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2 years ago

1 mole of ribose (C5H10O5(s)) has a greater entropy than 1 mole of sucrose (C12H22O11(s)) True or false

Chemistry

1 answer:

makkiz [27]2 years ago

7 0

Answer:

False

Explanation:

Hello,

In this case, it is possible to substantiate that the statement is false, since the entropy of sucrose is greater than the entropy of ribose due to the fact that molar mass of the sucrose (342 g/mol) is greater than the molar mass of ribose (150 g/mol) and the general rule states that the greater the molar mass the greater the entropy.

Best regards.

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Bonds between two atoms that are equally electronegative are _____. bonds between two atoms that are equally electronegative are

Anna [14]

Bonds of two atoms of equal electronegativity are nonpolar covalent bonds.

Your second sentence is identical to the first sentence; I'll bet the second sentence is "Bonds between two atoms that are unequally electronegative are polar covalent bonds."

4 0

2 years ago

A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of

allochka39001 [22]

Answer:

The answer to your question is C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

44 g of CO₂ --------------- 12 g of C

52.8 g --------------- x

x = (52.8 x 12)/44

x = 633.6/44

x = 14.4 g of C

12 g of C ------------------ 1 mol

14.4 g of C --------------- x

x = (14.4 x 1)/(12)

x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

18 g of H₂O --------------- 2 g of H

21.6 g of H₂O ------------- x

x = (21.6 x 2) / 18

x = 2.4 g of H

1 g of H -------------------- 1 mol of H

2.4 g of H ----------------- x

x = (2.4 x 1)/1

x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

= 6.4 g

16 g of O ---------------- 1 mol

6.4 g of O -------------- x

x = (6.4 x 1)/16

x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

C₃H₆O

8 0

2 years ago

You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number

irina1246 [14]

Answer: $9.11\times 10^{-31}kg$

Explanation:

Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

All zero’s preceding the first integers are never significant.

Thus $9.11\times 10^{-31}kg$ has three significant figures

7 0

2 years ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

2 years ago

What is the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide? hints

aleksklad [387]

The answer:

<span>the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide:
</span>CaBr2 + AgNO3-----------> AgBr2 + CaNO3, so among the given choices, the
true answe is
<span>agno3(aq)+cabr2(aq) (as reactants)</span>

4 0

2 years ago