CHEMISTRY COLLEGE

Which of the following is an example of physical change?

Answers

Answer 1

Answer:

Answer:

The glass cup falling from the counter

Explanation:

the glass isn't changing in any chemical way. it's still made of the same material, just broken apart.

Answer 2

Answer:

Final answer:

Physical changes involve the alteration of the state or appearance of matter, without changing the composition. An example is solid wax turning into liquid wax when heated, or steam condensing inside a cooking pot.

Explanation:

The question asks for an example of a physical change. Physical changes involve alterations in the state or appearance of matter, without changing its composition. For example, solid wax turning into liquid wax when heated is a physical change. The wax is still the same substance, it's just in a different state. Similarly, steam condensing inside a cooking pot is also a physical change. The water vapor turns back into liquid water, but it's still water. These are distinguished from chemical changes, which transform one substance into a different substance.

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If there is direct variation and y=75 when x=25, find x when y=48.A. x=16

B. x=10

C. x=12

D. x=14

Answers

If there is direct variation and y=75 when x=25,when y=48 X = 16. Therefore, option A is correct.

What is direct variation ?

link between two variables that can be described mathematically by an equation where one variable equals a constant multiplied by the other. For instance, the constant of variation is k = = 3 if y varies straight as x and y = 6 when x = 2. Consequently, y = 3x is the equation that describes this directvariation.

When x is not equal to zero, an equation of the form y = kx describes the linear function known as direct variation. When x is not equal to zero and k is a nonzero real number constant, the equation of the form xy = k describes the nonlinear function known as inverse variation.

x = ky

where k is constant

25 = k × 75

25 / 75 = k

k = 1 / 3

now when y = 48 then substitute the values

x = ky

x = 1 ÷ 3 × 48

x = 16

Thus, option A is correct.

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x = ky where k is a constant

first u find k so :
25 = k * 75
25/75 = k
k = 1/3

now when y = 48 just substitute the values
x = ky
x = 1/3 * 48
x = 16

the answer is A) x=16

What reaction conditions most effectively conver a cabocxylic acid to a methly ester?

Answers

Answer:

Esterification reaction

Explanation:

When we have to go from an acid to an ester we can use the esterification reaction. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).

In this case, we need the methyl ester, therefore we have to choose the appropriate alcohol, so we have to use the methanol as reactive if we have to produce the methyl ester.

How many atoms does 6H2O contain

Answers

Answer:

two atoms of oxygen. For H2O, there is one atom of oxygen and two atoms of hydrogen.

You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol,MW=60.05 g/mol, pKa=4.76), pKa=4.76), 3.00 M NaOH,3.00 M NaOH, and water. Answer the questions regarding the preparation of the buffer. 1. How many grams of acetic acid will be needed to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

Answers

The quantity of acetic acid that is needed to prepare the 500 mL buffer is 9.0075 grams.

Given the following data:

Volume of acetate buffer = 500 mL to L = 0.5 L.

Molarity of acetate buffer = 0.300 M.

pH = 4.90.

MW = 60.05 g/mol.

pKa = 4.76.

How to calculate the mass of acetic acid.

First of all, we would write the equilibrium chemical reaction for acetate-acetic acid as follows:

$CH_3COOH \rightleftharpoons CH_3COO^(-)+H^+$

Next, we would calculate HA by applying Henderson-Hasselbalch equation:

$pH =pka+ log_(10) (A^-)/(HA)$

Where:

HA is acetic acid.

$A^-$ is acetate ion.

Substituting the given parameters into the formula, we have;

$4.90 =4.76+ log_(10) (A^-)/(HA)\n\n4.90 -4.76+ log_(10) (A^-)/(HA)\n\n(A^-)/(HA)=1.38\n\nA^- = 1.38[HA]$

For the concentration of both acids, we have:

$[HA]+[A^-]=0.300M\n\n[HA]+1.38[HA]=0.300M\n\n2.38[HA]=0.300M\n\nHA = 0.126$

For acetate ion:

$A^- = 1.38[HA] = 1.38 * 0.126\n\nA^- =0.174$

At a volume of 0.5 liters, we have:

$HA = 0.5 * 0.126\n\nHA = 0.063 \;moles$

$A^- = 0.5 * 0.174\n\nA^- =0.087 \;moles$

By stoichiometry:

Total moles = $0.063 + 0.087$ = 0.15 moles.

$Mass = number \;of \;moles * molar\;mass\n\nMass =0.15 * 60.05$

Mass = 9.0075 grams.

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Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀ $([A^(-)])/([HA])$

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀ $([A^(-)])/([HA])$

1,38 = $([A^(-)])/([HA])$ (1)

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] (2)

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid ≡ 0,15 moles of acetic acid × $(60,05 g)/(1mol)$ = 9,0 g of acetic acid

I hope it helps!

A volume of 40.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2SO4). What was the molarity of the KOHKOH solution if 16.2 mLmL of 1.50 MM H2SO4H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Answers

Answer: The concentration of KOH solution is 1.215 M

Explanation:

For the given chemical equation:

$2KOH(aq.)+H_2SO_4(aq.)\rightarrow K_2SO_4(aq.)+2H_2O(l)$

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\nM_1=1.50M\nV_1=16.2mL\nn_2=1\nM_2=?M\nV_2=40.0mL$

Putting values in above equation, we get:

$2* 1.50* 16.2=1* M_2* 40.0\n\nM_2=(2* 1.50* 16.2)/(1* 40.00)=1.215M$

Hence, the concentration of KOH solution is 1.215 M

A student was asked to prepare exactly 250 mL of a 0.500 M aqueous potassium hydroxide solution. What mass of potassium hydroxide (molar mass = 56.10 g/mol) must the student dissolve in the 250 mL of solution? 1. 28.1 g 2. 3.0 g 3. None of these 4. 14.0 g 5. 7.01 g 6. 56.1 g

Answers

Answer:

Option 5 → 7.01 g

Explanation:

Molarity . volume (L) = Moles

This can help us to determine the moles of KOH that are in the solution.

We convert the volume from mL to L → 250 mL . 1L / 1000mL = 0.250 L

0.5 mol /L . 0.250L = 0.125 moles of KOH

Now, we only have to convert the moles to mass, by the molar mass:

Moles . molar mass = mass → 0.125 mol . 56.1 g/mol = 7.01 g

Answer:

We need 7.01 grams of KOH (option 5)

Explanation:

Step 1: Data given

Volume aqueous KOH solution = 250 mL = 0.250 L

Molarity = 0.500 M

Molar mass of KOH = 56.10 g/mol

Step 2: Calculate moles KOH

Moles KOH = molarity * volume

Moles KOH = 0.500 M * 0.250 L

Moles KOH = 0.125 moles

Step 3: Calculate mass of KOH

Mass KOH = moles KOH * molar mass KOH

Mass KOH = 0.125 moles * 56.10 g/mol

Mass KOH = 7.01 grams

We need 7.01 grams of KOH

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