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netineya [11]
2 years ago
14

What is the mass in grams of 6.00 × 10¹⁵ molecules of CS₂?

Chemistry
1 answer:
Elenna [48]2 years ago
3 0

6,160.506

Explanation:

That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively.

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A chemistry student needs of acetone for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student dis
galben [10]

Answer:

The answer is "7.90 \ g".

Explanation:

Please finds the complete question in the attached file.

Volume = 10.0 \ mL \\\\Density =0.79 g\  cm^3

\text{Density}=\frac{Mass}{Volume}

Mass = Density \times Volume

         =(0.790 g\ cm^3) \times 10.0 \ mL\\\\=7.90 \ g

The mass of acetone to be weighed is 7.90 \ g

8 0
2 years ago
V1 = l; P1=40Pa; P2= 100 kpa; V2= 1.0 L please help!
blsea [12.9K]

Answer:

V1 = 2500 L

Explanation:

V1P1 = V2P2

x * 40 = 100.000 * 1.0

==> x =2500 L

6 0
3 years ago
Calculate the number of milliliters of 0.440 M KOH required to precipitate all of the Fe2+ ions in 187 mL of 0.692 M FeSO4 solut
EleoNora [17]

Answer:

588.2 mL

Explanation:

  • FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)

First we <u>calculate how many Fe⁺² moles reacted</u>, using the given <em>concentration and volume of FeSO₄ solution</em> (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):

  • moles = molarity * volume
  • 187 mL * 0.692 M = 129.404 mmol Fe⁺²

Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:

  • 129.404 mmol Fe⁺² * \frac{2mmolKOH}{1mmolFeSO_4} = 258.808 mmol KOH

Finally we<u> calculate the required volume of KOH solution</u>, using <em>the given concentration and the calculated moles</em>:

  • volume = moles / molarity
  • 258.808 mmol KOH / 0.440 M = 588.2 mL
6 0
2 years ago
VA 19.75-g sample was heated by 12.35 calories. The specific heat of the sample is 0.125 cal/g°C. What was the initial temperatu
SOVA2 [1]

Answer:

31.9 °C  

Explanation:

The formula for the heat q absorbed by an object is

q = mCΔT where ΔT = (T₂ - T₁)

Data:

q = 12.35 cal

m = 19.75 g

C = 0.125 cal°C⁻¹g⁻¹

T₂ = 37.0 °C

Calculations

(a) Calculate ΔT

q = mCΔT

12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT

12.35 = 2.406ΔT °C⁻¹  

ΔT  = 12.35/(2.406 °C⁻¹) = 5.13 °C

(b) Calculate T₂

ΔT = T₂ - T₁

T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C

The original temperature was 31.9 °C.

 

6 0
2 years ago
Why would Magnesium Phosphate (Mg3(PO4)2) not make an aqueous solution? <br><br> Please help!
netineya [11]

Answer:

Basically, all phosphates except Sodium phosphates, Potassium phosphates and Ammonium phosphates are insoluble in water. That, of course, includes Magnesium phosphate.

Explanation:

Hope this helped!

5 0
3 years ago
Read 2 more answers
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