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3 years ago

The half-life of Radium-226 is 1590 years. If a sample contains 100 mg, how many mg will remain after 4000 years?

1 answer:

4 0

We can calculate the amount left after 4000 years by using the half-life equation. It is expressed as:

A = Ao e^-kt

where A is the amount left at t years, Ao is the initial concentration, and k is a constant.

From the half-life data, we can calculate for k.

1/2(Ao) = Ao e^-k(1590)
k = 4.36 x 10^-4

A = 100 e^-<span> 4.36 x 10^-4(4000)
</span>A = 17.48 mg

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What mass of aluminum chloride could be made from 8.1 g of aluminum and 4.2 L of chlorine at STP?

posledela

In this problem Al metal is a limiting reactant as it is present in less amount as compared to chlorine gas, Hence, controls the formation of ALCl3. So, the amount of AlCl3 produced is 40.05 grams. Solution is as follow,

3 0

3 years ago

Please review the attachment

astra-53 [7]

Answer: The correct answer is -297 kJ.

Explanation:

To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:

2SO3 —> O2 + 2SO2 (196 kJ)

Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.

Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:

SO3 —>1/2O2 + SO2 (98 kJ)

S + 3/2O2 —> SO3 (-395 kJ)

Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.

S + O2 —> SO2

Now, we must add the enthalpies together to get our final answer.

-395 kJ + 98 kJ = -297 kJ

Hope this helps!

8 0

3 years ago

Read 2 more answers

Sodium hydroxide reacts with aluminum and water to produce hydrogen gas: 2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H

lianna [129]

Answer:

The mass of hydrogen gas formed is 0.205 grams

Explanation:

<u>Step 1:</u> Data given

Mass of 1.83 grams of Al

Mass of NaOH = 4.30 grams

Molar mass of Al = 26.98 g/mol

Molar mass of NaOH = 40 g/mol

<u>Step 2:</u> The balanced equation:

2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)

<u>Step 3:</u> Calculate moles of Al

Moles Al = mass Al / Molar mass Al

Moles Al = 1.83 grams / 26.98 g/mol

Moles Al = 0.0678 moles

<u>Step 4:</u> Calculate moles of NaOH

Moles NaOH = 4.30 grams / 40 g/mol

Moles NaOH = 0.1075 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles of Al, we need 2 moles of NaOH

Aluminium is the limiting reactant. It will completely be consumed ( 0.0678 moles)

NaOH is in excess. There will react 0.0678 moles

There will remain 0.1075 - 0.0678 = 0.0397 moles

<u>Step 6</u>: Calculate moles of hydrogen

For 2 moles of Al, we need 2 moles of NaOH, to produce 3 moles of hydrogen

For 0.0678 moles of Al, there is produced 0.0678 *3/2 = 0.1017 moles of H2

<u>Step 7</u>: Calculate mass of H2

Mass of H2 = Moles H2 * Molar mass of H2

Mass of H2 = 0.1017 moles * 2.02 g/mol

Mass of H2 = 0.205 grams

The mass of hydrogen gas formed is 0.205 grams

6 0

2 years ago

The layer of the Earth that the lithospheric plates move on top of is called the

ikadub [295]

Answer:

A. Lithosphere

4 0

3 years ago

A engineer measures the peak power output is 0.3227MW what is the peak power output in kilowatts ?

Lana71 [14]

<h3>Answer:</h3>

322.7 kW

<h3>Explanation:</h3>

Power refers to the rate at which work is done.
Therefore; Power = Work done ÷ time
It is measured in joules per seconds or Watts

In this case, we are required to convert 0.3227 MW to kilowatts

We need to know that;

10^6 watts = 1 Megawatts(MW)
10^3 Watts = 1 kilowatts (kW)

Therefore;

10^3 kW = 1 MW

Therefore, the suitable conversion factor is 10^3kW/MW

Hence;

0.3227 MW is equivalent to;

= 0.3227 MW × 10^3kW/MW

= 322.7 kW

Thus, the peak power output is 322.7 kW

3 0

2 years ago