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katovenus [111]
3 years ago
15

Technetium-104 has a half life of 18.0 minutes. how much of a 165.0 g sample remains after 90.0 minutes?

Chemistry
2 answers:
11Alexandr11 [23.1K]3 years ago
8 0

The mass of 165.0 g sample that remains after 90.0 minutes is 5.16 grams

calculation

lambda㏑2/18= 0.0385

m(t)= 165 x e( 0.0385 x90)  =5.16g

UkoKoshka [18]3 years ago
4 0

The mass of 165.0 g sample that remains after 90.0 minutes is 5.16 grams

calculation

lambda㏑2/18= 0.0385

m(t)= 165 x e( 0.0385 x90)  =5.16g

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2.3g of sodium metal is burned completely in oxygen of air.
Nina [5.8K]

Answer:

mass of Na2O= 3.1 grams

volume of O2= 0.56 L

Explanation:

First what you need to do is establish the balanced equation for the reaction that they're giving you. In this case would be:

4Na + O2 > 2Na2O

And then, you can read the stequiometry of the reaction:

4 moles of Na react with one mole of oxygen to produce 2 moles of sodium oxide.

So now you know that you have 2.3 grams of sodium so you have to obtain the quantity of moles that are present in 2.3 grams of sodium. The molar mass the (the grams by which you will have exactly one mole) so all you have to do is divide 2.3g ÷ 23g/mol= 0.1 mol

So here you have 0.1 mol of sodium, and you know for a fact that for each 4 moles of sodium, one mole of oxygen reacts, so you simply divide the quantity of moles of sodium you have by the 4 moles of oxygen present in the balanced equation.

0.1 moles ÷ 4 moles = 0.025.

So for each 0.1 moles of sodium that react, 0.025 moles of Oxygen react.

Now what you need to do is to find the limitant reactor (the reactor that's according to the comparison to the balanced equation is in a smaller ratio to the other element, that would be in excess.

In this case, both elements are in a ratio that respects the balanced equation so neither of them is in excess or is limitant.

Now, to determine the amount of Na2O, you simply know that one mole of oxygen react to produce to moles of sodium oxide, So all you have to do is establish the factor of conversion:

0.025 moles O2 x 2 moles Na2O/ 1 mol O2= 0.05 moles of Na2O

Then you convert that to grams by multiplying 0.05 by the molar mass of Sodium oxide and it will be:

0.05 mol Na2O x 62 g/mol= 3.1 grams of Na2O

Now what you need to do is find the volume by which O2 is reacting in the reaction so you can use the equation of the ideal gas law.

PV=nRT at standard temperature and pressure (273 K and 1 Atm)

P=Pressure

V=Volume

n=Number of atoms

R=constant of gases (0.082 Atm L/ mol K)

T=Temperature

V=nRT/P (because you need to find the volume)

V= (0.025 mol)(0.082)(273 K)/1 atm

V=0.56 L

And there you go :)

4 0
3 years ago
I need help with this science!
Eddi Din [679]

Answer:

All living organisms share several key characteristics or functions: order, sensitivity or response to the environment, reproduction, growth and development, regulation, homeostasis, and energy processing. When viewed together, these characteristics serve to define life.

3 0
2 years ago
Read 2 more answers
Which of the following chemical substances has a triple
Otrada [13]

Answer:

c. carbon monoxide

Explanation:

no one bothers to read why anyways so look at the picture

3 0
3 years ago
Which subatomic particles have a positive charge and are located in the nucleus?
d1i1m1o1n [39]

Answer: Atomic Particles

Explanation:

toms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged).

8 0
2 years ago
How many liters are in 5.98 moles of nitrogen gas at STP?
Shtirlitz [24]

Answer:

The correct answer is c) 134L

Explanation:

We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol.

1 atm x   V = 5, 98 mol x 0, 082 l atm / K mol x 273 K

V = 5, 98 mol x 0, 082 l atm / K mol x 273 K / 1 atm

V = 133, 86828 l

8 0
3 years ago
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