Which of the following is considered to be a living organism?

The half-life of tritium is 12.3 years. If 48.0 mg of tritium is released from a nuclear power plant during the course of a mish

Thepotemich [5.8K]

Half life is defined as the time when the final concentration reduced to half of the initial value.

Thus, for first half life, the remaining percentage of the reactant is 50 percent that is $\frac{50}{100}=\frac{1}{2}$

Now, for second half life, the remaining percentage will be 50 percent of the remaining 50 percentage that is $\frac{50}{2}=25$ $\frac{25}{100}=\frac{1}{4}$ or $\frac{1}{2^{2}}$

Thus, general formula for n half lives = $\frac{1}{2^{n}}$

Now, 98.4 years is equal to eighth times 12.3 years that is $\frac{1}{2^{8}}$ = $\frac{1}{256}$ remaining material.

Therefore, mass of the nuclide will remain after 98.4 years= $\frac{48.0 mg}{256}$

= $0.1875 mg$

Mass of remaining nuclide after 98.4 years = $0.1875 mg$

4 0

4 years ago

..can anyone tell why can't we text anything in indian brainly <br>3724221217<br>12345<br>

Mademuasel [1]

Answer:

because its not international language

Explanation:

because its not international language like English

4 0

3 years ago

Read 2 more answers

Cuales son los colores del arcoíris

WINSTONCH [101]

Rojo, naranja, amarillo, verde, cian, azul y violeta

3 0

3 years ago

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly

trapecia [35]

Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of $Ag_2O$ = 25.0 g

Mass of $C_{10}H_{10}N_4SO_2$ = 50.0 g

Molar mass of $Ag_2O$ = 231.7 g/mole

Molar mass of $C_{10}H_{10}N_4SO_2$ = 250.3 g/mole

Molar mass of $AgC_{10}H_{9}N_4SO_2$ = 357.1 g/mole

First we have to calculate the moles of $Ag_2O$ and $C_{10}H_{10}N_4SO_2$ .

$\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles$

$\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)$

From the balanced reaction we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 1 mole of $Ag_2O$

So, 0.1998 moles of $C_{10}H_{10}N_4SO_2$ react with $\frac{0.1998}{2}=0.0999$ moles of $Ag_2O$

From this we conclude that, $Ag_2O$ is an excess reagent because the given moles are greater than the required moles and $C_{10}H_{10}N_4SO_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgC_{10}H_9N_4SO_2$

From the reaction, we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 2 mole of $AgC_{10}H_9N_4SO_2$

So, 0.1998 mole of $C_{10}H_{10}N_4SO_2$ react with 0.1998 mole of $AgC_{10}H_9N_4SO_2$

Now we have to calculate the mass of $AgC_{10}H_9N_4SO_2$

$\text{ Mass of }AgC_{10}H_9N_4SO_2=\text{ Moles of }AgC_{10}H_9N_4SO_2\times \text{ Molar mass of }AgC_{10}H_9N_4SO_2$

$\text{ Mass of }AgC_{10}H_9N_4SO_2=(0.1998moles)\times (357.1g/mole)=71.35g$

Therefore, the mass of silver sulfadiazine produced can be, 71.35 grams.

8 0

3 years ago

What is the total amount of heat required to vaporize 1.00 gram of H2O (l) at 100 c and 1 atmosphere

Firlakuza [10]

The equation to be used to calculate the heat required to vaporize 1 gram of water at 100 degrees eclsius and 1 atmosphere is the enthalpy. It is represented as H = mCpT where H is the heat released after phase change, m is the mass of the substance, Cp is the specific heat capacity of the substance and T is the change in temperature.

H = mCpT

H = (1g/1000g/kg)(4.184kJ/kg-K)(100 - 20)

H = 0.335 kJ

8 0

3 years ago