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jonny [76]
2 years ago
12

2. Which of the following elements does not lose an electron easily? (a) Na (6) F (c) Mg (d) Al​

Chemistry
1 answer:
vaieri [72.5K]2 years ago
3 0

Answer:

Answer: (b) F

Explanation:

Sodium has 1, magnesium has 2 and Aluminium has 3 electrons in its outermost shell whereas Fluorine has 7 electrons in its outermost shell hence Fluorine does not lose electrons easily.

The electronic configuration of fluorine is 2,7.

Fluorine is the ninth element with a total of 9 electrons.

The first two electrons will go in the 1s orbital.

The next 2 electrons for F go in the 2s orbital.

The remaining five electrons will go in the 2p orbital. Therefore the F electron configuration will be 1s22s22p5.

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An unknown metal has a mass of 86.8 g. When 5040 J of heat are added to the sample, the sample temperature changes by 64.7 ∘ C .
grandymaker [24]

Answer: The specific heat of the unknown metal is 0.897J/g^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=5040 Joules

m= mass of substance = 86.8 g

c = specific heat capacity = ?

Initial temperature of the water = T_i

Final temperature of the water = T_f

Change in temperature ,\Delta T=T_f-T_i=(64.7)^0C

Putting in the values, we get:

5040=86.8\times c\times 64.7^0C

c=0.897J/g^0C

The specific heat of the unknown metal is 0.897J/g^0C

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3 years ago
List examples of organisms from the movie that represent each of the six kingdoms. Monera (Archaebacteria/Eubacteria): Protista:
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Paracanthurus- Dory
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2.0 L of Carbon dioxide is heated from -25.0 °C to Standard Temperature.
RideAnS [48]

The final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L.

<h3>How to calculate volume?</h3>

The volume of a given gas can be calculated using the Charles law equation as follows:

V1/T1 = V2/T2

Where;

  • V1 = initial volume
  • V2 = final volume
  • T1 = initial temperature
  • T2 = final temperature

  • V1 = 2L
  • V2 = ?
  • T1 = -25°C + 273 = 248K
  • T2 = 273K

2/248 = V2/273

273 × 2 = 248V2

546 = 248V2

V2 = 546/248

V2 = 2.2L

Therefore, the final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L

Learn more about volume at: brainly.com/question/11464844

5 0
2 years ago
When you burned copper, the product copper
MatroZZZ [7]

Answer:

Oxygen.

Explanation:

The copper must be combined with something in the air.

6 0
3 years ago
Read 2 more answers
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0
2 years ago
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