A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with distilled water. the final molarity of the dilute solution is 0.102 M.
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 5 mL
Molarity of stock solution (M₁) = 5.103 M
Volume of diluted solution (V₂) = 250 mL
Molarity of diluted solution (M₂) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5.103 × 5 = M2 × 250
25.515 = M2 × 250
Divide both side by 250
M2 = 25.515 / 250
M2 = 0.102 M
Thus, the molarity of the diluted solution is 0.102 M.
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Answer:
0.102 M.
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 5 mL
Molarity of stock solution (M1) = 5.103 M
Volume of diluted solution (V2) = 250 mL
Molarity of diluted solution (V2) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M1V1 = M2V2
5.103 × 5 = M2 × 250
25.515 = M2 × 250
Divide both side by 250
M2 = 25.515 / 250
M2 = 0.102 M
Thus, the molarity of the diluted solution is 0.102 M.
Answer:
The electrons are lost from the valence shell (outermost electron shell) of the atom.
Explanation:
This is able to be inferred not only because valence electrons being lost first is a trend but also because the atom in question has actually 3 valence electrons (13-2-8 = 3).
Answer:
The ionic bond in NaCl are stronger than the stronger than the dispersion forces in HCl.
The hydrogen bonds in H2O are stronger than the dispersion forces in H2Se
Hydrogen bonds in NH3 are stronger than the dipole-dipole attractions in PH3.
Hydrogen bonds in HF are stronger than the dispersion forces in F2
Explanation:
Ionic bonds occur in molecules with high differences in their electronegative value where there are actual transfer of electrons. HCl has a bond which is involved in the sharing of electrons.
Hydrogen bonds are present in H2O which is stronger than the dispersion forces.
PH3 is a larger molecule with greater dispersion forces than ammonia, NH3 has very polar N-H bonds leading to strong hydrogen bonding. This dominant intermolecular force results in a greater attraction between NH3 molecules than there is between PH3 molecules.
F2 is a non-polar molecule, therefore they have London dispersion forces between molecules while HF has a hydrogen bond because F is highly electronegative.
Ionic bonds are stronger than dispersion forces in HCl, while hydrogen bonds are stronger than dispersion forces in H2Se, PH3, and F2.
In the given sentences, the blanks represent the types of intermolecular forces. The options given are ionic bonds, hydrogen bonds, dispersion forces, and dipole-dipole attractions. Ionic bonds are stronger than the dispersion forces in HCl. Hydrogen bonds are stronger than the dispersion forces in H2Se. Hydrogen bonds are stronger than the dipole-dipole attractions in PH3. Hydrogen bonds are stronger than the dispersion forces in F2.
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Answer:
Graham's law of effusion was formulated by Scottish physical chemist Thomas Graham in 1848. Graham found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles. This formula can be written as: , where: Rate₁ is the rate of effusion for the first gas.
Explanation:
r1 = rate of effusion for gas 1
r2 = rate of effusion for gas 2
M1 = molar mass of gas 1
M2 = molar mass of gas 2
Answer:
35.6 g of W, is the theoretical yield
Explanation:
This is the reaction
WO₃ + 3H₂ → 3H₂O + W
Let's determine the limiting reactant:
Mass / molar mass = moles
45 g / 231.84 g/mol = 0.194 moles
1.50 g / 2 g/mol = 0.75 moles
Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.
Let's make rules of three:
1 mol of tungsten(VI) oxide needs 3 moles of H₂
Then 0.194 moles of tungsten(VI) oxide would need (0.194 .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)
3 moles of H₂ need 1 mol of WO₃ to react
0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles
It's ok. I do not have enough WO₃.
Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.
Let's convert the moles to mass (molar mass . mol)
0.194 mol . 183.84 g/mol = 35.6 g
b) All bases
c) all solvents
d) nonpolar solvents
Answer:
d) Non-polar solvents
Answer:
Is it Group?
Explanation: