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2 years ago

3.4 moles of solid CuSO4 is added to 1.8 L of water and allowed to dissolve. Will all the solid dissolve?

Chemistry

1 answer:

ludmilkaskok [199]2 years ago

8 0

Answer:

Yes, all the solid will dissolve.

Explanation:

To solve this problem we will first find the solubility of CuSO₄ at given temperature (in this case we will assume the temperature to be 20 °C. According to reported data 37.8 g of CuSO₄ can be dissolved in 100 ml of H2O at 20 °C.

While we are provided with moles of CuSO₄ and volume is given in Liters. So, we will convert moles to mass as,

Mass = Moles × Molar Mass

Mass = 3.4 mol × 159.60 g/mol

Mass = 542.64 g

Secondly, we will convert Liters to milliliters as,

Milliliters = Liters × 1000

Milliliters = 1.8 × 1000

Milliliters = 1800 ml

Therefore,

According to reference data,

37.8 g CuSO₄ dissolves in = 100 ml of water

So,

542.64 g CuSO₄ will dissolve in = X ml of water

Solving for X,

X = 542.64 g × 100 ml / 37.8 g

X = 1435.55 ml of Water

This means that we are provided with greater amount of water (i.e. 1800 m). Hence, the given amount of CuSO₄ will completely dissolve in 1.8 L of water and the solution formed is unsaturated and can dissolve further CuSO₄.

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A new student is planning to use thin layer chromatography (TLC) for his research project. After setting up the apparatus the st

tamaranim1 [39]

Answer:

The open system evaporates the solvent in the solution

Explanation:

An open system is a system in which exchange of materials and energy can occur. If a TLC set up is left open, then the set up constitutes an open system.

During TLC, the sample is dotted on the plate and inserted into a suitable solvent. The solvent moves up the plate and achieves the required separation of the mixture.

Most of these solvents used used TLC are volatile organic compounds. Therefore, if the TLC set up is left open, the solvent will evaporate leading to poor results after running the TLC experiment.

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1 year ago

Identify which of the following two reactions you would expect to occur more rapidly: (1) addition of HBr to 2-methyl-2-pentene

arlik [135]

Answer:

(1) addition of HBr to 2-methyl-2-pentene

Explanation:

In this case, we will have the formation of a carbocation for each molecule. For molecule 1 we will have a tertiary carbocation and for molecule 2 we will have a secondary carbocation.

Therefore the most stable carbocation is the one produced by the 2-methyl-2-pentene. So, this molecule would react faster than 4-methyl-1-pentene. (See figure)

3 0

2 years ago

Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of O

Kipish [7]

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = $\frac{15.0 g}{40 g/mol}=0.375 mol$

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

$Molarity=\frac{Moles}{Volume(L)}$

$n=0.250 M\times 0.150 L=0.0375 mol$

$NaOH+HNO_3\rightarrow NaNO_3+H_2O$

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

$\frac{1}{1}\times 0.0375 mol$ of NaOH

Moles of NaOH left unreacted in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

$NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)$

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

$1\times 0.3375 moles=0.3375 mol$ of hydroxide ion

The molarity of hydroxide ion in solution ;

$=\frac{0.3375 mol}{0.150 L}=2.25 M$

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

3 0

2 years ago

Which of the following is a reasonable ground-state electron configuration?

Thepotemich [5.8K]

Answer:

The answer to your question is: letter A.

Explanation:

To solve this question, we must consider the total number of electrons that can be placed in each orbital.

Orbital Number of electrons

s 2

p 6

d 10

f 14

Options given

A. 1s² 2s² 2p⁶ 3s² This electron configuration is correct because

is in agreement with the number of electrons

allowed in each orbital and the correct order

of them.

B. 1s² 2s² 2p⁶ 3s² 3d⁴ This electron configuration is incorrect because

3d⁴ goes after 3p.

C. 1s² 2s² 2d¹⁰ 2p³ This option is incorrect because 2d¹⁰ does not

exist.

D. 1s² 2s^s 2p³ 2d¹⁰ This option has 2 mistakes, s as a power does

not exist and 2d¹⁰ is also incorrect.

3 0

2 years ago

Which substance can be decomposed by chemical means?

liubo4ka [24]

Methane, answer c, can be decomposed by chemical means because methane is a compound. Methane is made up of one atom of carbon and four atoms of hydrogen. Compounds can be broken down. Cobalt, krypton, and zirconium are all elements, and elements cannot be chemically decomposed.

5 0

1 year ago

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