Answer:
2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP? STP is a common abbreviation for "standard temperature and pressure." You have to recognize that five values are given in the problem and the sixth is an x. Also ... 273 1. A gas has a volume of 800.0 mL at minus 23.00 °C and 300.0 torr.
Explanation:
An osmolarity of saline solution is 308 mosmol/L.
m(NaCl) = 9 g; the mass of sodium chloride
V(solution) = 1 L; the volume of the saline solution
n(NaCl) = 9 g ÷ 58.44 g/mol
n(NaCl) = 0.155 mol; the amount of sodium chloride
number of ions = 2
Osmotic concentration (osmolarity) is a measure of how many osmoles of particles of solute it contains per liter.
The osmolarity = n(NaCl) ÷ V(solution) × 2
The osmolarity = 0.154 mol ÷ 1 L × 2
The osmolarity = 0.154 mol/L × 1000 mmol/m × 2
The osmolarity of the saline solution = 308 mosm/L.
More about osmolarity: brainly.com/question/13258879
#SPJ4
Answer:
pH = 2.46
Explanation:
Hello there!
In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

Whereas the moles of the salt are computed as shown below:

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:
![[salt]=0.01428mol/0.0276L=0.517M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D0.01428mol%2F0.0276L%3D0.517M)
Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

Whose equilibrium expression is:
![Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BC_6H_5NH_2%5D%5BH_3O%5E%2B%5D%7D%7BC_6H_5NH_3%5E%2B%7D)
Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

Whereas x is:

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

Regards!
From the given balanced equation we have find out the amount (in gm) of Ag formed from 5.50 gm of Ag₂O.
2Ag₂O(s) → 4Ag (s) + O₂ (g)
We know, molecular mass of Ag₂O= 231.7 g/mol, and atomic mass of Ag= 107.8 g/mol. Given, mass of Ag₂O=5.50 gm. Number moles of Ag₂O=
= 0.0237 moles.
From the balanced chemical reaction we get 2 (two) moles of Ag₂O produces 4 (four) moles of Ag. So, 0.0237 moles of Ag₂O produces
moles=0.0474 moles of Ag= 0.0474 X 107.8 g of Ag=5.11g Ag.
Therefore, 5.50 g Ag₂O produces 5.11 g of Ag as per the given balanced chemical reaction.