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koban [17]
3 years ago
10

A bottle of acetone (fingernail polish remover) contains 55.0 mL of acetone. If the acetone has a mass of 43.5 g, what is the de

nsity of acetone?
Chemistry
1 answer:
mihalych1998 [28]3 years ago
5 0
If you were to use the formula to find density which is D=M/V and plug in your numbers M which stands for mass and and plug in your numbers for V which what’s for Volume and should be a number in mL and then divide them you should get your answer. So 43.5g divided by 55.0mL and your density would = 0.79g/mL
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A solution has a pH = 4.63. What is the [H3O+] concentration?
LUCKY_DIMON [66]

Explanation:

The pH of a solution can you be found by using the formula

pH = - log [ { H_3O}^{+}]

Since we are finding the [H3O+] , substitute the value of the pH and find it's antilog

We have

4.63 =  -  log[ { H_3O}^{+}] \\ [ { H_3O}^{+}]   =  {10}^{ - 4.63}  \\   \\  = 2.344 \times  {10}^{ - 5} mol {dm}^{ - 3}

Hope this helps you

7 0
3 years ago
What is the energy required to remove the electron from a hydrogen atom in the n = 11 state?
zaharov [31]

Based on the data given, the energy required to remove an electron from a hydrogen atom in the n = 11 state is -0.112 eV

<h3>What is ionization energy?</h3>

Ionization energy is the energy requiredto remove an electron from a neutral atom or a cation in its gaseous state.

To calculate the energy required to remove the electron from a hydrogen atom in the n = 11 state, the formula below is used:

  • E_n = \frac{E_0 × Z^{2}}{n^{2}}

where

E_0  \: is  \: ground  \: state  \: energy  \: of \:  hydrogen \:  atom \:  = -13.6 eV \\

  • Z = 1
  • n = 11

substituting the values:

E_n = \frac{ - 13.6 × 1^{2}}{11^{2}} \:  =  - 0.112 \: eV

Therefore, the energy required to remove an electron from a hydrogen atom in the n = 11 state is -0.112 eV

Learn more about ionization energy at: brainly.com/question/1445179

5 0
2 years ago
Help with questions
gulaghasi [49]

1) This type of energy transfer is called conduction.

2) Hydroelectricity

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8 0
4 years ago
What would be the freezing point of a solution that has a molality of 1.468 m which was prepared by dissolving biphenyl (C12H10)
Readme [11.4K]

Answer:

Freezing point solution = 70.131 °C

Explanation:

Step 1: Data given

Molality = 1.468 molal

A solution is created by dissolving biphenyl (C12H10) into naphthalene

Biphenyl is a non-electrolyte

Freezing point of naphthalene = 80.26 °C

Step 2: Calculate the freezing point depression

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = TO BE DETERMINED

⇒with i = the van't Hoff factor of biphenyl = 1

⇒with Kf = the freezing point depression constant of naphthalene = 6.90 °C/m

⇒with m = the molality = 1.468 molal

ΔT = 1 * 6.90 °C/m * 1.468 °C

ΔT = 10.13 °C

Step 3: Calculate the freezing point of the solution

ΔT = 10.13 °C

Freezing point solution = freezing point naphthalene - 10.13 °C

Freezing point solution = 80.26 °C - 10.129 °C

Freezing point solution = 70.131 °C

5 0
4 years ago
What is an example
Paha777 [63]

Answer:

NaF sodium flouride

Explanation:

7 0
3 years ago
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