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Reika [66]
3 years ago
10

A student increases the temperature of a 556 cm3 balloon from 278 K to 308 K. Assuming constant pressure, what should the new vo

lume of the balloon be? A --376cm3
B --462cm3
C --924cm3
D --417cm3
Chemistry
1 answer:
Mandarinka [93]3 years ago
3 0
The answer is:  [D]:  " 417 cm³ " .
_____________________________________________________
Explanation:  Use the formula:

V₁ /T₁= V₂ /T₂  ;

in which:  V₁ = initial volume = 556 cm³ ;
                T₁ = initial temperature = 278 K ;
                V₂ = final ("new") temperature = 308 K
                T₂ = final ("new:) volume = ?

Solve for  "V₂" ;

Since:  V₁ /T₁= V₂ /T₂ ;

We can rearrange this "equation/formula" to isolate "V₂" on one side of the equation; and then we can plug in our know values to solve for "V₂" ;
_______________________________________________________
       V₁ /T₁= V₂ /T₂  ;  Multiply EACH side of the equation by "T₂ " :

          →  T₂ (V₁ /T₁) = T₂  (V₂ /T₂) ;
______________________________
to get:

↔ T₂  (V₂ /T₂) = T₂ (V₁ /T₁) ;

     →  V₂ = T₂ (V₁ /T₁) ;
______________________________
Now, plug in our known values, to solve for "V₂" ;
______________________________
    →  V₂ = T₂ (V₁ /T₁) ;
______________________________
    →  V₂ = 308 K ( 556 cm³ /278 K)  ;
             → The units of "K" cancel to "1" ; and we have:
________________________________________________________
    →  V₂ = 308*( 556 cm³ / 278 ) = [(208 * 556) / 278 ] cm³ ;
Note:  We will keep the units of volume as:  "cm³ ";  since all the answer choices given are in units of:  "cm³ " ; {that is, "cubic centimeters"}.

   →  [(208 * 556) / 278 ] cm³ = [ (115,648) / (278) ] cm³ ;
                                        
              → For the "(115,648)" ;  round to "3 (three significant figures)" ;
                        → "(115,648)" → rounds to:  "116,000" ;
____________________________________________________
              →      (116,000) / (278) = 417.2661870503597122  ;
                                 → round to 3 significant figures; → "417 cm³ " ;
                                               → which corresponds with "choice [D]".
______________________________________________________
The answer is:  [D]:  "417 cm³ " .
______________________________________________________

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Answer: 2 moles

Explanation:

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Since we know that CO₂ has a 44.8 L container, we can use that to find the moles of oxygen.

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5 0
3 years ago
How many grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate?
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Answer: There is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.

Explanation:

Chemical equation depicting reaction between sodium phosphate and aluminum carbonate is as follows.

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As this equation contains same number of atoms on both reactant and product side. So, this equation is a balanced equation.

According to the equation, 2 moles of sodium phosphate is giving 3 moles of sodium carbonate.

Therefore, sodium carbonate formed by 5.3 moles of sodium phosphate is as follows.

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As number of moles is the mass of substance divided by its molar mass. So, mass of sodium carbonate ( molar mass = 105.98 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\7.95 mol = \frac{mass}{105.98 g/mol}\\mass = 842.54 g

Thus, we can conclude that there is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.

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