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sashaice [31]
3 years ago
13

A 6.26 g sample of a solid containing ni is dissolved in 20.0 ml water. a 5.00 ml aliquot of this solution is diluted to 100.0 m

l and analyzed in the lab. the solution was determined to contain 5.28 ppm ni. determine the molarity of ni in the 20.0 ml solution.
Chemistry
1 answer:
saveliy_v [14]3 years ago
5 0
Concentration of Ni in 20mL = 5.28ppm x dilution factor = 5.28 x 100/5 = 105.6 ppm = 105.6 mg/L 

molar mass of Ni = 58.6934 g
<span>Molarity of Ni = 100.40 x 10^{-3} / 58.6934 = 1.71 x 10^{-3} M = 1.71 mM. </span>
You might be interested in
How many moles of sodium hydroxide are there in 1.0mL of 2.0M NaOH
kvasek [131]
To find the moles, you can use the following formula

moles= Molarity x Liters

Molarity= 2.0 M
Liters= 0.0010 Liters  ---------------->>>>>>>>>> 1.0 mL= 0.0010 Liters

moles= 2.0 M x 0.0010 Liters= 0.0020 moles
4 0
2 years ago
The complete orbital notation diagram of an atom is shown.
Elan Coil [88]

values of the quantum numbers: -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6

location of the electron: In the 7th energy level away from the nucleus.

Explanation:

From the description of the problem, the magnetic number is given is as -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6 and the electron is located in the 7th energy level away from the nucleus. Basically, the problem is testing for the understanding of the principal quantum numbers which gives the location of electrons and the magnetic quantum number that shows the spatial orientation of the orbitals.

           The orbital designation of the describe electron is 7d

  1. Magnetic quantum number is limited by the azimuthal quantum number which is the quantum number describing the possible shapes. The azimuthal is given as L= n-1. "n" is the principal quantum number which is 7. Therefore L is 6 and the magnetic quantum numbers are -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6
  2. The position of the electron is given by the principal quantum number which represents the main energy level in which the orbital is located or the average distance from the nucleus. Here it is 7.

Learn more:

brainly.com/question/9288609

#learnwithBrainly

5 0
3 years ago
A gas at 300 k and 4.0 atm is moved to a new loacation with a temperature of 250 k. The volume changes from 5.5 L to 2.0 L. What
alexgriva [62]
Answer is: <span>the pressure of the gas is 9,2 atm.
</span>p₁ = 4,0 atm.
T₁ = 300 K.
V₁ = 5,5 L.
p₂ = ?
T₂ = 250 K.
V₂ = 2,0 L.
Use combined gas law - the volume of amount of gas is proportional to the ratio of its Kelvin temperature and its pressure.<span> 
</span>p₁V₁/T₁ = p₂V₂/T₂.
4 atm · 5,5 L ÷ 300 K  = p₂ · 2,0 L ÷ 250 K.
0,0733 = 0,008p₂.
p₂ = 9,2 atm.

4 0
3 years ago
A student has a large number of coins of different diameters, all made by the same metal. She wishes to find the density of the
Elenna [48]

Answer:

A. In a graduated cylinder, put some quantity of water and measure the initial volume. Then put a coin and measure the volume. To find the volume of the coin, simply subtract the initial volume (water only) from the ending volume (water + coin). To measure the mass, take a dry coin and place it on an electronic scale. Density = mass / volume, so divide the mass by the volume to calculate the density of the coin.

B. When measuring the volume, make sure to look at the graduated cylinder at eye level and read from the bottom of the meniscus.

6 0
2 years ago
1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to
Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant,  NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

MM:        17.03    32.00     30.01

              4NH₃  +  5O₂ ⟶ 4NO + 6H₂O

Mass/g:    1.5        1.85

2. Calculate the moles of each reactant  

\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

\text{Moles of NO} =  \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}

7. Calculate the mass of NH₃ reacted

\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0
2 years ago
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