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Firdavs [7]
3 years ago
12

Nombor proton bagi atom K,L,M dan N adalah 8,10,5 dan 17 masing-masing. Susunan elektron bagi atom Z ialah 2.8.3. Antara unsur b

erikut, yang manakah boleh membentuk sebatian ion dengan atom Z? *
A. K dan L
B. K dan N
C. L dan M
D. M dan N​
Chemistry
2 answers:
Crazy boy [7]3 years ago
8 0

Answer:

I don't understand your language please

dimulka [17.4K]3 years ago
3 0
I rasa jwp D, kerana Z donate 3 to M and dia bole donate juga kpd N

5+3=8
17+3=20

Maksudnya stable.
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The percentage of oxygen in sulphur lv oxide (s=32, o=16)​
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Answer:

66 percent that was my best guess

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Which element in group 1 has the greatest tendency to lose an electron?
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A. cesium. bacause as the elements go down a group of the atoms, the sizes of the atom will increase . we know that tge tendency to lose an electron is based on the sizes of the atoms of group 1.
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3 years ago
How many grams of O2 are present in 44.1 L of O2 at STP?
ycow [4]

Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

First of all, the STP conditions refer to the standard temperature and pressure, where the values ​​used are: pressure at 1 atmosphere and temperature at 0°C. These values ​​are reference values ​​for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that occupies.
  • T is its temperature.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

Then, in this case:

  • P= 1 atm
  • V= 44.1 L
  • n= ?
  • R= 0.082 \frac{atmL}{molK}
  • T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 \frac{atmL}{molK}× 273 K

Solving:

n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}

n=1.97 moles

Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:

1.97 molesx\frac{32 g}{1 mole}= 63.04 g ≈ <u><em>63 g</em></u>

Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

Learn more about the ideal gas law:

  • <u>brainly.com/question/4147359?referrer=searchResults</u>
7 0
3 years ago
A nugget of gold has a mass of 28 grams and a volume of 1.45 cubic centimeters. What is the density?
Travka [436]

Answer:

0.01931034

Explanation:

Steps:

ρ =  m/v

       =  

28 gram

1.45 cubic meter

=  19.310344827586 gram/cubic meter

=  0.019310344827586 kilogram/cubic meter

7 0
3 years ago
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