All categories

3 years ago

A rock is an

Chemistry

2 answers:

Gekata [30.6K]3 years ago

6 0

Answer:

mineral

Explanation:

OLga [1]3 years ago

6 0

Answer:

1. Aggregate

2). Color

3). Mineral

Explanation:

1. A rock is an aggregate of minerals

2. The color of a rock is a result of its minerals composition

3 most rocks contain more than one type of minerals

You might be interested in

How much positive charge is in 1.4 kg of oxygen? The atomic weight (15.9994 g) of oxygen contains Avogadro’s number of atoms, wi

Cerrena [4.2K]

Answer:

$6.7511\times 10^7\ C$

Explanation:

The atomic weight of oxygen = 15.9994 g

This mass corresponds to 1 mole of the oxygen atoms.

Thus,

15.9994 g mass of oxygen contains $6.02\times 10^{23}$ atoms of oxygen.

1.4 kg = 1400 g ( 1 kg = 1000 g)

So,

1400 g mass of oxygen contains $\frac {6.02\times 10^{23}}{15.9994}\times 1400$ atoms of oxygen.

Number of atoms in 1400 g of oxygen = $526.769754\times 10^{23}$

Also, 1 atom of oxygen contains 8 protons

Charge of 1 proton = + $1.602\times 10^{-19}\ C$

So, Charge on 1 atom of oxygen = $8\times 1.602\times 10^{-19}\ C$

Thus,

Charge on $526.769754\times 10^{23}$ atoms of oxygen = $526.94476\times 10^{23}\times 8\times 1.602\times 10^{-19}\ C=6.7533\times 10^7\ C$

Thus, positive charge in 1.4 kg of oxygen = $6.7511\times 10^7\ C$

5 0

3 years ago

How can you separate a mixture containing two different liquids? filtration evaporation distillation magnetic attraction

maxonik [38]

Distillation I believe.....

7 0

4 years ago

Calculate the equilibrium concentration of H 3 O H3O in a 0.20 M M solution of oxalic acid. Express your answer to two significa

Black_prince [1.1K]

Answer: The equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

Explanation:

We are given:

Molarity of oxalic acid solution = 0.20 M

Oxalic acid $(H_2C_2O_4)$ is a weak acid and will dissociate 2 hydrogen ions.

The chemical equation for the first dissociation of oxalic acid follows:

$H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)$

Initial: 0.20

At eqllm: 0.20-x x x

The expression of first equilibrium constant equation follows:

$Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}$

We know that:

$Ka_1\text{ for }H_2C_2O_4=0.059$

Putting values in above equation, we get:

$0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = x = 0.083 M

The chemical equation for the second dissociation of oxalic acid:

$HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)$

Initial: 0.083

At eqllm: 0.083-y 0.083+y y

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}$

We know that:

$Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}$

Putting values in above equation, we get:

$6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = y = 0.0000639 M

Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M

Hence, the equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

7 0

3 years ago

Which substance can be described as a strong base

Shalnov [3]

Lithium Hydroxide or LiOH
Basically anything with a hydroxide (OH) is a strong base

7 0

3 years ago

Read 2 more answers

A mixture contains only NaCl and Al2(SO4)3. A 1.45-g sample of the mixture is dissolved in water, and an ex- cess of NaOH is add

chubhunter [2.5K]

Answer:

The mass percent of aluminum sulfate in the sample is 16.18%.

Explanation:

Mass of the sample = 1.45 g

$Al_2SO_3+6NaOH\rightarrow 2Al(OH)_3+3Na_2SO_4$

Mass of the precipitate = 0.107 g

Moles of aluminum hydroxide = $\frac{0.107 g}{78 g/mol}=0.001372 mol$

According to reaction, 2 moles of aluminum hydroxide is obtained from 1 mole of aluminum sulfate .

Then 0.001372 moles of aluminum hydroxide will be obtained from:

$\frac{1}{2}\times 0.001372 mol=0.000686 mol$

Mass of 0.000686 moles of aluminum sulfate :

= 0.000686 mol × 342 g/mol = 0.2346 g

The mass percent of aluminum sulfate in the sample:

$=\frac{ 0.2346 g}{1.45g}\times 100=16.18\%$

5 0

3 years ago