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dangina [55]
3 years ago
6

Copper sulfate can crystalize as an anhydrate (no water in the crystal) or as three different hydrates (varying amounts of water

incorporated into the crystal structure). All four forms of copper sulfate are naturally occurring minerals: bonattite, boothite, chalcanthite, and chalcocyanite. Please suggest two factors that might affect which copper sulfate mineral occurs at a given location, including brief explanations.
Chemistry
1 answer:
Maurinko [17]3 years ago
8 0

Answer:

Two factors that might have a affect of which copper sulphate mineral will occur at a given location  is:

A. Copper sulphate high solubility in water

B. Also it binds nicely with the sediments  or the crystal.

Explanation:

As it is mentioned here that copper sulphate can be crystallized as an anhydrate which means that their  is no waterin those crystals or can be as of those three different hydrates whose crystal structure varies with the amount of water present in it.

The four forms are also given of the copper sulphate are:

  • Bonatite
  • Boothite
  • Chalcanthite
  • Chalcocyanite

So, the two factors that might give an affect which type of copper sulphate  mineral willoccur at a given location is:

A. The copper sulphate high solubility in water.

B. It binds extremely nicely with the sediments or say to the crystal. It is  also regulated by plants.

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Answer:

Temperature is one of the major factors that affects the rate of change of the liquid colour, this is because ; like when if you freeze hot water the ice formed will be clear transparent, while on the other hand, if we freeze cold water it would be foggy inside the ice. This change occurs because of the temperature difference of the cold and hot water.

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6 0
3 years ago
Use the information below to answer the following questions.
Slav-nsk [51]

Answer:

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3 0
3 years ago
The density of silver is 10.5 g/cm3. What is the mass (in kilograms) of a cube of silver that measures 0.94 m on each side?
Rzqust [24]

Answer:

8721.132 Kg.

Explanation:

The following data were obtained from the question:

Density (D) of silver = 10.5 g/cm³

Length (L) of silver = 0.94 m

Mass (m) of silver =.?

Next, we shall determine the volume of silver. This can be obtained as follow:

Volume = Length × Length × Length

Volume = L × L × L

Volume = L³

Length (L) of silver = 0.94 m

Volume (V) of silver =?

Volume (V) of silver = 0.94³

Volume (V) of silver = 0.830584 m³

Next, we shall convert 0.830584 m³ to cm³. This can be obtained as follow:

1 m³ = 1×10⁶ cm³

Therefore,

0.830584 m³ = 0.830584 m³ / 1 m³ × 1×10⁶ cm³

0.830584 m³ = 830584 cm³

Therefore, 0.830584 m³ is equivalent 830584 cm³.

Thus, the volume of the silver is 830584 cm³.

Next, we shall determine the mass of the silver. This can be obtained as follow:

Density (D) of silver = 10.5 g/cm³

Volume of the silver = 830584 cm³.

Mass of silver =.?

Density = mass /

10.5 = mass of silver /830584

Cross multiply

Mass of silver = 10.5 × 830584

Mass of silver = 8721132 g

Finally, we shall convert the mass of silver, 8721132 g to kilogram (kg). This can be obtained as follow:

1000 g = 1 Kg

Therefore,

8721132 g = 8721132 g / 1000 g × 1 Kg

8721132 g = 8721.132 Kg

Therefore, the mass of each cube of silver is 8721.132 Kg

3 0
3 years ago
What is the formula for hydrogen iodide trihydrate
Natali [406]

HI.3H2O. This is the answer

6 0
3 years ago
A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH
Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5    → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0
3 years ago
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