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Drupady [299]
2 years ago
7

Brainless so come in help a bro out

Chemistry
1 answer:
Ludmilka [50]2 years ago
8 0
The answer should be D
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PLEASE HELP IM SO CONFUSED How does potassium need to be modified either on site or in a factory to make it useful. This should
3241004551 [841]


Potassium is not found  free in nature but is found in the form of potash. Potash is the ore of potassium and this ore is mined from deep down the earth or can sometimes  be found on the surface. Potash was mostly formed as sea water receded and left deposits.

Potash is usually in the form of potassium salts such potassium chloride and  potassium sulphate.  The potash  is mined then taken to the factory where it is crushed and purified  by removing such impurities as clay.

The now purified potassium salts are subjected to a process called electrolysis where  potassium metal is obtained from its salt. 

5 0
3 years ago
Which pair of ions can form an ionic bond with each other and why? A.Cu and Ag; They are both metal ions.B. S and O; They have l
kap26 [50]
Opposite pairs form ionic bonds, due to this the answer is D - Li and Br; They have unlike charges.
5 0
3 years ago
Read 2 more answers
Write the Ka expression for an aqueous solution of hydrofluoric acid: (Note that either the numerator or denominator may contain
damaskus [11]

Answer:

Ka = ( [H₃O⁺] . [F⁻] ) / [HF]

Explanation:

HF is a weak acid which in water, keeps this equilibrium

HF (aq)  +  H₂O (l)  ⇄  H₃O⁺ (aq)  +  F⁻ (aq)      Ka

2H₂O (l)  ⇄  H₃O⁺ (l)  +  OH⁻ (aq)   Kw

HF is the weak acid

F⁻ is the conjugate stron base

Let's make the expression for K

K = ( [H₃O⁺] . [F⁻] ) / [HF] . [H₂O]

K . [H₂O] = ( [H₃O⁺] . [F⁻] ) / [HF]

K . [H₂O] = Ka

Ka, the acid dissociation constant, includes Kwater.

4 0
3 years ago
How many milliliters of a 1.5 m h2so4 are needed to neutralize 35ml sample of a 1.5 m solution?
DochEvi [55]

Answer:

1) 17.5 mL

Explanation:

Hello,

In this case, the reaction between sulfuric acid and potassium hydroxide is:

H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O

In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:

2*n_{acid}=n_{base}

And in terms of concentrations and volumes:

2*M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the volume of acid:

V_{acid}=\frac{M_{base}V_{base}}{2*M_{acid}} =\frac{35mL*1.5M}{2*1.5M} \\\\V_{acid}=17.5mL

Best regards.

5 0
3 years ago
The image shows an example of a sedimentary rock.
Pani-rosa [81]
Can I see the image?
5 0
3 years ago
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