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2 years ago

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A substance has a density of 4.5 g/cm3. The substance is heated and its density recalculated. What is the newly recorded density

?

1 answer:

Dmitry_Shevchenko [17]2 years ago

6 0

Answer:

Aluminum

Explanation:

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How much energy is required to heat 0.24 KG lutetium from 296.2K to 373.5 K? The specific heat for lutetium is 0.154 J/g-K

Olenka [21]

I’m not sure I need help with this question

6 0

2 years ago

A student has 100. mL of 0.400 M CuSO4 (aq) and is asked to make 100. mL of 0.150 M CuSO4 (aq) for a spectrophotometry experimen

GenaCL600 [577]

Answer:

We have to take 37.5 mL of a 0.400 M solution

Explanation:

Step 1: Data given

Stock volume = 100 mL = 0.100L

Stock concentration 0.400 M

Volume of solution he wants to make = 100 mL = 0.100L

Concentration of solution he wants to make = 0.150 M

Step 2: Calculate the volume of 0.400 M CuSO4 needed

C1*V1 = C2*V2

⇒with C1 = the stock concentration = 0.400M

⇒with V1 = the volume of the stock = TO BE DETERMINED

⇒with C2 = the concentration of the solution he wants to make = 0.150 M

⇒with V2 = the volume of the solution made = 0.100 L

0.400 M * V1 = 0.150M * 0.100L

V1 = (0.150M*0.100L) / 0.400 M

V1 = 0.0375 L = 37.5 mL

We have to take 37.5 mL of a 0.400 M solution

5 0

2 years ago

Read 2 more answers

Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f

Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0

2 years ago

10.000g of boron (B) combines with hydrogento form 11.554g of a pure compound. What is the empirical formula of this compound?

Citrus2011 [14]

Answer:

B3H5

Explanation:

The law of conservation of mass states that matter in an closed system is neither created nor destroyed by physical transformations or chemical reactions but changes from one form to the other.

That is, the sum of masses of the reactants = The sum of masses of the product

10.00g of Boron + x grams of Hydrogen = 11.55g of the product

Mass of hydrogen = 11.55 - 10.00 = 1.55g

Molar mass of Boron = 10.811g

Molar mass of Hydrogen = 1.00784g

Number of moles of Boron = (mass of Boron)/(molar mass of Boron) = 10/10.811 = 0.9249 mols

Number of moles of Hydrogen = (mass of Hydrogen)/(molar mass of Hydrogen) = 1.55/1.00784 =1.5379mols

0.9249 mols of Boron combines with 1.5379mols of Hydrogen

Dividing both sides mols by 0.9249 gives

1 mole of Boron combines with 1.66266 mols of Hydrogen

Converting 1.66266 to fractions we have 1.66266 approximately 5/3

or 1 mole of Boron combines with 5/3 moles of Hydrogen

Multiplying both sides by 3 we have

3 moles of Boron combines with 5 moles of Hydrogen

Molecular formula of the compound is

B3H5

4 0

2 years ago

In the citric acid cycle, after the acetyl-group (2C) from acetyl-CoA is transferred to oxaloacetate (4C) to produce citrate (6C

Alborosie

Answer:

The carbons of the acetyl group oxidize which generate CO2, and in turn H2O.

Explanation:

The pyruvic acid that is generated during glycolysis enters the mitochondria. Inside this organelle, the acid molecules undergo a process called oxidative decaborxylation in which an enzyme of several cofactors is involved, one of which is coenzyme A. Pyruvic acid is transformed into an acetyl molecule and these are been introduced to the begining of the Krebs Cycle where the acetyl-group (2C) from acetyl-CoA is transferred to oxaloacetate (4C) to produce citrate (6C). As the molecule cycles the two carbons of the acetyl oxidize and are released in the form of CO2. Then the energy of the Krebs cycle becomes sufficient to reduce three NAD +, which means that three NADH molecules are formed. Although a small portion of energy is used to generate ATP, most of it is used to reduce not only the NAD + but also the FAD which, if oxidized, passes to its reduced state, FADH2

8 0

2 years ago