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3 years ago

What does it mean when a mineral has a definite chemical composition?

Chemistry

1 answer:

mote1985 [20]3 years ago

4 0

Answer:

For a substance to classify as a mineral, it must lie within certain parameters. It should be an inorganic solid, that is naturally occurring in nature (not synthesized), with an ordered internal structure and a definite chemical composition.

By definite chemical composition, geologists mean that the mineral must be have chemical constituents that have an unvarying chemical composition, or a chemical composition that oscillates withing a very limited and specific range.

An example is the mineral, halite. It has a chemical composition of one sodium atom and one chloride atom, represented as NaCl and is unchanging in this composition throughout nature.

<h3>Hope this helps</h3>

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A monatomic ideal gas that is initially at 1.50 * 105 Pa and has a volume of 0.0800 m3 is compressed adiabatically to a volume o

lubasha [3.4K]

Answer:

300000Pa or 3×10^5 Pa

Explanation:

Since the problem involves only two parameters of volume and pressure, the formula for Boyle's law is suitably used.

Using Boyle's law

P1V1 = P2V2

P1 is the initial pressure = 1.5×10^5Pa

V1 is the initial volume = 0.08m3

P2 is the final pressure (required)

V2 is the final volume = 0.04 m3

From the formula, P2 = P1V1/V2

P2 = 1.5×10^5 × 0.08 ÷ 0.04

= 300000Pa or 3×10^5 Pa.

8 0

3 years ago

A 0. 462 g sample of a monoprotic acid is dissolved in water and titrated with 0. 180 m koh. what is the molar mass of the acid

Len [333]

The molar mass of the acid if 28. 5 ml of the koh solution is required to neutralize the sample is 90.23g/mol.

<h3>For the calculation of molarity of solution</h3>

Molarity = (moles of solute/volume of solution) × 1000

Given,

Molarity of KOH solution = 0.180 M

Volume of solution = 28.5 mL

0.180 = (moles of KOH/ 28.5) × 1000

Moles = (0.18× 28.5)/1000

= 0.00513 mol

<h3>Chemical equation for the reaction</h3>

HA + KOH ------- KA + H2O

1 moles of KOH reacts with 1 moles of HA.

So, 0.00513 moles of KOH react with 0.00513 moles of HA.

<h3>To calculate the molar mass for given number of moles</h3>

Number of moles= given mass/ Molar mass

Given,

Mass of HA = 0.462 g

Moles of HA = 0.00512 mol

0.00512 = 0.462/ Molar mass

Molar mass = 90.23 g/ mol.

Thus the molar mass of HA required to neutralize the 28.5 mL of KOH is 90.23g/mol.

learn more about molar mass or moles:

brainly.com/question/26416088

#SPJ4

3 0

1 year ago

The International Union of Chemistry (IUC) developed a specific set of rules for chemists to follow. What were these rules calle

vampirchik [111]

im just seeing something

sorry egnor me ok?

8 0

3 years ago

Read 2 more answers

How many grams of ammonia (NH3) are produced when you react 4.76 grams of hydrogen (H2)

Andrews [41]

Answer: hre

Explanation:

N2(g) + 3H2-> 2NH3(g) This is the balanced equation

Note the mole ratio between N2, H2 and NH3. It is 1 : 3 : 2 This will be important.

moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present

moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present

Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2.

moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced

grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced

NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.

6 0

4 years ago

Read 2 more answers

In a Dam, If we doubled the depth of the dam the hydrostatic force will be ?

AnnZ [28]

Answer: Option (A) is the correct answer.

Explanation:

Force acting on a dam is as follows.

F = $\frac{1}{2}\rho g\omega H^{2}$ .......... (1)

Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.

F' = $\frac{1}{2}\rho g\omega (2)^{2}$

F' = $\frac{1}{2}\rho g\omega 4$ ........... (2)

Now, dividing equation (1) by equation (2) as follows.

$\frac{F}{F'}$ = $\frac{\frac{1}{2}\rho g\omega H^{2}}{\frac{1}{2}\rho g\omega 4}$

Cancelling the common terms we get the following.

$\frac{F}{F'}$ = $\frac{1}{4}$

4F = F'

Thus, we can conclude that if doubled the depth of the dam the hydrostatic force will be 4F.

4 0

3 years ago