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ololo11 [35]
2 years ago
7

How many grams of sulfur are required in the preparation of 35.7 moles of sulfur dioxide Sg + O2 SO2​

Chemistry
1 answer:
lidiya [134]2 years ago
6 0
  • S+O_2-->SO_2

Its balanced already

So

  • 1 mol SO_2 require 1 mole sulphur
  • 35.7moles require 35.7mol sulphur

Mass of S

  • 35.7(32)
  • 1142.4g
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Why does water flow decrease in your shower when someone flushes the toilet ​
Rashid [163]

Answer:

A toilet requires a large amount of cold water to flush. So, when the toilet flushes while you're in the shower, it's stealing from your shower's cold water supply. When the pressure-balancing valve senses the drop in cold water pressure, it responds by restricting the hot water pressure.

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3 years ago
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How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2
Leya [2.2K]

Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g  ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1  mol of KNO₃  gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4  * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

3 0
3 years ago
By friends leaving this app​
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Answer:

Whyyyyyyyyyy

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Determine the molar mass of a compound that has a density of 0.1633 g/L at STP.<br> (show work)
hodyreva [135]

Answer:

                     M.Mass  =  3.66 g/mol

Data Given:

                  M.Mass  =  M = ??

                  Density  =  d  =  0.1633 g/L

                  Temperature  =  T  =  273.15 K (Standard)

                  Pressure  =  P  =  1 atm (standard)

Solution:

              Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.

                                             P V = n R T    ---- (1)

Also, we know that;

                       Moles  =  n  =  mass / M.Mass

Or,                                   n  =  m / M

Substituting n in Eq. 1.

                                             P V = m/M R T   --- (2)

Rearranging Eq.2 i.e.

                                             P M = m/V R T   --- (3)

As,

                     Mass / Volume = m/V = Density = d

So, Eq. 3 can be written as,

                                             P M = d R T

Solving for M.Mass i.e.

                                             M = d R T / P

Putting values,

M  =  0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm

M  =  3.66 g/mol

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