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100.0 mL of Ca(OH)2 solution is titrated with 5.00 x 10–2 M HBr. It requires 36.5 mL of the acid solution for neutralization. What is the number of moles of HBr used, and the concentration of the Ca(OH)2 solution, respectively?

Answers

Answer 1

Answer:

The number of moles of HBr and the concentration of the Ca(OH)2 solution is:

The number of moles HBr is = 0.001825

The concentration of Ca(OH)2 is= 0.009125 M

What is the Acid solution for neutralization?

Data given as per question:

The Volume of the Ca(OH)2 is = 100.0 mL = 0.100 L

Then, Molarity of HBr is = 5.00 * 10^-2 M

After that Volume of HBR is = 36.5 mL = 0.0365 L

When The balanced equation is:

Then, Ca(OH)2 + 2HBr → CaBr2 + 2H2O

Then the Calculate molarity of Ca(OH) 2

After that b*Va* Ca is = a * Vb*Cb

Then ⇒with b = the coefficient of HBr is = 2

Now, ⇒with Va = the volume of Ca(OH)2 is = 0.100 L

After that ⇒with ca is = the concentration of Ca(OH)2 = TO BE DETERMINED

Now, ⇒with a = the coefficient of Ca(OH)2 = 1

Then ⇒with Vb is = the volume of HBr = 0.0365 L

Now, ⇒with Cb is = the concentration of HBr = 5.00 * 10^-2 = 0.05 M

Then 2 * 0.100 * Ca = 1 * 0.0365 * 0.05

Now, Ca is = (0.0365*0.05) / 0.200

Therefore, Ca is = 0.009125 M

After that, we Calculate moles HBr

Moles HBr = concentration HBr * volume HBr

Moles HBr = 0.05 M * 0.0365 L

Moles HBr = 0.001825 moles

Find more information Acid solution for neutralization here:

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Answer 2

Answer:

Answer:

The number of moles HBr = 0.001825

The concentration of Ca(OH)2 = 0.009125 M

Explanation:

Step 1: Data given

Volume of the Ca(OH)2 = 100.0 mL = 0.100 L

Molarity of HBr = 5.00 * 10^-2 M

Volume of HBR = 36.5 mL = 0.0365 L

Step 2: The balanced equation

Ca(OH)2 + 2HBr → CaBr2 + 2H2O

Step 3: Calculate molarity of Ca(OH) 2

b*Va* Ca = a * Vb*Cb

⇒with b = the coefficient of HBr = 2

⇒with Va = the volume of Ca(OH)2 = 0.100 L

⇒with ca = the concentration of Ca(OH)2 = TO BE DETERMINED

⇒with a = the coefficient of Ca(OH)2 = 1

⇒with Vb = the volume of HBr = 0.0365 L

⇒with Cb = the concentration of HBr = 5.00 * 10^-2 = 0.05 M

2 * 0.100 * Ca = 1 * 0.0365 * 0.05

Ca = (0.0365*0.05) / 0.200

Ca = 0.009125 M

Step 4: Calculate moles HBr

Moles HBr = concentration HBr * volume HBr

Moles HBr = 0.05 M * 0.0365 L

Moles HBr = 0.001825 moles

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Be sure to answer all parts. Consider both 5-methyl-1,3-cyclopentadiene (A) and 7-methyl-1,3,5-cycloheptatriene (B). Which labeled H atom is most acidic? Hb is most acidic because its conjugate base is aromatic. Hc is most acidic because its conjugate base is antiaromatic. Ha is most acidic because its conjugate base is antiaromatic. Hd is most acidic because its conjugate base is aromatic. Which labeled H atom is least acidic? Ha is least acidic because its conjugate base is aromatic. Hb is least acidic because its conjugate base is antiaromatic. Hd is least acidic because its conjugate base is aromatic. Hc is least acidic because its conjugate base is antiaromatic.

