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Serggg [28]
3 years ago
7

What is chemistry n laboratory​

Chemistry
1 answer:
grin007 [14]3 years ago
3 0
a laboratory for research in chemistry. chem lab, chemistry lab. lab, laboratory, research lab, research laboratory, science lab, science laboratory - a workplace for the conduct of scientific research.
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What is the square root of 10<br><br> Can you guys subscribe to my friend pls, it would mean a lot
Korolek [52]

Answer:

3.16228

Explanation:

3 0
2 years ago
List 3 household items. Identify 2 physical properties of each item.
ioda

Answer:

Water

Explanation:

Solid, liquids, and gases: Water can exist in several states, including ice (solid), water (liquid), and water vapor (gas). color (intensive)

density (intensive)

volume (extensive)

mass (extensive)

boiling point (intensive): the temperature at which a substance boils

melting point (intensive): the temperature at which a substance melts

3 0
3 years ago
Read 2 more answers
Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine
Yuliya22 [10]

Answer:

  • <u>194 g/mol</u>

Explanation:

<u>1) Content of C:</u>

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

  • Mass of C in 1.813 mg of CO₂

       Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

       12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

        ⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

  • Number of moles of C

      number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

<u>2) Content of H</u>

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

  • Mass of H in 0.4639 mg of H₂O

       Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

       2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

        ⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

  • Number of moles of H

      number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

<u>3) Content of N</u>

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

  • Mass of N in 0.2885 mg of N₂ is 0.2885 mg

  • Number of moles of N

      number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

<u>4) Content of O</u>

The mass of O is calculated by difference:

  • Mass of O = mass of sample - mass of C - mass of H - mass of N

       Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

     Mass of O = 0.1648 mg

  • Moles of O =  0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

<u>5) Ratios</u>

Divide every number of mililmoles by the smallest number of milimoles:

  • C:  0.041195 / 0.01030 = 4
  • H: 0.051501 / 0.01030 = 5
  • N: 0.020597 / 0.01030 = 2
  • O: 0.01030 / 0.01030 = 1

  • C: 4
  • H: 5
  • N: 2
  • O: 1

<u>6) Empirical formula:</u>

  • C₄H₅N₂O₁

<u>7) Calculate the approximate mass of the empirical formula:</u>

  • 4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 =  97 g/mol

So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

Thus, the estimate is 194 g/mol

7 0
3 years ago
What is the pressure of 1.27 L of a gas at 288°C, if the gas had a volume of 875 ml at
lyudmila [28]

The pressure of 1.27 L of a gas at 288°C, if the gas had a volume of 875 ml at 145 kPa and 176°C is 1.195 atm.

<h3>What is ideal gas equation?</h3>

Ideal gas equation of any gas will be represented as:
PV = nRT, where

P = pressure

V = volume

n = moles

R = universal gas constant

T = temperature

First we calculate the moles of gas, when the volume of gas 875 ml at

145 kPa and 176°C as:

n = (1.431atm)(0.875L) / (0.082L.atm/K.mol)(449.15K)

n = 1.252 / 36.83 = 0.033 moles

Now we measure the pressure of 0.033 moles of gas of 1.27 L of a gas at 288°C as:

P = (0.033mol)(0.082L.atm/K.mol)(561K) / (1.27L) = 1.195 atm

Hence required pressure of gas is 1.195 atm.

To know more about ideal gas equation, visit the below link:
brainly.com/question/555495

#SPJ1

3 0
2 years ago
Percent composition of Mg(OH)2​
LuckyWell [14K]

Answer:

The composition of Mg(OH)2 is

41.67 % Mg,

54.87 % O

3.457 % H

Explanation:

Thats the percent composition for each one in your question

4 0
2 years ago
Read 2 more answers
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