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lyudmila [28]
3 years ago
7

A balloon contains a fixed amount of helium. If the initial conditions are 1.12 liters, 3.00 atmospheres, and 19 degrees Celsius

, what will the final volume be when the balloon is exposed to 1.00 atmosphere pressure and 22 degrees Celsius?
Chemistry
1 answer:
Semmy [17]3 years ago
4 0

Answer:

The final volume will be 3.39 liters

Explanation:

P1V1/T1 = P2V2/T2

P1 = 3 atm

T1 = 19( degrees celsius) + 273 = 292 K

V2 = TO BE DETERMINED

P2 =1 atm

T2 = 22( degrees celsius) + 273 = 295 k

P1V1/T1 = P2V2/T2

V2 = P1V1T2/P2T1 = (3)(1.12)(295)/(1)292) = 3.39 Liters

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kirill [66]

6400 cubic centimeters


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3 years ago
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ANSWER FAST PLEASE!<br> Balance the following equation:<br> __Hg + __O₂ --&gt; __HgO
ICE Princess25 [194]

Answer:

4Hg+2O2=4HgO

four Mercury + four oxygen

3 0
3 years ago
PLEASE HELP!!!!!!!
dsp73

Answer: 1. 2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu

2. 3 moles of CuCl_2 : 2 moles of Al

3. 0.33 moles of CuCl_2 : 0.92 moles of Al

4. CuCl_2 is the limiting reagent and Al is the excess reagent.

5. Theoretical yield of AlCl_3 is 29.3 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles

\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles

The balanced chemical equation is:

2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu  

According to stoichiometry :

3 moles of CuCl_2 require = 2 moles of Al

Thus 0.33 moles of CuCl_2 will require=\frac{2}{3}\times 0.33=0.22moles  of Al

Thus CuCl_2 is the limiting reagent as it limits the formation of product and Al is the excess reagent.

As 3 moles of CuCl_2 give = 2 moles of AlCl_3

Thus 0.33 moles of CuCl_2 give =\frac{2}{3}\times 0.33=0.22moles  of AlCl_3

Theoretical yield of AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3

Thus 29.3 g of aluminium chloride is formed.

5 0
2 years ago
Help please please please!
LenaWriter [7]

Answer:

I say the second one, im not rlly sure tho

Explanation:

5 0
3 years ago
How much heat is evolved in converting 1.00 mol of steam at 155.0 ∘c to ice at -50.0 ∘c? the heat capacity of steam is 2.01 j/(g
Ne4ueva [31]
When the specific heat capacity of the water is 4.18 J/g.°C so, we are going to use this formula to get the heat for cooling three  phases changes from steam to liquid and from liquid to ice (solid) :

when Q = M*C*ΔT 

Q is the heat in J

and M is the mass in gram = 1 mol H2O * 18 g/mol(molar mass) = 18 g

C is the specific heat J/g.°C

ΔT is the change in temperature

Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice  * ΔT ice)]

    = 18 g * [(2.01 * (155-100°C)) + (4.18 * (100-0°C)) + (2.09 * (0 - 55 °C))]

∴Q = 7444.8 J

and when we know that the heat of fusion for water = 334J/g

and heat of vaporization for water =  2260J/g


∴Q for the two phases changes = M * (2260+334) 

                                                      = 18 * (2260+334)

                                                      = 46692 J 

∴ Q total = 7444.8 + 46692 = 54136.8 J
5 0
3 years ago
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