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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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For evaporation you need a warm, preferably humid temperature. The rate of evaporation increases with an increase in temperature. A windy climate is best, as wind helps to remove the evaporated water vapour, and therefore creating a better scope for evaporation to continue. The speed of wind is important for evaporation because the wind pulls in dry air, increasing the rate of evaporation.

Short Answer- Hot and humid temperature, lots of wind speed.

4 0
2 years ago
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Hexanal would be soluble in water? True or false?
telo118 [61]

Answer:

False.

Explanation:

Hexanal is a non-polar compound while water is a polar solvent.

We have the role "Like dissolves like".

So, hexanal is insoluble in water.

5 0
3 years ago
A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from
sweet-ann [11.9K]

Answer:

T_i~=163.1 ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT

Q_A_u=m_A_u*Cp_A_u*deltaT

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT

Now we can <u>put the values into the equation</u>:

60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)

Now we can <u>solve for the initial temperature of gold</u>, so:

T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20

T_i~=163.1 ºC

I hope it helps!

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3 years ago
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gtnhenbr [62]

Answer: A golfer hitting a golf ball.

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The atomic particles move more in this option than the others.

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Use these images to identify each state of matter.
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Answer:

Just too clarify its actually

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