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3 years ago

How many millimeters of 1.25 M Copper(2) sulfate, CuSO4, will contain 55.75g CuSo4

Chemistry

1 answer:

Shkiper50 [21]3 years ago

8 0

Answer:

280 mL

Explanation:

Given data:

Mass of CuSO₄ = 55.75 g

Molarity of CuSO₄ = 1.25 M

Volume of CuSO₄ = ?

Solution:

First of all we will calculate the number of moles.

Number of moles:

Number of moles = mass / molar mass

Number of moles = 55.75 g/ 159.6 g/mol

Number of moles = 0.35 mol

Volume of CuSO₄:

Molarity = number of moles / volume in litter

1.25 M = 0.35 mol / volume in litter

Volume in litter = 0.35 mol / 1.25 M

Volume in litter = 0.28 L

Volume in milliliter:

0.28 L × 1000 mL/ 1 L = 280 mL

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Benzene exists as a resonance hybrid and its bonds exhibit characteristics that are halfway between single and double bonds. Res

gregori [183]

Answer:

Explanation:

We can only talk about resonance hybrid for a compound in which more than one structure is possible based on its observed chemical properties.

There are compounds whose chemical properties can not be satisfactorily explained on the basis of a single chemical structure. In the case of such compounds, we invoke the idea of resonance.

A resonance hybrid is a single structure drawn to represent a given chemical specie which exhibits resonance behaviour and can otherwise be represented on paper in the form of an average of two or more chemical structures separated each from the next by a double-headed arrow.

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3 years ago

A saturated solution of manganese(II) hydroxide was prepared, and an acid–base titration was performed to determine its KspKsp a

trapecia [35]

Answer:

10.945 x 10^-4

Explanation:

Balanced equation:

Mn(OH)2 + 2 HCl --> MnCl2 + H2O

it takes 2 moles HCL for each mole Mn(OH)2

Next find the molarity of the Mn(OH)2 solution

= (1 mole Mn(OH)2 / 2 mole HCl) X (0.0020 mole HCl / 1000ml) X (4.86 ml)

= 4.86 x 10^-3 mole

this is now dissolved in (70 + 4.86) = 74.86 ml or 0.07486 L

thus [Mn(OH)2] = 4.86 x 10^-3 mole / 0.07486 L = 0.064921 M

Ksp = [Mn2+][OH-]^2 = 4x^3 = 4(0.064921)^3 = 10.945 x 10^-4

6 0

3 years ago

How many elements are in N2 and how many atoms are in N2?

Lady bird [3.3K]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

7. Suppose 1.01 g of iron (III) chloride is placed in a 10.00-mL volumetric flask with a bit of water in it. The flask is shaken

Nana76 [90]

<u>Answer:</u> The molarity of Iron (III) chloride is 0.622 M.

<u>Explanation:</u>

Molarity is defined as the number of moles present in one liter of solution. The equation used to calculate molarity of the solution is:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Or,

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of iron (III) chloride = 1.01 g

Molar mass of iron (III) chloride = 162.2 g/mol

Volume of the solution = 10 mL

Putting values in above equation, we get:

$\text{Molarity of Iron (III) chloride}=\frac{1.01g\times 1000}{162.2g/mol\times 10mL}\\\\\text{Molarity of Iron (III) chloride}=0.622M$

Hence, the molarity of Iron (III) chloride is 0.622 M.

3 0

3 years ago

Given the following UNBALANCED reaction: NH3 (g) <--> N2 (g) + H2 (g) If 1

Yakvenalex [24]

Answer:

C. 1.35

Explanation:

2NH3 (g) <--> N2 (g) + 3H2 (g)

Initial concentration 2.2 mol/0.95L 1.1 mol/0.95L 0

change in concentration 2x x 3x

-0.84 M +0.42M +1.26M

Equilibrium 1.4 mol/0.95L=1.47M 1.58 M 1.26 M

concentration

Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M

Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M

Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M

K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M

8 0

3 years ago