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2 years ago

Calculate the molecular mass of a 2.66g sample of gas at a temperature of pressure of 845torr and occupying a volume of 0.70L.

1 answer:

8 0

The molecular mass of a 2.66g sample of gas at a temperature of pressure of 845torr and occupying a volume of 0.70L is 84.16g/mol.

<h3>How to calculate molecular mass?</h3>

The molecular mass can be calculated by first calculating the number of moles using the following formula:

PV = nRT

Where;

P = pressure
V = volume
n = no of moles
R = gas law constant
T = temperature

1.112 × 0.7 = n × 0.0821 × 300

0.7784 = 24.63n

n = 0.032mol

molecular mass = 2.66g ÷ 0.032mol

molecular mass = 84.16g/mol

Therefore, the molecular mass of a 2.66g sample of gas at a temperature of pressure of 845torr and occupying a volume of 0.70L is 84.16g/mol.

Learn more about molar mass at: brainly.com/question/14122402

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Plan a chromotography experiment to investigate the colours in an ink

Usimov [2.4K]

Answer:Step 1: Gather the Needed Materials (coffee Filter, Pencil, Beaker, Water Soluble Pen/marker, Water) Step 2: Cut a Strip of Filter Paper From the Coffee Filter and Using the Marker Draw a Thick Line Near the Bottom of the Filter Paper- About 1/4 Inch From the Bottom. Step 3: Allow the Water to Move Up the Paper for Approx. 5 Minutes and Then Remove the Strip From the Water Step 4: Students Should Identify the Individual Colors in the Chromatogram That Make Up the Markers Unique Mix of Ink. Step 5: Once Students Have Mastered the Technique Its Time to Put Their Skills to the Test in "The Case of the Celebrity Dog-napping" Step 7: Write the Ransom Note on a Piece of Filter Paper in One of the Suspect Inks.

Explanation: Chromatography is the physical separation of a mixture into its individual components. This lesson integrates mixtures, solutions & solubility into an activity where students will identify the solute & the solvent. Because chromatography can be used to separate the components of inks & dyes, students are able to identify the unique "recipe" that makes up the mixture of a specific brand of pen. After the students are taught the ink chromatography technique, the fun can continue with a mini forensic investigation, "The Case of the Celebrity Dog-napping".

hopes this helps

6 0

3 years ago

Calculate the molarity of a solution consisting of 60.0 g of NaOH in 1.50 L of solution.

Trava [24]

Explanation:

here's the answer to your question

7 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago

What is the molar mass, in grams, of a mole of an element equal to?

insens350 [35]

Answer:

B. the atomic weight of the element

Explanation:

since it talks about mass i think weight

5 0

4 years ago

** worth 20 points + brainliest **

Bad White [126]

Answer : The correct option is, (D) $83^oC$

Solution :

Formula used :

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

where,

Q = heat released = -1300 J

m = mass of water = 40 g

c = specific heat of water = $4.18J/g^oC$

$\Delta T=\text{Change in temperature}$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $91.0^oC$

Now put all the given values in the above formula, we get the final temperature of water.

$-1300J=40g\times 4.18J/g^oC\times (T_{final}-91.0^oC)$

$T_{final}=83.2^oC\approx 83^oC$

Therefore, the final temperature of the water is, $83^oC$

5 0

3 years ago