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Aleksandr-060686 [28]
4 years ago
9

Rank the following bonds in order of increasing ionic character ​

Chemistry
1 answer:
wolverine [178]4 years ago
8 0
It’s B then A then C then D
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What is the pressure exerted by a 0.500 mol sample of nitrogen in a 10.0 L container at 20°C?
lidiya [134]
The answer would be D.120 kPa
6 0
4 years ago
Say I grab a bunch of aluminum foil (9.5g) and ball it up. Aluminum has a specific heat of .9J/g*C. How much would the aluminum’
Brrunno [24]

Answer:

ΔT = 0.78 °C

Explanation:

Given data:

Mass of Al = 9.5 g

Specific heat capacity of Al = 0.9 J/g.°C

Temperature change = ?

Heat added = 67 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

67 J = 9.5 g × 0.9 j/g.°C × ΔT

67 J = 85.5 j/°C × ΔT

ΔT = 67 J  /  85.5 j/°C

ΔT = 0.78 °C

7 0
3 years ago
Someone please help me
Fudgin [204]
The answer is ultraviolet
6 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
Which scientist performed the cathode ray experiment leading to the discovery of electrons?
Mrrafil [7]

Answer:

E. Thomson

Explanation:

J.J. Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons.

https://www.khanacademy.org/science/chemistry/electronic-structure-of-atoms/history-of-atomic-structure/a/discovery-of-the-electron-and-nucleus

8 0
4 years ago
Read 2 more answers
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