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siniylev [52]
3 years ago
8

If a substans is a mineral,how could you identify what type of mineral it is?

Chemistry
1 answer:
lidiya [134]3 years ago
4 0
You can use color, hardness, luster, streak, breakage pattern, magnetism, effervescense, specific gravity, and even taste or smell to identify minerals.
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We can also use the equation for enthalpy change for physical phase changes. Consider the phase change H2O(l) → H2O(g). Calculat
BlackZzzverrR [31]
<span>44.0 kJ is the answer</span>
4 0
3 years ago
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PLEASE!!! I NEED HELP ASAP!!
Lana71 [14]
W=m₁/m₀=2^(-t/T)

t=4.6·10⁹ years
T=5·10¹⁰ years

w=2^(-4.6·10⁹/5·10¹⁰)

w=0.9382
w=93.82%

7 0
3 years ago
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Please help!
liberstina [14]

Answer:

To help determine what type of rock it is

Explanation:

Geologists can use information such as color, hardness, grain size, texture and other aspects of the rock to figure out the classification of a rock. for example, a light blue rock with no visible grain that is translucent and has a hardness of 9 is most likely going to be a saphire. hope this helps!!!!

8 0
2 years ago
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What claim is being made?
dangina [55]
The correct answer is c
7 0
3 years ago
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What is the maximum amount of HCl, in grams, that can be produced if 37.5 g of BCl3 and 60.0 g of H2O are reacted according to t
Rus_ich [418]
Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.

First, how many moles of each substance are there

the molar mass of BCl3 is <span>117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.

Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles.

But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.

.96 x 36.46 = ~35 g</span>
6 0
3 years ago
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