04.04.2022

What mass of feso4^2- x 6h20 (molar mass=260g/mol) is required to produce 500 ml of a .10m iron (ii) sulfate solution.
a.) 9g
b.) 13g
c.) 36g
d.) 72g

. 0

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30.05.2023, solved by verified expert
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The correct option is: B. 13g

Explanation:

Given: Molar mass of iron (II) sulfate: m = 260g/mol,

Molarity of iron (II) sulfate solution: M =  0.1 M,

Volume of iron (II) sulfate solution: V = 500 mL = 500 × 10⁻³ = 0.5 L           (∵ 1L = 1000mL)

Mass of iron (II) sulfate taken: w = ? g

Molarity: What mass of feso4^2- x 6h20 (molar mass=260g/mol), №16482452, 04.04.2022 20:23

Here, n- total number of moles of solute, w - given mass of solute, m- molar mass of solute, V- total volume of solution in L

∴ Molarity of iron (II) sulfate solution:  What mass of feso4^2- x 6h20 (molar mass=260g/mol), №16482452, 04.04.2022 20:23

⇒  What mass of feso4^2- x 6h20 (molar mass=260g/mol), №16482452, 04.04.2022 20:23

⇒  What mass of feso4^2- x 6h20 (molar mass=260g/mol), №16482452, 04.04.2022 20:23

⇒  mass of iron (II) sulfate taken: What mass of feso4^2- x 6h20 (molar mass=260g/mol), №16482452, 04.04.2022 20:23

Therefore, the mass of iron (II) sulfate taken for preparing the given solution is 13 g.

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Chemistry
Step-by-step answer
P Answered by Specialist

The correct option is: B. 13g

Explanation:

Given: Molar mass of iron (II) sulfate: m = 260g/mol,

Molarity of iron (II) sulfate solution: M =  0.1 M,

Volume of iron (II) sulfate solution: V = 500 mL = 500 × 10⁻³ = 0.5 L           (∵ 1L = 1000mL)

Mass of iron (II) sulfate taken: w = ? g

Molarity: M = \frac{n}{V (L)} = \frac{w}{m\times V(L)}

Here, n- total number of moles of solute, w - given mass of solute, m- molar mass of solute, V- total volume of solution in L

∴ Molarity of iron (II) sulfate solution:  M = \frac{w}{m\times V(L)}

⇒  w = M\times m\times V(L)

⇒  w = (0.1 M)\times (260g/mol)\times (0.5L)

⇒  mass of iron (II) sulfate taken: w = 13 g

Therefore, the mass of iron (II) sulfate taken for preparing the given solution is 13 g.

Chemistry
Step-by-step answer
P Answered by Master
Answer: 0.0045 mol
Explanation: Convert 30 ml to l: 30 mL = 0.03 L
Molarity = mol/l
mol = molarity * L
mol = 0.15 * 0.03 = 0.0045 mol
Answer: 0.0045 mol
Explanation: Convert 30 ml to l: 30 mL = 0.03 L
Molarity = mol/l
mol = molarit
Chemistry
Step-by-step answer
P Answered by PhD

Answer:

52.6 gram

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Chemistry
Step-by-step answer
P Answered by PhD

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N ×328 K

Solving:

Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

Chemistry
Step-by-step answer
P Answered by PhD
15 moles.Explanation:Hello,In this case, the undergoing chemical reaction is:Clearly, since carbon and oxygen are in a 1:1 molar ratio, 15 moles of carbon will completely react with 15 moles of oxygen, therefore 15 moles of oxygen remain as leftovers. In such a way, since carbon and carbon dioxide are also in a 1:1 molar ratio, the theoretical yield of carbon dioxide is 15 moles based on the stoichiometry:Best regards.
Chemistry
Step-by-step answer
P Answered by PhD
Answer: 25 g
Explanation: Given:
Original amount (N₀) = 100 g
Number of half-lives (n) = 11460/5730 = 2
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^2 × 100
N = 0.25 × 100
N = 25 g
Chemistry
Step-by-step answer
P Answered by PhD
Answer: 7.8125 g
Explanation: Given:
Original amount (N₀) = 500 g
Number of half-lives (n) = 9612/1602 = 6
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^6 × 500
N = 0.015625 × 500
N = 7.8125 g
Chemistry
Step-by-step answer
P Answered by PhD
Answer: The product formed is potassium chloride.
Explanation:
Precipitation reaction is defined as the chemical reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

The chemical equation for the reaction of potassium phosphate and magnesium chloride follows (look at the picture)

2 moles of aqueous solution of potassium phosphate reacts with 3 moles of aqueous solution of magnesium chloride to produce 1 mole of solid magnesium phosphate and 6 moles of aqueous solution of potassium chloride.
Answer: The product formed is potassium chloride.
Explanation:
Precipitation reaction is defined a
Chemistry
Step-by-step answer
P Answered by PhD
Answer: B. carbon tetrachloride, CCI4
Explanation: The other options are incorrect. Let's write the correct formulas:
A. Diarsenic pentoxide - As2O5
C. Sodium dichromate - Na2Cr2O7
D. magnesium phosphide - Mg3P2

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