Answers

Due to the conjugate base of the hydrogen atom is aromatic, Hb is regarded as the most acidic. Because the conjugate base of the hydrogen atom Hc is anti-aromatic, it is the least acidic.

The correct options are:

(A) - (a)

(B) - (d)

What are the most and the least acidic hydrogen atom?

The hydrogen connected at the heptatriene's tertiary position (at the 7-methyl) would be particularly acidic, as its removal would leave a positive charge that could be transported around the ring via resonance.

The hydrogen connected to the pentadiene (5-methyl) at the tertiary position would not be acidic, as removing it would result in an anti-aromatic structure.

Thus, the least acidic H atom is Hc and the most acidic H atom is Hb.

Learn more about hydrogen atom, here:

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I don’t have a picture but I can describe it to you.

The hydrogen that is attached at the tertiary position on the heptatriene (at the 7-methyl) would be very acidic, as removal would leave a positive charge that could be moved throughout the ring through resonance. This would mean that the three double bonds would be participating in resonance, and the deprotonated structure would be aromatic, thus making this favorable.

The hydrogen that is attached at the tertiary position on the pentadiene (5-methyl) would NOT be acidic, as removal would cause an antiaromatic structure.

Any other hydrogens would NOT be acidic. Those vinylic to their respective double bonds would seriously destabilize the double bond if removed, and hydrogens attached to the methyl group jutting off the ring have no incentive to leave the carbon.

Hope this helps!

If a large marshmallow has a volume of 2.75 in3 and density of 0.242 g/cm3, how much would it weigh in grams? 1 in3=16.39 cm3.

Answers

• Volume of the marshmallow:

V = 2.75 in^3 (but, 1 in^3 = 16.39 cm^3)

V = 2.75 × 16.39 cm^3

V = 2.75 × 16.39 cm^3

V = 45.0725 cm^3

• Density:

d = 0.242 g/cm^3

• Mass:

m = d × V

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = 10.907545 g

m ≈ 10.9 g <——— this is the answer.

I hope this helps. =)

Answer:

9.92g

Explanation:

2.50 in31×16.39 cm31 in3×0.242 gcm3=9.92 g

Need help ASAP!!!!!!!

Answers

My Heroes is the time of my time I am in college now that I will do it again tomorrow night and you can do something to me about it and then I will never go back again I hope that I can do this again I don’t want

A substance that cannot be chemically broken down into simpler substances is a an electron. b a heterogeneous mixture. c an element. d a homogeneous mixture. e a compound.

Answers

Answer:

c. an element.

Explanation:

An element -

It refers to the substance , which has same type of atoms , with exactly same number of protons , is referred to as an element .

In term of chemical species , elements are the smallest one , and can not be bifurcated down to any further small substance by the means of any chemical reaction .

Hence , from the given information of the question ,

The correct term is an element .

Answer:

C. an element.

Explanation:

What is the mass, in grams, of 1.50 mol of iron (III) sulfate?Express your answer using three significant figures.

Answers

Answer : The mass of 1.50 mole of iron(III) sulfate is, $5.99* 10^2g$

Explanation : Given,

Moles of iron(III) sulfate = 1.50 mole

Molar mass of iron(III) sulfate = 399.88 g/mole

The formula of iron(III) sulfate is, $Fe_2(SO_4)_3$

Formula used :

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3* \text{Molar mass of }Fe_2(SO_4)_3$

Now put all the given values in this formula, we get:

$\text{Mass of }Fe_2(SO_4)_3=1.50mole* 399.88g/mole$

$\text{Mass of }Fe_2(SO_4)_3=599.82g=5.99* 10^2g$

Therefore, the mass of 1.50 mole of iron(III) sulfate is, $5.99* 10^2g$

The following reaction was monitored as a function of time: A→B+C A plot of ln[A] versus time yields a straight line with slope −4.3×10−3 /s. If the initial concentration of A is 0.260 M, what is the concentration after 225 s?