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photoshop1234 [79]
3 years ago
5

Which element is extracted from bauxide ore​

Chemistry
2 answers:
FinnZ [79.3K]3 years ago
8 0
The answer is Aluminum
lorasvet [3.4K]3 years ago
5 0

Answer:

ALUMINUM

✧\(>o<)ノ✧

...............................

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10. A sample of an unknown composition was tested in a laboratory. The sample could not be broken down by
damaskus [11]

Answer:

it would be an element because its an element

Explanation:

8 0
3 years ago
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What are some visible signs of an acid-base reaction?
marta [7]
I think one of the signs is <span>water and  salt are formed 

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6 0
3 years ago
You have 0.14 moles of sodium chloride (NaCl). How many grams do you have? ( 1 mole of NaCl = 58 grams NaCl) *
Step2247 [10]

Answer:

Option D is correct = 8.12 grams of NaCl

Explanation:

Given data:

Moles of sodium chloride = 0.14 mol

Mass of sodium chloride = ?

Solution:

Formula:

Number of moles = mass of NaCl / Molar mass of NaCl

Molar mass of NaCl = 58 g/mol

Now we will put the values in formula.

0.14 mol = Mass of NaCl / 58 g/mol

Mass of NaCl = 0.14 mol  ×  58 g/mol

Mass of NaCl = 8.12 g of NaCl

Thus, 0.14 moles of NaCl contain 8.12 g of NaCl.

4 0
3 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
3 years ago
Can someone plz help me???? will mark as brainliest!!!!!!!!!!!
Nookie1986 [14]

1=c    2=a     3=b

i hope these answer your questions


7 0
3 years ago
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