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2 years ago

Describe the stages of water cycle

Chemistry

2 answers:

lora16 [44]2 years ago

6 0

Evaporation, Condensation, precipitation and collection would be stages

love history [14]2 years ago

4 0

There are four main stages in the water cycle. They are evaporation, condensation, precipitation and collection.

Hope this helps!

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An element with 7 valence electrons will most likely

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Answer:

An element with 7 valence electrons will most likely be a halogen and gain an electron

Explanation:

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2 years ago

According to the passage, which of the following would cause milk to go bad?

olga_2 [115]

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the answer D) will cause milk to go bad

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2 years ago

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If gas in a sealed container has a pressure of 50 kPa at 300 K, what will the pressure be if the temperature rises to 360 K?

VLD [36.1K]

The answer is 60 kpa, also did you try look it up because that is what i did but i did not copy and paste it.

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6 0

2 years ago

a. Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g

Tanya [424]

Answer:

(A) $\Delta H^{\circ }_{r}= -144 kJ$

(B) $\Delta H^{\circ }_{r}= - 2552kJ$

Explanation:

(A) 2NO(g) + O₂(g) → 2NO₂(g)

$1/2 N_{2}(g)+O_{2}(g)\rightarrow NO_{2}(g), \Delta H^{\circ }_{a}=33.2 kJ....equation (a)$

$1/2N_{2}(g)+1/2O_{2}(g)\rightarrow NO(g), \Delta H^{\circ }_{b}=90.2 kJ ....equation (b)$

Now, multiplying equation (a) with 2:

⇒ $N_{2}(g)+2 O_{2}(g)\rightarrow 2 NO_{2}(g)....equation (a)$

Then equation b is reversed and multiplied with 2:

$2 NO(g)\rightarrow N_{2}(g)+ O_{2}(g)....equation (b)$

Now by adding the equation (a) and equation (b), we get:

⇒ $2 NO(g)+ \bcancel N_{2}(g)+\bcancel 2 O_{2}(g)\rightarrow 2 NO_{2}(g) +\bcancel N_{2}(g)+ \bcancel O_{2}(g)$

⇒ 2NO(g) + O₂(g) → 2NO₂(g)

Therefore, the enthalpy of the reaction:

$\Delta H^{\circ }_{r}= 2\times \Delta H^{\circ }_{a} - 2\times \Delta H^{\circ }_{b}$

$= (2\times33.2)- (2\times90.2)=66.4 - 180.4= -144 kJ$

(B) 4B(s)+3O₂(g) → 2B₂O₃(s)

$B_{2}O_{3}(s)+3H_{2}O(g)\rightarrow 3O_{2}(g)+B_{2}H_{6}(g), \Delta H_{a }^{\circ }=+2035 kJ...equation (a)$

$2B(s)+3H_{2}(g)\rightarrow B_{2}H_{6}(g), \Delta H_{b }^{\circ }= +36 kJ...equation (b)$

$H_{2}(g)+1/2O_{2}(g)\rightarrow H_{2}O(l), \Delta H_{c }^{\circ }= -285 kJ...equation (c)$

$H_{2}O(l)\rightarrow H_{2}O(g), \Delta H_{d }^{\circ }=+44 kJ...equation (d)$

Now multiplying equation (b) with 2, reversing equation (a) and multiplying with 2. Reversing equation (c) and (d) and multiplying both with 6.

$6O_{2}(g)+2B_{2}H_{6}(g)\rightarrow 2B_{2}O_{3}(s)+6H_{2}O(g)...equation (a)$

$4B(s)+6H_{2}(g)\rightarrow 2B_{2}H_{6}(g)...equation (b)$

$6H_{2}O(l)\rightarrow 6H_{2}(g)+3O_{2}(g)...equation (c)$

$6H_{2}O(g)\rightarrow 6H_{2}O(l)...equation (d)$

Now by adding the equations (a), (b), (c), (d); we get:

4B(s)+3O₂(g) → 2B₂O₃(s)

Therefore, the enthalpy of the reaction:

$\Delta H^{\circ }_{r}= -2\times \Delta H^{\circ }_{a} + 2\times \Delta H^{\circ }_{b} - 6 \times \Delta H_{c }^{\circ } - 6 \times \Delta H_{d }^{\circ }$

$= -2\times (+2035 kJ)+ 2\times (+36 kJ) - 6 \times (-285 kJ)- 6 \times (+44 kJ) = -4070 + 72 + 1710 - 264 = - 2552kJ$

4 0

2 years ago

4.053g 4.2 mL = Please help

butalik [34]

Answer: 2

Explanation:

5 0

1 year ago