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andriy [413]
3 years ago
15

A 10.00 mL sample of 8.50 M HNO3 solution is diluted to a new volume of 65.0 mL. What is the concentration of the dilute solutio

n
Chemistry
1 answer:
loris [4]3 years ago
4 0

Answer: 1.31M

Explanation:

V1 = 10mL

C1 = 8.5M

V2 = 65mL

C2 =?

C1V1 = C2V2

10 x 8.5 = C2 x 65

C2 = (10 x 8.5 ) /65

C2 = 1.31M

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Chris the chemist was working in the and he wanted to sou what would happen he put copper chloride into a Bunsen flameAs soon as
Reptile [31]

The question is incomplete, the complete question is;

Chris the Chemist was working in the lab, and he wanted to see what would happen if he put copper chloride into a Bunsen burner flame. As soon as he put it in the flame, it began to emit a green color. Which of these BEST describes this outcome?

A) The heat split the nucleus and created a new element.

B) As the copper chloride was heated, electrons were turned into neutrons.

C) As the electrons moved to a higher energy level, a photon of light is emitted.

D) When excited electrons return back to the ground state, a photon of light is emitted.

Answer:

D) When excited electrons return back to the ground state, a photon of light is emitted.

Explanation:

According to Bohr's model of the atom, electrons may absorb energy and move up to higher energy levels.

These electrons quickly return to ground state from such higher energy excited states thereby emitting the excess energy absorbed during excitation.

If the higher energy level is E2 and the ground state is E1 then the energy of the photon emitted when the electron descends from energy level E2 to E1 is;

ΔE = E2 - E1

This photon emitted accounts for the green colour of the copper salt observed when it was heated.

4 0
3 years ago
We can know what elements make up a star by studying the star's position in the
Amanda [17]
The answer is "night sky"
hope i helped :)
7 0
3 years ago
Isotope b has a half-life of 3 days. a scientist measures out 100 grams of this substance. after 6 days has passed, the scientis
Brilliant_brown [7]
So half life is the time taken for a sample to decay to half its original mass, its a constant and applies to any original mass, it could be 5g or 1kg, it will take the same amount of time for the original mass to half. In this case the half life is 3 days.

After 3 days the sample will be at half its original mass, now 50g. 

Now we can treat the 50g as if its a new sample. After another 3 days (6 days in total) there will be half of 50g left, = 25g. 


6 0
3 years ago
The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma
Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C=p_o= 187.54 mmHg

Vapor pressure of the solution at 65 °C= p_s

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water=n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol

Moles of ethylene glycol=n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol

\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}

\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}

p_s=173.83 mmHg

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0
3 years ago
Write a balanced equation for the complete oxidation reaction that occurs when ethane burns in air
damaskus [11]
<h3>Balanced equation : 2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)</h3><h3>Further explanation</h3>

Alkanes are saturated hydrocarbons that have single bonds in chains

General formula for alkanes :

\tt \large{\bold{C_nH_{2n+2}}

Hydrocarbon combustion reactions (specifically alkanes)

\large {\box {\bold{C_nH _ (_2_n _ + _ 2_) + \dfrac {3n + 1} {2} O_2 \Rightarrow nCO_2 + (n + 1) H_2O}}}

So that the burning of ethane with air (oxygen):

\tt C_2H_6+\dfrac{7}{2}O_2\rightarrow 2CO_2+3H_2O

2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)

or we can use mathematical equations to solve equilibrium chemical equations by giving the coefficients for each compound involved in the reaction

C₂H₆ (g) + aO₂ (g) ⟶ bCO₂ (g) + cH₂O (ℓ)

C : left 2, right b ⇒ b=2

H: left 6, right 2c⇒ 2c=6⇒ c= 3

O : left 2a, right 2b+c⇒ 2a=2b+c⇒2a=2.2+3⇒2a=7⇒a=7/2

6 0
3 years ago
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