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True [87]
2 years ago
5

1) If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature?

Chemistry
2 answers:
sergey [27]2 years ago
6 0

Answer:

204.8 K

Explanation:

We use the ideal gas equation:

PV = nRT

where R is the gas constant (0.082 L.atm/K.mol).

We have the following data:

n= 4 moles

P = 5.6 atm

V = 12 L

So, we introduce the data in the ideal gas equation to calculate the temperature (T):

T = PV/nR = (5.6 atm x 12 L)/(4 mol x 0.082 L.atm/K.mol) = 204.8 K ≅ -68 °C

ss7ja [257]2 years ago
5 0

Answer: 204.63 K

Explanation: Because it makes the most sense

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43 inHg = 43 inHg*2.54cm/in = 109.22cmHg * 10 mm/cm = 1092.2 mmHg


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3 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
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Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

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Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
An atom of an element forms a 2+ ion. In which group on the Periodic Table could this element be located?
Mashutka [201]
To form a 2+ ion, you need to lose two negatively charged electrons. Because metals have relatively empty outer shells, they lose electrons. Metals are on the left hand side of the table, and in order to lose two electrons, you need to have two electrons to start with. The groups of the table are based on the number of electrons an element has in the outer shell, so Group 2 metals will lose two electrons to gain a 2+ charge
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Solve this questions please​
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Answer:

A

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