The mass of electrons , i just did the same test and got it right
The molar mass is 242 g/mol (<span>241.8597 g/mol). There is a very good online service to calculate molar masses:
http://www.webqc.org/mmcalc.php
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Answer:
Q = 270 Joules (2 sig. figs. as based on temperature change.)
Explanation:
Heat Transfer Equation of pure condensed phase substance => Q = mcΔT
Mixed phase (s ⇄ l melting/freezing, or l ⇄ g boiling/condensation) heat transfer equation => Q = m∙ΔHₓ; ΔHₓ = phase transition constant
Since this is a pure condensed phase (or, single phase) form of lead (Pb°(s)) and not melting/freezing or boiling/condensation, one should use
Q = m·c·ΔT
m = mass of lead = 35.0g
c = specific heat of lead = 0.16J/g°C
ΔT = Temp change = 74°C - 25°C = 49°C
Q = (35.0g)(0.16J/g·°C )(49°C) = 274.4 Joules ≅ 270 Joules (2 sig. figs. as based on temperature change.)
If Ka for HCN is 6. 2×10^−10 at 25 °C, then the value of Kb for cn− at 25 °C is 1.6 × 10^(-5).
<h3>What is base dissociation constant? </h3><h3 />
The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 6.2× 10^(-10)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{6.2×10^(-10) }
= 1.6× 10^(-5)
Thus, the value of base dissociation constant at 25°C is 1.6 × 10^(-5).
learn more about base dissociation constant :
brainly.com/question/9234362
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Answer:
14 mL
Explanation:
To prepare a solution by a concentrated solution, we must use the equation:
C1xV1 = C2xV2, where <em>C</em> is the concentration, <em>V</em> is the volume, 1 is the initial solution and 2 the final solution.
The final solution must have 2 mL and a concentration of 350 pg/mL, and the initial solution has a concentration of 50 pg/mL.
Then:
50xV1 = 350x2
50xV1 = 700
V1 = 700/50
V1 = 14 mL
