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Sliva [168]
2 years ago
14

Warm Up: Compete this conversion. Put just the number below *

Chemistry
1 answer:
Gwar [14]2 years ago
5 0

Answer:

4 1/2

Explanation:

Use a ratio to find your answer

 4          6

-----  =  -------

 3           x

Cross multiply to solve for x.

4x = 18

x = 18/4

x = 4 2/4  which is the same as 4 1/2

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Using the reaction below determine the amount of Sulfur proceduced in grams (this is a limiting reaction question) 2 H2S + SO2 -
marusya05 [52]

Answer:

<u>= 2.2 g pf S. produced</u>

Explanation:

Balanced Reaction equation:

2H_{2} S + SO_{2} →  3S + 2H_{2} O

1 mole of H2S - 34.1g

? moles - 3.2g

= 3.2/34.1 =<u> 0.09 moles of H2S</u>

Also,

1 mole of S02 - 64.07 g

? moles - 4.42g

= 4.42/64.07 <u>= 0.069 moles of SO2</u>

<u />

<em>Meaning SO2 is the limiting reagent</em>

Finally, 3 moles of S -  32g of sulphur

0.069 mole = ? g of Sulphur

= 0.069 x 32

<u>= 2.2 g pf S.</u>

7 0
3 years ago
How are new kinds of matter formed in a chemial change?
shtirl [24]
Because the reactants react with each other and chemically react to produce a different product like with fire when it reacts with the wood it burns and the product left behind is a new different substance from the reactants in this case the product is ash
7 0
3 years ago
Read 2 more answers
Write the chemical formula, condensed formula, and Lewis structure for each of the following hydrocarbons:
son4ous [18]

Explanation:

Here I send you all the 3 elements that you are asking for.

Notice that first on the left you will find lewis structure, then condensed and finally chemical formula of each of the compound you enlisted.

5 0
2 years ago
Consider that calcium metal reacts with oxygen gas in the air to form calcium oxide. suppose we react 7.43 mol calcium with 4.00
Natali [406]
The balanced chemical equation for the above reaction is as follows;
2Ca + O₂ --> 2CaO
stoichiometry of Ca to O₂ is 2:1
this means that 2 mol of Ca reacts with 1 mol of O₂.
If O₂ is the limiting reactant, 
4 mol of O₂ should react with (4x2) - 8 mol of Ca
however only 7.43 mol of Ca is present. Therefore Ca is the limiting reactant.
7.43 mol of Ca reacts with - 7.43/2 = 3.715 mol of O₂
therefore there's excess O₂₂ remaining after the reaction
Since Ca is the limiting reactant, it is fully used up in the reaction and there is no Ca remaining after the reaction is completed.


4 0
2 years ago
A closed vessel system of volume 2.5 L contains a mixture of neon and fluorine. The total pressure is 3.32 atm at 0.0°C. When th
gayaneshka [121]

Answer:

moles Ne = 0.154 mol

moles F₂ = 0.217 mol

Explanation:

Step 1: Data given

Volume of the vessel system = 2.5 L

Total pressure = 3.32 atm at 0.0 °C

The mixture is heated to 15.0 °C

The entropy of the mixture increases by 0.345 J/K

The heat capacity of monoatomic gas = 3/2R and that for a diatomic gas = 5/2R

Step 2: Define the gas

Neon is a monoatomic gas, composed of Ne atoms

 ⇒ Cv(Ne) ≅ (3/2)R

Fluorine is a diatomic gas, composed of F₂ molecules.  

⇒ Cv(F₂) ≅ (5/2)R

Step 3: Calculate moles of gas

p*V = n*R*T

⇒ with p = the total pressure = 3.32 atm

⇒ with V = the total volume = 2.5 L

⇒ with n = the number of moles of gas

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 273.15 Kelvin

n(total) = p*V/RT = (3.32 atm*2.5 L)/(0.08206 L*atm/mol•K*273.15) = 0.3703 mol

Step 4: Calculate moles of Ne and F2

For one mole heated at constant volume,  

∆S = Cv*ln(288.15/273.15) = 0.05346*Cv

⇒ ∆S for 0.3703 mol,  

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

 ⇒ Cv = 17.43 J/mol*K for the Ne/F₂ mixture.

For pure Ne, Cv = (3/2)R = 1.5*8.314 J/mol*K = 12.471 J/mol*K

For pure F₂, Cv = (5/2)R = 2.5 * 8.314 J/mol*K = 20.785 J/mol*K

if X is the mole fraction of Ne, we can find X by:

17.43 J/mol*K = X* 12.471 J/mol*K + (1 – X) * 20.785 J/mol*K

 ⇒ 20.875 – 8.314 * X = 17.43

X = 0.415 , 1 – X = 0.585

moles Ne = (0.415)(0.3703 mol) = 0.154 mol

moles F₂ = (0.585)(0.3703 mol) = 0.217 mol

4 0
3 years ago
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