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2 years ago

Why is water in the gas phase when it comes out of a volcano?

Chemistry

2 answers:

Lostsunrise [7]2 years ago

6 0

Lava heats the water where water vapor come

umka2103 [35]2 years ago

4 0

Due to lava which heats water where water vapour is comes

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A potential energy diagram is shown.

Vesna [10]

Answer:

25kJ

Explanation:

Given the initial energy to be 30kJ

The energy change from the initial energy to the peak energy = (65-30) kJ

= 35kJ

since the second energy change was a drop in energy it is regarded negative

= (55-65)

= -10kJ

Therefore total energy change

= (35-10)kJ

= 25kJ

5 0

2 years ago

Which distinguishes an atom of one element from an atoms of a different element?

dalvyx [7]

Answer:

The number of protons you welcome

Explanation:

7 0

2 years ago

What is the difference between ionization energy and ionization potential

rjkz [21]

Answer:

They are similarly

Explanation:

3 0

3 years ago

This is for a study guide, I can't figure it out!

miskamm [114]

<span>The answer is "D" where the number of collisions per unit area is reduced by one-half. Drawing back on the piston means the volume is increased. The pressure is reduced. There are fewer collisions when the pressure is reduced.</span>

3 0

3 years ago

When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C

natka813 [3]

Answer:

$\Delta_{r}U$ of the reaction is -6313 kJ/mol

$\Delta_{r}H$ of the reaction is -6312 kJ/mol

Explanation:

$Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C$

$q_{cal}= C \times \Delta T$

$=5.861 \times 3.596 = 21.076\,kJ$

$q_{rxn}= -q_{cal}= -21.076\,kJ$

$\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol$

Therefore, $\Delta_{r}U$ of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter is as follows.

$C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)$

$Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5$

$\Delta_{r}H=\Delta E+ \Delta n. RT$

$=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol$

Therefore, $\Delta_{r}H$ of the reaction is -6312 kJ/mol.

3 0

3 years ago