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denis23 [38]
2 years ago
15

What is specific gravity in minerals?

Chemistry
2 answers:
natima [27]2 years ago
4 0

Specific gravity is the "heaviness" of a mineral. It is defined as a number that expresses the ratio between the weight of a mineral and the weight of an equal volume of water.

fomenos2 years ago
4 0

Answer:

The ratio of it's mass of an equal volume of water.

Explanation:

You might be interested in
1. 100gm of a 55% (M/M) nitric acid solution is to be diluted to 20% (M/M) nitric acid.
stealth61 [152]

Answer:

The volume of water to be added is 0.175 liters of water

Explanation:

The given concentration of the nitric acid = 55% (M/M)

The mass of the nitric acid solution = 100 gm

The concentration solution is to diluted to = 20% (M/M)

The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution

Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get

Let "x" represent the volume of the resulting solution, we have;

20% of x = 55 g of nitric acid

∴ 20/100 × x = 55 g

x = 55 g × 100/20 =  275 g

The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid

The mass of extra water to be added = 275 g - 100 g = 175 g

Volume = Mass/Density

The density of water ≈ 1 g/ml

∴ The volume of water to be added that gives 175 g of water =  175 g/(1 g/ml) = 175 ml. = 0.175 l

The volume of water to be added = 0.175 liters of water.

3 0
2 years ago
31) To extend a length of 1.00 inches, 2.54 x 10^8 average sized atoms would have to be placed in a straight line (in other word
Phoenix [80]

Answer:

Here's the conversion factor you need:

1 Kilometer   =   39370 inches

So, for your question we want to go 405,696 km....

405696 km   x   39370 inches/ 1 km  =   15972283464 inches

                       

15972283464 inches   x   2.54 x10^8 atoms/1 inch   =   4.05 x 10^18 atoms

                                     

4 0
2 years ago
Do you know any of these?
bulgar [2K]
The second one is no
6 0
3 years ago
Read 2 more answers
If you have a 1500 g aluminum pot, how much heat energy is needed to raise its temperature by 100°C?
Nataly [62]

The heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ.

The heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

Explanation:

The heat energy required to raise the temperature of any body can be obtained from the specific heat formula. As this formula states that the heat energy required to raise the temperature of the body is directly proportional to the product of mass of the body, specific heat capacity of the material and temperature change experienced by the material.

So in this problem, the mass of the aluminium is given as m = 1500 g, the specific heat of the aluminium is 0.900 J/g °C. Then as it is stated that the temperature is raised by 100 °C, so the pots are heat to increase by 100 °C from its initial temperature. This means the difference in temperature will be 100°C (ΔT = 100°C).

Then, the heat energy required to raise the temperature will be

q = m*c*del T = 1500 * 0.900 * 100 = 135000 = 135 kJ

Thus, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100 °C is 135 kJ.

Similarly, the mass of copper pot is given as 1500 g, the specific heat capacity of copper is 0.385 and the difference in temperature is 100  °C.

Then, the heat energy required to raise its temperature will be

q = m*c*del T = 1500 * 0.385 * 100 = 57750 = 57.75 kJ

And the heat energy required to raise the temperature of 1500 g of copper pot by 100°C is 57.75 kJ.

So, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ. And the heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

8 0
3 years ago
How to balance the equation Mg2+HO2=Cu3O+H2<br><br>​
Mariulka [41]
I think the answer is 2Mg + H2O4 = Cu12O4 + 2H

I’m really not sure though so it might be wrong… I’m not the best at balancing equations lol
6 0
2 years ago
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