B. 0.9 <span>
</span>D.Light intensity has no effect on whether electrons are emitted or not.
and
A. X=1.9eV,Y=0.2eV
I already took the gizmo so I know these are right. The first one I got wrong b/c there was no graph and the last one I didn't understand. Basically for the last one you calculate the work function for the metals and find their difference.
Hope this helps.
Answer:It is not an element because elements are the purest form of a substance; hence, they are no longer broken down by heating
Explanation:
Equilibrium equation is
<span>Ag2CO3(s) <---> 2 Ag+(aq) + CO32-(aq) </span>
<span>From the reaction equation above, the formula for Ksp: </span>
<span>Ksp = [Ag+]^2 [CO32-] = 8.1 x 10^-12 </span>
<span>You know [CO32-], so you can solve for [Ag+] as: </span>
<span>(8.1 x 10^-12) = [Ag+]^2 (0.025) </span>
<span>[Ag+]^2 = 3.24 x 10^-10 </span>
<span>[Ag+] = 1.8 x 10^-5 M </span>
Answer:
- <u>259,000 g of chalk.</u>
Explanation:
<u>1) Data:</u>
a) 2000 boxes
b) 175 g / box
c) % yield = 74%
<u>2) Formula: </u>
- % yield = (theoretical yield / actual yield) × 100
<u>3) Solution:</u>
a) Calcualte the actual yield:
- mass of product = 2000 box × 175 g/ box = 350,000 g
b) Solve for the theoretical yield from the % yield formula:
- % yield = (theoretical yield / actual yield) × 100
⇒ theoretical yield = % yield × actual yield / 100
theoretical yield = 74% × 350,000g / 100 = 259,000 g
Answer:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
