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2 years ago

What is the general formula for a straight-chain alkane?

Chemistry

1 answer:

TEA [102]2 years ago

6 0

Answer:

The general formula for a straight chain alkane is CnH(2n+...so correct option is C and D

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Studentka2010 [4]

Answer:

Explanation:

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6 0

2 years ago

1. How much heat is required to melt 25.0 g of ice at 0°C?

tigry1 [53]

Answer: The heat required to melt 25.0 g of ice at $0^0C$ is 8350 Joules

Explanation:

Heat of Fusion tells us how much energy is needed to convert 1g of a solid to a liquid at the same temperature.

$Q=m\times L$

Q = Heat absorbed = ?

m = mass of ice = 25.0 g

L = Latent heat of fusion of ice = 334 J/g

Putting in the values, we get:

$Q=25.0g\times 334J/g=8350J$

Thus heat required to melt 25.0 g of ice at $0^0C$ is 8350 Joules

8 0

2 years ago

1. What Volume of HCl is generated if 3.44 g of Cl2 are reacted at STP? 2. What volume does 4.87 mol of Kr have at STP? 3. What

IgorLugansk [536]

Answer:

1. 2.17 dm3 VOLUME OF HCl IS PRODUCED WHEN 3.44 g OF Cl2 REACT AT STP

2. 4.87 MOLE OF Kr AT STP CONTAINS 109.088 dm3 .

3. THE PRESSURE WHEN 3.44 g OF Cl2 are reacted at 4.55 L AT 455 K IS 0.77 atm

Explanation:

1 Volume of HCl if 3.44 g of Cl2 are reacted at STP?

Equation for the reaction:

H2 + Cl2 ---------> 2HCl

1 mole of Cl2 reacts to form 2 mole of HCl

At STP, 1 mole of a gas is equal to the molar mass of the gas sample

35.5 * 2 g of Cl2 reacts to form 2 mole of HCl

3.44 g of Cl2 will react to form ( 3.44 * 2 / 71 ) mole of HCl

= 0.0969 mole of HCl

1 mole of HCl = 22.4 dm3

0.0969 mole of HCl = ( 22.4 * 0.0969 / 1)

= 2.17056 dm3

The volume of HCl is 2.17 dm3 when 3.44 g of Cl2 are reacted at STP.

2. What volume does 4.87 mol of Kr have at STP?

1 mole of a substance is 22.4 dm3 of the sample

1 mole of Kr = 22.4 dm3

4.87 mole of Kr = 4.87 * 22.4

= 109.088 dm3

4.87 mole of Kr at STP contains 109.088 dm3 volume

3. Whta pressure of HCl is generated if 3.44 g of Cl2 are reacted at 4.55 L at 455 K

Using the formula:

PV = nRT

V = 4.55 L

R = 0.082 L atm/ mol K

T = 455 K

m = 3.44 g

n = mass / molar mass

molar mass = ( 1 + 35.5) = 36.5 g/mol

n = 3.44 g / 36.5 g/mol

n = 0.094 mole

P = nRT / V

P = 0.094 * 0.082 * 455 / 4.55

P = 3.50714 / 4.55

P = 0.7708 atm

The pressure of HCl if 3.44 g of Cl2 are reacted at 4.55 L and 455 K is 0.7708 atm.

8 0

2 years ago

Consider the mechanism. Step 1: A+B↽−−⇀CA+B↽−−⇀C equilibrium Step 2: C+A⟶DC+A⟶D slow Overall: 2A+B⟶D2A+B⟶D Determine the rate la

tatuchka [14]

Answer:

rate = k[A][B] where k = k₂K

Explanation:

Your mechanism is a slow step with a prior equilibrium:

$\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}$

(The arrow in Step 1 should be equilibrium arrows).

1. Write the rate equations:

$-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]$

2. Derive the rate law

Assume k₋₁ ≫ k₂.

Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.

In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is

rate = k[A][B] where k = k₂K

5 0

3 years ago

The smallest part of a compound that still acts like that compound is ____.

Leokris [45]

It is a molecule of the compound.

3 0

2 years ago