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Damm [24]
3 years ago
10

Which of the following would form an anion? (2 points) Al O Ar K

Chemistry
1 answer:
gavmur [86]3 years ago
7 0
O or Oxygen because it needs electrons to be stable causing it to be an anion with a 2- charge to have a complete shell while Aluminum would be an cation with a 3+ charge, Potassium or K with a 1+ charge, and Argon wouldn’t be an anion because it is a noble gas and is already stable and don’t need electrons due to a full shell.
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What is the name of this molecule?
Neko [114]
The triple bond is in the first place, not the 3rd because we have to choose the smallest possible number. That means the name will be: ‘but-1-yne’ or ‘1-butyne’
3 0
3 years ago
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What is the volume of SO3 in 835 g SO3?
Pie
233.856 , sorry if i’m wrong
7 0
3 years ago
Discuss FOUR ways in which one's acceptance of responsibility can influence
andrew11 [14]

Answer:

being responsible is owing up to the possible

consequences of your decision whether it was right or wrong.

communication is a two way process in which there is a messenger and receiver

one of the ways by which a person could be responsible is by saying 'I'

in all the things he would say, instead of 'we'

for example , " We don't like the way you acted a while ago " , change it to i don't like tye way you acted a while ago.

here u r establishing ownership on the things you want to say to the person

8 0
3 years ago
Compute diffusion coefficients for the interdiffusion of carbon in both (a) α-iron (BCC) and (b) γ-iron (FCC) at 800.˚C. Assume
Ne4ueva [31]

Answer:

a)1.404 E-10 m^2/s

b)1.438 E-12 m^2/s

Explanation:

Diffusion Coefficients can be calculated at different temperatures taking into account the energy of the diffusive process, remember that this process takes place in the interstices of the solid, and gets with higher temperatures due to the higher kinetic energy of the particles.

D=D_{o} *exp(-Q/RT)

a) For α-iron (BCC)

D=1.10*10^{-6} m^{2}/s *exp(-80.0kJ/mol/8.314*10^{-3}kJ/mol*k*1073.15K)= 1.404*10^{-10} m^{2}/s

b) For γ-iron (FCC)

D=2.30*10^{-5} m^{2}/s *exp(-148.0kJ/mol/8.314*10^{-3}kJ/mol*k*1073.15K)= 1.438*10^{-12} m^{2}/s

5 0
3 years ago
Calculate the empirical formula for a compound containing 67.6% Hg 10.8% S 21.6% O
pashok25 [27]

Answer:

HgSO₄

Explanation:

% => g => moles => ratio => reduce => empirical ratio

%Hg = 67.6% => 67.6g/201g/mol = 0.34mol

%S    = 10.8% => 10.8g/32g/mol = 0.34mol

%O   = 21.6% => 21.6g/16g/mol = 1.35mol

Hg:S:O => 0.34:0.34:1.35

Reduce to whole number ratio by dividing by the smaller mole value...

Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4

∴ Empirical Formula is HgSO₄

3 0
3 years ago
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