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3 years ago

What would you need to know in order to calculate how many pre-1982 pennies and post-1982 pennies are in the mixed sample?

1 answer:

8 0

<span>In 1982, the physical composition of the penny was changed; the ratio of the zinc to copper was essentially reversed. As a result, a pre-82 penny has a different density from a post-82 penny. It would help to know several things: first, the average mass and density of both a single pre and post 82 penny. Then you need to know how many pennies in total are in the mixed sample. You would also need to know the mass and density measures of "pure" samples, that is, if we have a mixed sample of 20 pennies, it would be useful to know the mass and density measurements of 20 pre-82 and 20 post-82 pennies. You can then figure out your estimate based on the variances in the density measurements.</span>

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What do the elements in each pair have in common?

disa [49]

The elements in each pair have in common it that they are metaloids

7 0

2 years ago

How many grams of a stock solution that is 87.5 percent H2SO4 by mass would be needed to make 275 grams of a 55.0 percent by mas

dalvyx [7]

87+55=142
5+0=5
The answer is 142.5

8 0

2 years ago

This is for a study guide for my semester final, and I have no clue how to answer this, please help, and please at least try to

user100 [1]

Positron emission = emission of a positron and a neutrino when a proton is convert into a neutron. The total number of particles in the nucleus doesn't change, -1 proton +1 neutron

It's a spontaneous reaction for some nucleus.

eg:

Positron = e+

Neutrino=ve

O-15 --> N-15 + e+ +ve

Electron capture= A nucleus absorb an electron while a proton is convert in a neutron and emit a neutrino. The total number of particles in the nucleus doesn't change, -1 proton +1 neutron

eg:

Al-26 +e- --> Mg-26 + ve

Electron capture and positron emission are two mechanisms to explain the decay of some unstable isotopes. Electron capture is usually observed when the energy difference between the initial and final state is low. Mainly because of the larger amount of kinetic energy need for the expulsion two particles with the positron emission mechanism.

7 0

2 years ago

a chemical reaction produced 2.50 moles of nitrogen gas. what volume in liters, does this gas sample occupy at STP? (show your w

Harrizon [31]

The volume of N₂ at STP=56 L

<h3>Further explanation</h3>

Given

2.5 moles of N₂

Required

The volume of the gas

Solution

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, the volume per mole of gas or the molar volume-Vm is 22.4 liters/mol.

So for 2.5 moles gas :

$\tt 2.5\times 22.4=56~L$

5 0

2 years ago

Equal moles of H2, N2, O2, and He are placed into separate containers at the same temperature. Assuming each gas behaves ideally

lbvjy [14]

Answer:

They would all exhibit the same pressure.

Explanation:

Since the same number of mole of each gas is placed in different containers, it means the gas will occupy the same volume.

Now, the gases were observed at the same temperature. This means they will all have the same pressure as their volume is the same.

Now we can further understand this by doing a simple calculation as follow:

Assumptions:

For H2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, H2 has a pressure of 1 atm.

For N2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, N2 has a pressure of 1 atm

For O2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, O2 has a pressure of 1 atm

For He:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, He has a pressure of 1 atm.

From the above illustrations we can see that the gases have the same pressure since they have the same number of mole, volume and were observed at the same temperature.

4 0

3 years ago