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4 years ago

During which process of the water cycle does water change from a gas to liquid

Chemistry

2 answers:

maw [93]4 years ago

7 0

Answer:

condensation

Explanation:

condensation is when water vapor turns into liquid.

Kay [80]4 years ago

7 0

Answer:

B: Condensation

Explanation:

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Sulfur reacts with oxygen to form sulfur dioxide (SO2(g), Hf = –296.8 kJ/mol) according to the equation below.

Pavel [41]

Given that the reactants are the very consituents (elements) of the final compound, the enthalpy change of this reaction is the same heat of formation, Hf.

That means that the enthalpy chage for the reaction is -296.8 kj/mol

7 0

3 years ago

Read 2 more answers

What volume of carbon dioxide is produced when 0.489 mol of calcium carbonate reacts completely according to the following react

Vilka [71]

Answer:

11.0L of carbon dioxide is produced

Explanation:

Balanced equation: $CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)$

According to balanced equation, 1 mol of $CaCO_{3}$ produces 1 mol of $CO_{2}$

So, 0.489 mol of $CaCO_{3}$ produces 0.489 mol of $CO_{2}$

Let's assume $CO_{2}$ behaves ideally.

So, $P_{CO_{2}}V_{CO_{2}}=n_{CO_{2}}RT$

where P is pressure, V is volume , n is number of moles, R is gas constant and T is temperature in kelvin

Plug-in all the values in the above equation-

$(1atm)\times V_{CO_{2}}=(0.489mol)\times (0.0821L.atm.mol^{-1}.K^{-1})\times (273K)$

or, $V_{CO_{2}}=11.0L$

So, 11.0L of carbon dioxide is produced

4 0

3 years ago

The moon is about 400,000 km from the earth . Using Newton’s law of gravity how would the attraction between the earth and moon

Ray Of Light [21]

Answer:

The moon is 384472.282 Km from earth

Explanation:

6 0

4 years ago

In a chemical reaction, [_____] are the substances present after the reaction.

worty [1.4K]

Answer:

Products

Explanation:

5 0

3 years ago

2 SO2(g) + O2(g) 2 SO3(g) Assume that Kc = 0.0680 for the gas phase reaction above. Calculate the corresponding value of Kp for

son4ous [18]

Answer: The corresponding value of $K_p$ for this reaction at 84.5°C is 0.00232

Explanation:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

Relation of with is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

= equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = 0.0680

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $84.5^0C=(273+84.5)K=357.5K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=0.0680\times (0.0821\times 357.5)^{-1}\\\\K_p=0.00232$

Thus the corresponding value of $K_p$ for this reaction at 84.5°C is 0.00232

6 0

3 years ago