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viva [34]
3 years ago
11

For each of the salts on the left, match the salts on the right that can be compared directly, using Ksp values, to estimate sol

ubilities. (If more than one salt on the right can be directly compared, include all the relevant salts by writing your answer as a string of characters without punctuation, e.g, ABC.) 1. aluminum phosphate A. Mn(OH)2 2. magnesium hydroxide B. AgBr C. CuS D. MgCO3 Write the expression for K in terms of the solubility, s, for each salt, when dissolved in water. aluminum phosphate magnesium hydroxide Ksp
Chemistry
1 answer:
tigry1 [53]3 years ago
7 0

Answer:

1.AD

2.BC

Explanation:

For aluminum phosphate AlPO4 (s) <------> Al^3+(aq) + PO4^3-(aq)

Ksp= [Al^3+] [PO4^3-]

For magnesium hydroxide

Mg(OH)2(s) <--------> Mg^2+(aq) +. 2OH^-(aq)

Ksp= [Mg2+] [2OH^-]^2

For

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4 years ago
What is the likely impact of water's high specific heat on life on Earth? (3 points)
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Answer:

Marine life is least affected by the temperature fluctuations of the atmosphere.

Explanation:

The high specific heat capacity of water means that it takes much more energy to raise the temperatures of water by one (1) degree than land. This means that on a hot sunny day, land temperatures would increase dramatically while ocean temperatures would only rise slightly. Conversely,  at night, the land cools rapidly while oceans cool slowly hence the temperatures drop slightly.

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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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