Mass of X₄Z₅= 378.621 g
<h3>Further explanation</h3>
Given
3.89 x 10²³ molecules/formula units of X₄Z₅
Required
Mass of X₄Z₅
Solution
1 mol = 6.022 x 10²³ particles(molecules, atoms, ions)
N = n x No
No = Avogadro's number
n = mol
Conversion to mol :
n = N : No
n = 3.89 x 10²³ : 6.022 x 10²³
n = 0.646
MW X₄Z₅ :
= 4 x 76.15 + 5 x 56.3
= 304.6 + 281.5
= 586.1 g/mol
Mass :
= mol x MW
= 0.646 x 586.1
= 378.621 g
Answer:
1.4 × 10^-4 M
Explanation:
The balanced redox reaction equation is shown below;
5Fe2+ + MnO4- + 8H+ --> 5Fe3+ +Mn2+ + 4H2O
Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol
Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles
Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M
Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M
Volume of Fe^2+ (VA)= 20.0 ml
Let the concentration of MnO4^- be CB (the unknown)
Volume of the MnO4^- (VB) = 12.6 ml
Let the number of moles of Fe^2+ be NA= 5 moles
Let the number of moles of MnO4^- be NB = 1 mole
From;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CB= CAVANB/VBNA
CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5
CB = 1.4 × 10^-4 M
Answer: it is A.
Explanation:
Compare the electric force and gravitational force. ... Both forces are inversely proportional to the square of the distance between the objects, this is known as the inverse-square law. ... If there are two charges that are DIFFERENT, they attract.
Answer:
The correct option is A
Explanation:
Some amino acids, called glucogenic amino acids, when catabolized convert there carbon backbones to tricarboxylic acid (TCA) cycle intermediates. These intermediates can be subsequently metabolized into carbon dioxide and water with the release of ATP or the formation of glucose (known as gluconeogenesis.
<u>All amino acids (with the exception of leucine and lysine) are glucogenic and can thus generate the carbon backbones required for gluconeogenesis</u>. Thus, the correct option is a.
Answer:
D) 5.15
Explanation:
Step 1: Write the equation for the dissociation of HCN
HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)
Step 2: Calculate [H⁺] at equilibrium
The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.
α% = [H⁺]eq / [HCN]₀ × 100%
[H⁺]eq = α%/100% × [HCN]₀
[H⁺]eq = 0.0070%/100% × 0.10 M
[H⁺]eq = 7.0 × 10⁻⁶ M
Step 3: Calculate the pH
pH = -log [H⁺] = -log 7.0 × 10⁻⁶ = 5.15
