A sample of CO2(s) and a sample of CO2(g) differ in their

Hydrogen-3 has a half-life of 12.32 years. A sample of H-3 weighing 3.02 grams is left for 15.0 years. What will the final weigh

kaheart [24]

Answer:

$\large \boxed{\text{1.38 g}}$

Explanation:

Two important formulas in radioactive decay are

$(1) \qquad t_{\frac{1}{2}} = \dfrac{\ln 2}{k}\\\\(2) \qquad \ln \left(\dfrac{N_{0}}{N}\right) = kt$

1. Calculate the decay constant k

$\begin{array}{rcl}t_{\frac{1}{2}} &=& \dfrac{\ln 2}{k}\\\\\text{12.32 yr} &= &\dfrac{\ln 2}{k}\\\\k & = & \dfrac{\ln 2}{\text{12.32 yr}}\\\\& = & \text{0.056 26 yr}^{-1}\\\end{array}$

2. Calculate the mass remaining

$\begin{array}{rcl}\ln \left(\dfrac{A_{0}}{A}\right) &= &kt \\\\\ln \left(\dfrac{\text{3.20 g}}{A}\right) &= &\text{0.056 26 yr}^{-1}\times \text{15 yr} \\\\\ln \left(\dfrac{\text{3.20 g}}{A}\right) &= &0.8439 \\\\\dfrac{\text{3.20 g}}{A} &= &e^{0.8439} \\\\\dfrac{\text{3.20 g}}{A}&= &2.325 \\\\A &= &\dfrac{\text{3.20 g}}{2.325}\\\\&= & \textbf{1.38 g}\\\end{array}\\\text{The final mass of the sample will be $\large \boxed{\textbf{1.38 g}}$}$

8 0

3 years ago

Amines are ________. brønsted-lowry bases brønsted-lowry acids neutral in water solution unreactive

sammy [17]

Answer:

A.) Brønsted-Lowry bases

Explanation:

Amines have a lone pair of electrons.

Brønsted-Lowry bases donate a lone pair of electrons in exchange for a hydrogen ion.

Therefore, if exposed to an acid, amines will give up electrons in order to bond with a hydrogen. This makes them Brønsted-Lowry bases.

8 0

2 years ago

Is

BlackZzzverrR [31]

Answer:

here is your answer

enjoy

7 0

3 years ago

How many atoms of potassium, K, are in K3PO4

Rom4ik [11]

Answer:

1

Explanation:

4 0

3 years ago

You have 4 moles of a gas in a 50 L container held at 2 atm pressure. Currently the temperature is 27 ºC. R = 0.0821 L*atm/(mol*

Mariulka [41]

The formula used for determining gas pressure, volume and temperature interaction would be PV=nRT.

• What is the temperature in Kelvins?
You already right at this part. Kelvin temperature formula from celsius should be:
K= C+273.15=
K= 27 +273.15 = 300.15
It is important to remember that the formula in this question is using Kelvin unit at temperature, not Celcius or Fahrenheit.

• Assuming that everything else remains constant, what will happen to the pressure if the temperature decreases to -15 ºC?
In this case, the temperature is decreased from 27C into -15C and you asked the change in the pressure.
Using PV=nRT formula, you can derive that the temperature will be directly related to pressure. If the temperature decreased, the pressure will be decreased too.

 If you increase the number of moles to 6 moles, increase temperature to 400K and reduce the volume to 25 L, what will the new pressure be?
PV=nRT
P= nRT/V
P= 6 moles* 0.0821 L*atm/(mol*K) * 400K/25L= 7.8816 atm

3 0

3 years ago