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SIZIF [17.4K]
3 years ago
6

What is the molarity of a solution made by dissolving 1.25 mol of HCl in enough

Chemistry
1 answer:
Ronch [10]3 years ago
4 0

Answer:

Explanation:

C

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The nucleus of the atom contains the greatest concentration of mass of an atom. which arrow is pointing at the nucleus of nitrog
Mashutka [201]
The nucleus of an atom of any element is located at the very center. It contains subatomic particles, proton and neutron. That's why it contains the greater mass, because only the electrons orbit the nucleus. So, you would expect the arrow to point somewhere in the center. A picture of nitrogen's nucleus is shown in the attached picture.

5 0
2 years ago
For the reaction, 2Cr2+ + Cl2(g) ---> 2Cr3+ + 2Cl- Ecell (standard conditions) = 1.78V
Lapatulllka [165]

Answer:

Eºcell = -1.78 V

Explanation:

The Eº cell is an intensive property, i.e they do not depend on the quantity of material present and the desired reaction in our problem  is exactly half the reverse of the one given, the Eºcell will then  be the negative of the first then Eºcell is -1.78 V and the redox reaction will be non-spontaneous as opposed to the first.

5 0
3 years ago
Can some one plz help
blsea [12.9K]

The reaction uses B) 9.0 g Br₂.

 iron + bromine ⟶ product

2.0 g +     <em>x</em> g     ⟶   11.0 g

According to the <em>Law of Conservation of Mass</em>, the total mass of the reactants must equal the total mass of the products.

∴2.0 + <em>x</em> = 11.0

<em>x</em> = 11.0 – 2.0 = 9.0

The reaction uses 9.0 g Br₂.

5 0
2 years ago
Choose all the answers that apply. Which of the following has kinetic energy? rolling ball, moving car, book on a table, spinnin
Virty [35]
Spinning top, moving car, and rolling ball have kinetic energy I believe
5 0
3 years ago
The radius of an indium atom is 0.163 nm. If indium crystallizes in a face-centered unit cell, what is the length of an edge of
adoni [48]

<u>Answer:</u> The edge length of the unit cell is 0.461 nm

<u>Explanation:</u>

We are given:

Atomic radius of iridium = 0.163 nm

To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:

a=2\sqrt{2}R

Putting values in above equation, we get:

a=2\sqrt{2}\times 0.163=0.461nm

Hence, the edge length of the unit cell is 0.461 nm

8 0
2 years ago
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