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tigry1 [53]
3 years ago
8

I really need help for this chemistry question. Steps and how it was worked out would be helpful!

Chemistry
1 answer:
Aneli [31]3 years ago
6 0
Data:
V (volume) = 500.0 mL = 0.5 L
T (temperature) = 15.00ºC
(converting in Kelvin) → TK = TC + 273 → TK = 15 + 273 = 288 K
P (pressure) = 736.0 mmHg
R (constant) = 62.363 (mmHg*L/mol*K)
m (mass) = 2.688 g
M (Molar Mass) = ? (g/mol)

Formula: General Gas Equation
P*V = n*R*T \to \boxed{P*V =  \frac{m}{M} *R*T}

Solving:
P*V = \frac{m}{M} *R*T
736*0.5 =  \frac{2.688}{M} *62.363*288
368 = \frac{2.688}{M} *17960.544
Product of extremes equals product of means:
368*M = 2.688*17960.544
368M = 48277.94227
M =  \frac{48277.94227}{368}
M = 131.1900605\to \boxed{\boxed{M \approx 131.2\:g/mol}}

Therefore: <span>The gas found to have such a molar mass is xenon gas</span>



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Draw the products obtained from the reaction of 1 equivalent of HBr with 1 equivalent of 2,5-dimethyl-1,3,5-hexatriene. Draw the
fgiga [73]

Answer:

Here's what I get.

Explanation:

According to Markovnikov's rule, the H will add to a terminal carbon, generating three resonance stabilized carbocations.

The Br⁻ ion will add to any of the three carbocations.

There are three possible products:

  1. 5-bromo-2,5-dimethylhexa-1,3-triene (1)
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6 0
3 years ago
(a) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm3. (b) A student needs 15.0 g of ethanol fo
o-na [289]

<u>Answer:</u>

<u>For a:</u> The density of the sample of copper is 8.96g/cm^3

<u>For b:</u> The volume of ethanol needed is 19.0 mL

<u>For c:</u> The mass of mercury is 340. grams

<u>Explanation:</u>

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}      ......(1)

  • <u>For a:</u>

Mass of copper = 374.5 g

Volume of copper = 41.8cm^3

Putting values in equation 1, we get:

\text{Density of copper}=\frac{374.5g}{41.8cm^3}\\\\\text{Density of copper}=8.96g/cm^3

Hence, the density of the sample of copper is 8.96g/cm^3

  • <u>For b:</u>

Mass of ethanol = 15.0 g

Density of ethanol = 0.789 g/mL

Putting values in equation 1, we get:

0.789g/mL=\frac{15.0g}{\text{Volume of ethanol}}\\\\\text{Volume of ethanol}=\frac{15.0g}{0.789g/mL}=19.0mL

Hence, the volume of ethanol needed is 19.0 mL

  • <u>For c:</u>

Volume of mercury = 25.0 mL

Density of mercury = 13.6 g/mL

Putting values in equation 1, we get:

13.6g/mL=\frac{\text{Mass of mercury}}{25.0mL}\\\\\text{Mass of mercury}=(13.6g/mL\times 25.0mL)=340.g

Hence, the mass of mercury is 340. grams

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3 years ago
Which of the following is NOT a binary molecular compound?<br> KI <br> H2s<br> CO<br> CIF3
shutvik [7]
I believe KI is not a a binary molecule.
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