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Nataly [62]
3 years ago
10

Suppose there are two known compounds containing the generic elements X and Y. You have a 1.00-g sample of each compound. One sa

mple contains 0.25 g of X and the other contains 0.33 g of X. Identify plausible sets of formulas for these two compounds. Check all that apply.
Chemistry
1 answer:
Lapatulllka [165]3 years ago
8 0

I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y, for the following reason: 

If you look at the combining masses of X and Y in each of the two compounds, 

The first compound contains 0.25g of X combined with 0.75g of Y 
so the ratio (by mass) of X to Y = 1 : 3 

The second compound contains 0.33 g of X combined with 0.67 g of Y 
so the ratio (by mass) of X to Y = 1 : 2 

Now, you suppose to prepare each of these two compounds, starting with the same fixed mass of element Y ( I will choose 12g of Y for an easy calculation!) 

The first compound will then contain 4g of X and 12g of Y 
The second compound will then contain 6g of X and 12g of Y 

<span>The ratio which combined the masses of X and the fixed mass (12g) of Y
= 4 : 6 
<span>or 2 : 3 </span>

So, the ratio of MOLES of X which combined with the fixed amount of Y in the two compounds is also = 2 : 3 </span>

The two compounds given with the plausible formula must therefore contain the same ratio.

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If the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn
pishuonlain [190]
Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol. 

<span>[mW (m) - mW (g)]/mW (m) x 100% </span>

<span>(they want % error so, if it is negative, just get rid of the sign) </span>
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A negative △H means an exothermic reaction.<br><br> True<br> False
Nostrana [21]
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6 0
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A container of oxygen with a fixed volume has a pressure of 13.0 atm at a temperature of 20 °C. What will the pressure of the ox
Norma-Jean [14]

Given:

P1 = 13.0 atm

T1 = 20 °C

T2 = 102 °C

Required:

P2 of oxygen

Solution:

At constant volume, we can apply Gay-Lussac’s law of pressure and temperature relationship

P1/T1=P2/T2

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P2 = 66.3 atm

The answer is not in the choices given.

6 0
3 years ago
(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct
Zarrin [17]

Explanation:

The given data is as follows.

   \Lambda^{o}_{m}(NaCl) = 1.264 \times 10^{-2}

   \Lambda^{o}_{m}(H-O=C-ONO) = 1.046 \times 10^{-2}

   \Lambda^{o}_{m}(HCl) = 4.261 \times 10^{-2}

Conductivity of monobasic acid is 5.07 \times 10^{-2} S m^{-1}

     Concentration = 0.01 mol/dm^{3}

Therefore, molar conductivity (\Lambda_{m}) of monobasic acid is calculated as follows.

                 \Lambda_{m} = \frac{conductivity}{concentration}

                                  = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}

                                 = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}

                                 = 5.07 \times 10^{-3} S m^{2} mol^{-1}

Also, \Lambda^{o}_{m} = \Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}

                            = 4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}

                            = 4.043 \times 10^{-2} S m^{2} mol^{-1}

Relation between degree of dissociation and molar conductivity is as follows.

               \alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}

                             = \frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}

                             = 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

Putting the values into the above formula we get the following.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

                        = \frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}

                        = 0.017973 \times 10^{-2}

                       = 1.7973 \times 10^{-4}

Hence, the acid dissociation constant is 1.7973 \times 10^{-4}.

Also, relation between pK_{a} and K_{a} is as follows.

                 pK_{a} = -log K_{a}

                              = -log (1.7973 \times 10^{-4})

                              = 3.7454

Therefore, value of pK_{a} is 3.7454.

                             

3 0
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