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tresset_1 [31]
3 years ago
12

An unknown sample of mystery element T is analyzed. According to the data 7.42% of the element is ^6T and 92.58% is ^7T. The tru

e mass of ^6T is 6.02amu and 7.02amu for ^7T. Calculate the average atomic mass and identify the element.
Chemistry
1 answer:
Pepsi [2]3 years ago
8 0

Answer:

\boxed{\text{6.95 u; Li}}

Explanation:

1. Atomic mass of T

To get the atomic mass of T, you calculate the weighted average of the atomic masses of the isotopes.

That is, you multiply the atomic mass of each isotope by a number representing its relative importance (e.g.., its percent of the total).

\begin{array}{ccrcr}\textbf{Atom} & \textbf{Mass/u} &\textbf{Percent} & \textbf{Calculation}& \textbf{Result/u}\\^{6}\text{T}& 6.02 & 7.42 & 6.02 \times 0.0742 & 0.45\\^{7}\text{T}& 7.02 & 92.58 & 7.02 \times 0.9258& 6.50\\& & &\text{TOTAL } = &\textbf{6.95}\\\end{array}

2. Identity of T

Lithium has an atomic mass of 6.64 u.

Element T is lithium.

 

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If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?
kodGreya [7K]

Answer:

[Cu^{2+}]=0.041 M

Explanation:

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In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}

[Cu^{2+}]=0.041 M

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3 years ago
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Work done after rutherford's gold foil experiment demonstrated that the nucleus contained ________.
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Protons and neutrons

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A thermos bottle uses a ______ to keep heat in the thermos
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4 0
2 years ago
The average score for games played in the NFL is 21.1 and the standard deviation is 8.9 points. 46 games are randomly selected.
kaheart [24]

Answer:

A) P(21.4317 < ¯x < 22.7561) = 0.2975

B) Q1 for the ¯x distribution = 21.9844

Explanation:

The Central Limit theorem allows us to say that

Sample mean = Population mean = 21.1 points

Mean of sampling distribution = σₓ = (σ/√n)

σ = population standard deviation = 8.9 points

n = sample size = 46

σₓ = (8.9/√46) = 1.3122334098 = 1.3122

A) P(21.4317 < ¯x < 22.7561) =

This is a normal distribution problem

To find this probability, we will use the normal probability tables

We first normalize/standardize 21.4317 and 22.7561.

The standardized score of any value is that value minus the mean divided by the standard deviation.

For 21.4317

z = (x - μ)/σ = (21.4317 - 21.1)/1.3122 = 0.25

For 22.7561

z = (x - μ)/σ = (22.7561 - 21.1)/1.3122 = 1.26

The required probability

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

Checking the tables

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

= P(z < 1.26) - P(z < 0.25)

= 0.89617 - 0.59871

= 0.29746 = 0.2975 to 4 d.p.

B) Q1 for the distribution is the first quartile. The first quartile is greater than 25% of the distribution.

P(x > Q1) = 0.25

Let the z-score that corresponds to Q1 be z'

P(x > Q1) = P(z > z') = 0.25

But P(z > z') = 1 - P(z ≤ z') = 0.25

P(z ≤ z') = 1 - 0.25 = 0.75

From the normal distribution tables,

z' = 0.674

z' = (Q1 - μ)/σ

0.674 = (Q1 - 21.1)/1.3122

Q1 = 0.674×1.3122 + 21.1 = 21.9844228 = 21.9844 to 4 d.p.

Hope this Helps!!!

7 0
3 years ago
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