling point of the solution will be:

100 degrees + 0.133 degrees (100 degrees is the boiling point of water)

= 100.133 degree celcius is the boiling point of mixture formed.

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3 years ago
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4 years ago
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