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liberstina [14]
2 years ago
7

Help. I need to finish this in time for school tomorrow

Chemistry
1 answer:
zhenek [66]2 years ago
7 0
Hope this helps!

5 - Gas
4 - Weight
1 - Elements
3 - Atoms
7 - Mandeleev
2 - Compounds
6 - Carbon Dioxide
8- IUPAC

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In the laboratory you dissolve 12.3 g of chromium(ii) sulfate in a volumetric flask and add water to a total volume of 375. ml.
valentinak56 [21]
Hope it cleared your doubt.

4 0
3 years ago
Which of the following is a galvanic cell?
sladkih [1.3K]

C. Aluminum (Al) oxidized, zinc (Zn) reduced

<h3>Further explanation</h3>

Given

Metals that undergo oxidation and reduction

Required

A galvanic cell

Solution

The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.

\large {\boxed {\bold {E ^ ocell = E ^ ocatode -E ^ oanode}}}

or:  

E ° cell = E ° reduction-E ° oxidation  

For the reaction to occur spontaneously (so that it E cell is positive), the  E° anode must be less than the E°cathode

If we look at the voltaic series:

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>

The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)

<em />

From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.

7 0
3 years ago
Read 2 more answers
Q7-A graduated cylinder is filled to the 12.0 mL line with water. A solid with a mass of 14.52 g is placed in the graduated cyli
Tems11 [23]

Answer:

If it served you, give me 5 stars please, thank you!

<h3><u>c) 13.29 mL</u></h3>

6 0
2 years ago
Helppppppppppppppppppp me
masya89 [10]
The answer is A.
A pure substance is pure, so it cannot be separated in most cases.
8 0
3 years ago
Suppose of nickel(II) chloride is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of chl
stich3 [128]

Answer: Molarity of chloride anion = 0.32 M

<em>Note: the question is missing some values. The full question is given below;</em>

<em>Suppose 7.26 g of nickel(II) chloride is dissolved in 350 mL of a 0.50 M aqueous solution of potassium carbonate. Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the nickel(II) chloride is dissolved in it. Be sure your answer has the correct number of significant digits.</em>

Explanation:

Molarity or molar concentration is the number of moles (mol) of component per volume (liters) concentration of solution in mol/L or M

The mass of nickel (II) chloride is 7.26 g.

The volume of potassium carbonate is 350 mL = 0.35 L

The molarity of potassium carbonate solution is 0.50 M

The reaction of nickel (II) chloride and potassium carbonate is given below.

NiCl₂(aq) + KCO₃(aq) --------> KCl(aq) +NiCO₃(s)

The dissociation of nickel (II) chloride is given below.

NiCl₂   -----> Ni²⁺ + 2Cl⁻

The molar mass of nickel (II) chloride is  129.6 g/mol

The moles of nickel (II) chloride can be calculated by the formula given below;

No of moles  = mass(g) / molar mass (g/mol)

No of moles = 7.26 / 129.6 = 0.056 moles

Therefore, molarity of NiCl₂ = 0.056 moles/ 0.35 L = 0.16 M

The molarity of 1 mole nickel (ii) chloride is 0.16 m and according to dissociation of nickel (II) chloride, 1 mole of nickel (II) chloride gives 2 moles of chloride anion.

Therefore, the molarity of chloride anion = 0.16 * 2 = 0.32 M

3 0
3 years ago
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