Answer:
The concentration is 50,8 % w/v and radio strengths = 1,96.
Explanation:
Phenobarbital sodium is a medication that could treat insomnia, for example.
2,0 M of Phenobarbital sodium means 2 moles in 1L.
The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means 1g in <em>r</em> mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:
2 moles × 
= 508 g of Phenobarbital sodium.
1 L ×
= 1000 mL of solution
Thus, % w/v is:
× 100 = 50,8 % w/v
And radio strengths:
= 1,96. Thus, you have 1 g in 1,96 mL
I hope it helps!
Answer;
1.6 kg.
Solution;
The density is 1.36 g/ml;
The volume is 1.25 qt
However; 1 qt = 946.35 ml
Mass is given by; density × volume;
= 1.25 qts × 946.25 ml/qt × 1.36 g/ml =1609 g
but; 1 kg = 1000 g
Hence the mass = 1609/1000 = 1.609 Kg or 1.61 (sig figs)
Answer is: C. H₂, molecule of hydrogen, g is c<span>hemistry abbreviations or physical state symbol for gas.</span>
<span>
Lithium (Li) is solid (s) element (metal).
Water (H</span>₂O) is liquid (l) compound or molecule.
Lithium hydroxide (LiOH) is aqueous solution (aq). It dissociates in water on lithium cation (Li⁺) and hydroxy anion (OH⁻).
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</span>
Answer:
Ar < Cl - < S2-
Explanation:
All the species written above are isoelectronic. This means that they all possess the same number of electrons. All the species above possess 18 electrons, the noble gas electron configuration.
However, for isoelectronic species, the greater the atomic number of the specie, the smaller it is. This is because, greater atomic number implies that their are more protons in the nucleus exerting a greater attractive force on the electrons thereby making the specie smaller in size due to high electrostatic attraction.
Answer:
The equilibrium concentration of CH₃OH is 0.28 M
Explanation:
For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)
The equilibrium constant (Keq) is given for the following expresion:
Keq= 
=14.5
Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.
We have:
(CH3OH)= ?
(CO)= 0.15 M
(H2)= 0.36 M
So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:
14.5= 
14.5 x (0.15 M) x 
= (CH₃OH)
0.2818 M = (CH₃OH)
