<span>We are given the initial amount of 1 million carbon-14 atoms and the final amount which is 1/16 of the current atmospheric 14C levels. Also, the half life of carbon is </span>5,750 years. WE can use the decay formula
Aₓ = A₀e^-(ln2/t1/2)t
1,000,000(1/16) = (1,000,000)e^-(ln2/5750)t
t = 23,000 years
Election current because voltage is a measurement, information doesn't apply to all electrical devices and the wires within are usually copper bc it conducts and hardly ever will the wires be anything different because copper is cheap
Answer: Volume of the gas at STP is 22.53 L.
Explanation:
Given : Volume = 125 mL (as 1 mL = 0.001 L) = 0.125 L
Temperature = 
Pressure = 
According to the ideal gas equation, the volume of given nitrogen gas is calculated as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
Hence, volume of the gas at STP is 22.53 L.
17.8 mL NaOH
<em>Step 1.</em> Write the chemical equation
Fe^(2+) + 2NaOH → Fe(OH)2 + 2Na^(+)
<em>Step 2.</em> Calculate the moles of Fe^(2+)
Moles of Fe^(2+) = 500 mL Fe^(2+) × [0.0230 mmol Fe^(2+)]/[1 mL Fe^(2+)]
= 11.50 mmol Fe^(2+)
<em>Step 3.</em> Calculate the moles of NaOH
Moles of NaOH = 11.50 mmol Fe^(2+) × [2 mmol NaOH]/[1 mmol Fe^(2+)]
= 23.00 mmol NaOH
<em>Step 4.</em> Calculate the volume of NaOH
Volume of NaOH = 23.00 mmol NaOH × (1 mL NaOH/1.29 mmol NaOH)
= 17.8 mL NaOH
Answer:
Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.
Explanation:
The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:
![Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D%7D%7B%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D%7D)
.
Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both ![[\mathrm{SO_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D)
and ![[\mathrm{O_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D)
will increase if the pressure is increased through compression. However, because 
and 
have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient 
.
As a result, the increase in pressure will have no impact on the value of 
. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.
