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Firdavs [7]
3 years ago
8

Using stoichiometry, determine the mass of powdered drink mix needed to make 1.0 M solution of 100 mL

Chemistry
1 answer:
Rzqust [24]3 years ago
6 0

Answer:

34.23 g.

M = (no. of moles of solute)/(V of the solution (L)).

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An iron(iii) sulfate hydrate is 18.4% water. What is the formula of the hydrate? What is the name of the hydrate?
aleksley [76]

Answer:- Formula of the hydrate is Fe_2(SO_4)_3.5H_2O and it's name is Iron(III)sulfate pentahydrate.

Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.

Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.

We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:

81.6gFe_2(SO_4)_3(\frac{1mol}{399.88g})

= 0.204molFe_2(SO_4)_3

18.4gH_2O(\frac{1mol}{18.02g})

= 1.02molH_2O

Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.

Fe_2(SO_4)_3=\frac{0.204}{0.204}  = 1

H_2O=\frac{1.02}{0.204} = 5

There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is Fe_2(SO_4)_3.5H_2O and the name of the hydrate is Iron(III)sulfate pentahydrate.


3 0
3 years ago
Which excerpt from “Seventh Grade” is an example of an internal expectation?
alex41 [277]
A. Theresa is going to be my girl this year, he promised himself as he left the gym full of students in his new fall clothes.
6 0
3 years ago
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I will put the questions in the comments that i need help with*
Paha777 [63]
The answer is a ) electromagnetic waves
7 0
3 years ago
When balancing redox reactions under basic conditions in aqueous solution, the first step is to:________.
natta225 [31]

Answer:

When balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.

Explanation:

Oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The oxidation and reduction half-reactions, in a basic medium, adjust the oxygens and hydrogens as follows:

In the member of the half-reaction that presents excess oxygen, you add as many water molecules as there are too many oxygen. Then, in the opposite member, the necessary hydroxyl ions are added to fully adjust the half-reaction. Normally, twice as many hydroxyl ions, OH-, are required as water molecules have previously been added.

In short, you first adjust the oxygens with OH-, then you adjust the H with H₂O, and finally you adjust the charge with e-

So, <u><em>when balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.</em></u>

3 0
3 years ago
The first periodic table that was developed arranged elements in order of atomic number.
ziro4ka [17]
Heyyy!!


Your answer is True.


Please mark it as brainliest.


Hope it helps you.
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3 years ago
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