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4 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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How do you solve this? I am so confused. This is Chemistry 1 HON and its dimensional analysis. It would help me so much if someo

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3 years ago

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$

The rate constant is $0.4167\ \text{M}^{-1}\text{min}^{-1}$

3 0

3 years ago

If a hockey puck is made to slide along a straight line on a completely frictionless surface, which statement about the motion o

JulsSmile [24]

The hockey puck will continue in the same speed in a continues straight line

3 0

3 years ago

Read 2 more answers

(only reply if you know the answer or get reported) Please solve this ty

kolezko [41]

Answer:

7) 50 8)4

Explanation:

I answered the question what do you want to gave me

4 0

3 years ago

Balancing nuclear equations is slightly different than balancing chemical equations. The major difference is that in nuclear rea

vesna_86 [32]

Answer:

The correct answer is - 4.

Explanation:

As we known and also given that the total of the superscripts that is mass numbers, A in the reactants and products must be the same.The mass of products A can understand and calculated by this -

The sum of the product mass number of products = mass of reactant

237Np93 →233 Pa91 +AZX is the equation,

Solution:

Mass of reactants = 237

Mass of products are - Pa =233 and A = ?

233 + A = 237

A = 237 - 233

A = 4

So the equation will be:

237Np93 →233 Pa91 +4He2 (atomic number Z = 2 ∵ difference in the atomic number of reactant and products)

5 0

3 years ago