CHEMISTRY COLLEGE

What volume in milliliters of H20 you need during preparation of 100 milliliters of 14.00 M H2SO4 solution starting with bottle of 17.5 M H2SO4? (Round your answer to closest integer)

Answers

Answer 1

Answer:

Answer:

20 mL

Explanation:

We can determine the required volume of the concentrated (17.5 M) H₂SO₄ solution by using the C₁V₁=C₂V₂ formula:

17.5 M * V₁ = 14.00 M* 100 mL

V₁ = 80 mL

As out of the 100 mL of the final solution, 80 mL are from the concentrated H₂SO₄ solution, the remaining 20 mL are of water (H₂O).

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What is the most likely position for the hurricane indicated by the wind readings from the three weather stations shown?

Answers

Here's the answer, I remember doing this problem last year.

23.5 degrees north, 77 degrees west

The molecule XeF2 does not obey the octet rule. Draw its Lewis structure and state the type of octet-rule exception. Include lone pair electrons in the Lewis structure.

Answers

Answer: See attachment.

Explanation:

Exceptions to the octet rule fall into three categories:

An incomplete octet.
An odd number of electrons.
More than eight valence electrons around the central atom.

In addition to the 3s and 3p orbitals, xenon also has 3d orbitals that can be used in bonding. These orbitals enable xenon to form an expanded octet.

Can I+ (the iodine cation) becalled a Lewis base?
Fullyexlplain your answer.

Answers

Answer: No I+ cannot be called a Lewis base.

Explanation:

According to Lewis Theory, it defines an acid as an electron-pair acceptor and a base as an electron-pair donor.

In terms of Lewis basicity, Iodide ion (anions) has the more readily available lone pair electrons for donation since iodide ion is less electronegative .

With the help of the net electronic structure one can understand the answer of the question, because we need to study the I+ ion (cation) structure.

Lewis acid is therefore any substance, that can accept a pair of nonbonding electrons.

From the picture below I+ is most likely ready to accept electrons not to give from it 5s orbital to become stable.

Find the mass of oxygen in grams produced by the decomposition of 100.0 g of CO2

Answers

The balanced chemical equation is :

$2CO_2->2CO+O_2\n\n$

Moles of $CO_2$ ,

$n = (100\ g)/(44.01\ g/mol)\n\nn=2.27\ mol$

Now, by given chemical equation , we can see 2 mole of $CO_2$ react with 1 mole of $O_2$ .

So , 2.27 mole react with :

$N=(2.27)/(2)\ mol\n\nN=1.135\ mol$

Mass of oxygen is :

$M = N * 16\n\nM=1.135* 16\ g\n\nM =18.16\ g$

Therefore, mass of oxygen in grams produced is 18.16 g.

Hence, this is the required solution.

How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid? How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid? a. 37.50 mL
b. 50.00 mL
c. 75.00 mL
d. 100.00 mL
e. 25.00 mL

Answers

Answer:

We need 75 mL of 0.1 M NaOH ( Option C)

Explanation:

Step 1: Data given

Molarity of NaOH solution = 0.100 M

volume of 0.150 M CH3COOH = 50.00 mL = 0.05 L

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles of CH3COOH

Moles CH3COOH = Molarity * volume

Moles CH3COOH = 0.150 M * 0.05 L

Moles CH3COOH = 0.0075 moles

Step 4: Calculate moles of NaOH

For 1 mol of CH3COOH we need 1 mol of NaOH

For 0.0075 mol CH3COOH we need 0.0075 mole of NaOH

Step 5: Calculate volume of NaOH

volume = moles / molarity

volume = 0.0075 moles / 0.100 M

Volume = 0.075 L = 75 mL

We need 75 mL of 0.1 M NaOH

Object A has a molar heat of 31.2 J/mole∙°C and object B molar heat is 11.2 J/mole∙°C. Which object will heat up faster if they have the same mass and equal amount of heat is applied? Explain why.

Answers

Answer:

Substance B

Explanation:

Molar heat of A = 31.2J/mole.°C

Molar heat of B = 11.2 J/mole∙°C.

The molar heat of a substance is the amount of heat that must be added to a mole of a substance to raise the temperature by 1°C.

Substance B will heat up faster compared to A.
It has a smaller molar heat compared to A.
This suggests that it will require lesser heat to raise its temperature by 1°C.

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