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3 years ago

1. Is O2 a compound? Explain.

1 answer:

4 0

1. O2 is not a compound because it only contains one or more type of the same element atom.
2. O2 is a molecule because a molecule is one or more of the same element atom.
3. The law of conversion is that the mass of the system will stay the same when transfer takes place. Like if you had an equation O+H2—> H2O the mass will remain the same.

4. It will be equal to 10 because of law of conservation of matter.

5. One observation can be that the compound, reaction you’re observing, has change states.

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Latent heat is used to: 1. form chemical bonds. 2. change molecular structure. 3. change states of matter. 4. change the tempera

likoan [24]

Answer : The correct option is, (3) change states of matter.

Explanation :

Latent heat : It is defined as the heat required to convert the solid into liquid or vapor and a liquid into a vapor without changing the temperature.

There are two types of latent heat.

(1) Latent heat of fusion

(2) Latent heat of vaporization

Latent heat of fusion : It is defined as the amount of heat energy released or absorbed when the solid converted to liquid at atmospheric pressure at its melting point.

Latent heat of vaporization : It is defined as the amount of heat energy released or absorbed when the liquid converted to vapor at atmospheric pressure at its boiling point.

Hence, latent heat is used to change states of matter.

7 0

3 years ago

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the

emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

$k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}$

Rate factor in the presence of catalyst:

$k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}$

Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

$\dfrac{k_2}{k_1}={ \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}$

$\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}$

Thus;

$Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}$

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

$Ea_1-Ea_2 = 8.314 \ J/mol/K * 298 \ K * In (10^6)$

$Ea_1-Ea_2 = 34228.92 \ J/mol$

$\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}$

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

8 0

3 years ago

Brass is an alloy of

Gemiola [76]

Answer:

Cooper and zinc if I remember right

5 0

3 years ago

A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug

OverLord2011 [107]

Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 = log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] (1)

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M (2)

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = 0.035M

Thus: [CH₃COO⁻] = 0.100M - 0.035M = 0.065M

5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = 7.429x10⁻³ moles of CH₃COO⁻

Moles of CH₃COOH before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles + 2.646x10⁻³ moles = 8.071x10⁻³ moles of CH₃COOH. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = 0.296

4 0

3 years ago

R is the midpoint of FG. FR = 26 find RG

vichka [17]

Answer:

Explanation:

This is because if R is the midpoint of FRG, FR is half of FRG, so basically all you do it multiply by 2 to get the FRG

7 0

3 years ago