Question 3 (0.1 points)

Question 2(Multiple Choice Worth 4 points)

densk [106]

Answer:

B. Construct a hypothesis

6 0

3 years ago

Read 2 more answers

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

Mandarinka [93]

The answer is: [D]: " 417 cm³ " .
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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" ;
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       V₁ /T₁= V₂ /T₂ ; Multiply EACH side of the equation by "T₂ " :

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

↔ T₂  (V₂ /T₂) = T₂ (V₁ /T₁) ;

     → V₂ = T₂ (V₁ /T₁) ;
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Now, plug in our known values, to solve for "V₂" ;
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    → V₂ = T₂ (V₁ /T₁) ;
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    → V₂ = 308 K ( 556 cm³ /278 K) ;
             → The units of "K" cancel to "1" ; and we have:
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    → 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" ;
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              →      (116,000) / (278) = 417.2661870503597122 ;
                                 → round to 3 significant figures; → "417 cm³ " ;
                                               → which corresponds with "choice [D]".
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The answer is: [D]: "417 cm³ " .
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3 0

3 years ago

A balloon has a volume of 125mL with 10 pumps of gas. The balloon is reduced in volume to 88mL, how much gas is in the balloon?

blagie [28]

Answer:

7.04

Explanation:

i think

6 0

3 years ago

Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4

inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

$M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol$

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

So you can solve for the moles of the solute:

$n_{solute}=M*V_{solution}$

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

$n_{solute}=1mol/L*1.5L=1.5mol$