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Sergio [31]
3 years ago
7

Need Help with these 2. Will give Brainliest!!

Chemistry
1 answer:
Klio2033 [76]3 years ago
8 0

Answer:

1. b

2. a

Explanation:

1. The density of an object represents the mass per unit volume of the object. A density of 0.45 g/mL means that 1 mL of the object weighs 0.45 g, 1.000 g/mL means 1 mL weighs 1 g, etc.

<u>A density of 35,885 g/mL means that 1 mL of the object weighs 35,885 g. This is a ridiculously high amount of weight for an object with a volume of 1 mL and seems not reasonable. It is highly unlikely that such a substance exists in nature.</u>

2. <u>Considering the fact that only rock would sink in water of all the substances from a - d, it thus means that rock would have the greatest density.</u> Oil, oxygen, and ice will all float on water, meaning that they are less dense than water.

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NEED HELP! what is a orbital . describe
olchik [2.2K]

Answer:

In chemistry and quantum mechanics, an orbital is a mathematical function that describes the wave-like behavior of an electron, electron pair, or (less commonly) nucleons. An orbital can contain two electrons with paired spins and is often associated with a specific region of an atom.

Explanation:

6 0
3 years ago
Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured
yan [13]

Answer:

Water will boil at 76^{0}\textrm{C}.

Explanation:

According to clausius-clapeyron equation for liquid-vapor equilibrium:

                                         ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]

where, P_{1} and P_{2} are vapor pressures of liquid at T_{1} (in kelvin) and T_{2} (in kelvin) temperatures respectively.

Here, P_{1} = 760.0 mm Hg, T_{1} = 373 K, P_{2} = 314.0 mm Hg

Plug-in all the given values in the above equation:

                          ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]

                       or, T_{2}=349 K

So, T_{2}=349K=(349-273)^{0}\textrm{C}=76^{0}\textrm{C}

Hence, at base camp, water will boil at 76^{0}\textrm{C}

6 0
3 years ago
Which of the subshells in the electron configuration of hf behave as core orbitals?
Nesterboy [21]
<span>Of all the sub-shells shown ( 1s ,2s ,2p ,3s ,3p ,4s ,3d ,4p ,5s ,4d ,5p ,6s ,4f ,5d) the ones that act as core orbital of HF (Hydrogen Fluoride) is 6s and 5d</span>
3 0
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
tigry1 [53]

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+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

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
How many moles are in 25 grams of Li? How do you know?
Ipatiy [6.2K]

Answer:

3.6mol Li

Explanation:

To get the amount of moles from the mass we divide the mass by the molar mass.

25g ÷ 6.941g/mol = 3.6mol Li

8 0
3 years ago
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