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Daniel [21]
3 years ago
6

A Solute is the part of the mixture that is A. dissolved B.used to dissolve a substance

Chemistry
1 answer:
8_murik_8 [283]3 years ago
5 0
It is the thing being dissolved!!
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In the determination of iron in an ore, the ore (mostly Fe2O3) is first dissolved in HCl(aq). A slight excess of Sn2 is added to
katovenus [111]

Answer:

2Fe⁺³ + Sn₂ → 2Fe⁺² + 2Sn⁺²

Explanation:

A redox reaction occurs between a compound that loses electrons and others that gain an electron. The first is being oxidized, and the other is being reduced.

In this situation, in the compound Fe₂O₃, the iron, has an oxidation number equal to +3, so it's Fe⁺³, and it will gain 1 electron to become Fe⁺². Because it was first dissolved in HCl, we must use the ion at the equation. The other compound Sn₂ will be oxidized to Sn⁺², so it will need to lose 2 electrons.

So, it will be necessary 2 Fe⁺³ for this reaction happen:

2Fe⁺³ + Sn₂ → 2Fe⁺² + 2Sn⁺²

8 0
3 years ago
A student has 2 rubber balls of the same size and weight. Ball A is still and Ball B is rolling.Ball B hits Ball A, and Ball A s
Akimi4 [234]

Answer:

It starts to move because of Newtons first law "An object will not change it's motion unless a force acts on it"

Explanation:

Meaning ball B is that force acting upon ball A

6 0
3 years ago
An element X combines with oxygen to form a compound of formula XO2. If 24.0 grams of element X combines with exactly 16.0 grams
tensa zangetsu [6.8K]

Answer:

atomic mass of X is 48.0 amu

Explanation:

Let y be the atomic mass of X

Molar mass of O_2 is = 2×16 = 32 g / mol

X + O2 -----> XO_2

According to the equation ,

y g of X reacts with 32 g of O_2

24 g of X reacts with Z g of O_2

Z = ( 32×24) / y

But given that 24.0 g of X exactly reacts with 16.0 g of O_2

So Z = 16.0

⇒ (32×24) / y = 16.0

⇒ y = (32×24) / 16

y= 48.0

So atomic mass of X is 48.0 amu

4 0
3 years ago
(34.969amu)(0.7577) =<br> (36.966amu)(0.2423) =<br> +
Monica [59]

Answer:

26.4 960 for the first one

8.9569 for the second one

7 0
3 years ago
What is the final concentration of cl- ion when 250 ml of 0.20 m cacl2 solution is mixed with 250 ml of 0.40 m kcl solution? (as
serious [3.7K]

CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

<span>CaCl2 ---> Ca2+  +  2 Cl-</span>

KCl ---> K+ + Cl-

Therefore the total Cl- ion concentration would be coming from both salts. First, we calculate the Cl- from each salt by using stoichiometric ratio:

Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles Cl / 1 mole CaCl2)

Cl- from CaCl2 = 0.1 moles

 

Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

 

Therefore the final concentration of Cl- in the solution mixture is:

Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)

Cl- = 0.2 moles / 0.5 moles

<span>Cl- = 0.4 moles             (ANSWER)</span>

6 0
3 years ago
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