Let's use Boyle's Law here. P1*V1 = P2*V2
Given: (assuming that there are decimals at the end for Sig Figs)
P1 = 900.mmHg
P2 = 1140.mmHg
V1 = ???
V2 = 250.mL
900.mmHg* ??? = 1140.mmHg * 250.mL
??? = 1.27*250.mL
??? = 318.mL
Therefore, the original volume is 318mL.
Answer:
NaOH(aq)
Explanation:
NaOH(aq) is known to precipitate Mn^2+ ions according to the following reaction; Mn^2+(aq)+2OH^−(aq)↽−−⇀Mn(OH)2(s)
Hence, manganese(II) oxide reacts more readily with NaOH(aq) under ordinary conditions precipitating the metal hydroxide solid. This is one of the characteristic reactions of Mn^2+.
Answer:
B. their outer electron levels are filled
Answer: B
Explanation: Graph B compares the two temperatures on separate lines so that we can see the comoparison directly, as a function of time. Not only does the graph quickly answer which condition id most favorable to colony growth, but it also hints at some behaviors that may accelerate growth as time goes on. Graph C is a possible answer, if the <u>only</u> question is which promotes growth the fastest. But the questions asks "compare," which Graph B does not allow as well as Graph C.
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 :
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