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3 years ago

How do sand dunes move? Use the terms leeward and windward in your explanation.

1 answer:

7 0

As the sand mound grows, the point of maximum sand deposition on the leeward face moves closer to the summit, causing a steepening of the leeward face relative to the windward face. The steepening and growing dune now forces wind out over the top of the dune rather than down the leeward face.

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Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]$

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0

3 years ago

Which element cannot be the central atom of a lewis structure?.

zalisa [80]

Answer:

Hydrogen

Explanation:

Hydrogen can never be central atom despite its low electronegativity

5 0

2 years ago

I NEED THE ANSWER TO NUMBER 3 pleaseee

postnew [5]

Answer:

density = 2.769 g

Explanation:

4.26 times 0.65

3 0

3 years ago

The absolute temperature of a gas is increased four times while maintaining a constant volume. What happens

poizon [28]

Answer:

<u>It increases by a factor of four</u>

Explanation:

Boyle's Law : At constant temperature , the volume of fixed mass of a gas is inversely proportional to its pressure.

pV = K.......(1)

pV = constant

Charles law : The volume of the gas is directly proportional to temperature at constant pressure.

V = KT

or V/T = K = constant ....(2)

Applying equation (1) and (2)

$\frac{PV}{T}=K$

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

According to question ,

T2 = 4 (T1)

V2 = V1

Put the value of T2 and V2 , The P2 can be calculated,

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{1}}{4T_{1}}$

V1 and V1 cancel each other

T1 and T1 cancel each other

We get,

$P1=\frac{P2}{4}$

P2 = 4 P1

So pressure increased by the factor of four

5 0

3 years ago

Problem 4

Hunter-Best [27]

<h3>Answer:</h3>

1.93 g

<h3>Explanation:</h3>

<u>We are given;</u>

The chemical equation;

2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH = -3120 kJ

We are required to calculate the mass of ethane that would produce 100 kJ of heat.

From the equation given;

2 moles of ethane burns to produce 3120 Kilo joules of heat

Therefore;

Number of moles that will produce 100 kJ will be;

= (2 × 100 kJ) ÷ 3120 kJ)

= 0.0641 moles

But, molar mass of ethane is 30.07 g/mol

Therefore;

Mass of ethane = 0.0641 moles × 30.07 g/mol

= 1.927 g

= 1.93 g

Thus, the mass of ethane that would produce 100 kJ of heat is 1.93 g

3 0

3 years ago