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Kitty [74]
2 years ago
6

If the collected information showed that the temperature was 10°C and the wind speed was 75 mph, what type of weather would it

Chemistry
1 answer:
Shkiper50 [21]2 years ago
4 0

Answer:

cold and windy

Explanation:

if its 10 degrees outside then it must be cold. then 75 mph that is very windy

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Consider the following equilibrium systems: A⇌2B ΔH o =20.0 kJ/mol Reaction 1 A+B⇌C ΔH o =−5.4 kJ/mol Reaction 2 A⇌B ΔH o =0.0 k
galina1969 [7]

Answer:

Reaction 1: Kc increases

Reaction 2: Kc decreases

Reaction 3: The is no change

Explanation:

Let us consider the following reactions:

Reaction 1: A ⇌ 2B ΔH° = 20.0 kJ/mol

Reaction 2: A + B ⇌ C ΔH° = −5.4 kJ/mol

Reaction 3: 2A⇌ B ΔH° = 0.0 kJ/mol

To predict what will happen when the temperature is raised we need to take into account Le Chatelier Principle: when a system at equilibrium suffers a perturbation, it will shift its equilibrium to counteract such perturbation. This means that <em>if the temperature is raised (perturbation), the system will react to lower the temperature.</em>

Reaction 1 is endothermic (ΔH° > 0). If the temperature is raised the system will favor the forward reaction to absorb heat and lower the temperature, thus increasing the value of Kc.

Reaction 2 is exothermic (ΔH° < 0). If the temperature is raised the system will favor the reverse reaction to absorb heat and lower the temperature, thus decreasing the value of Kc.

Reaction 3 is not endothermic nor exothermic (ΔH° = 0) so an increase in the temperature will have no effect on the equilibrium.

8 0
3 years ago
What characteristics are used to clarify an area as a wetland?
Nadya [2.5K]
Soil covered, saturated, submerged, flooded w water, standing water
5 0
3 years ago
Which of the following is true about two neutral atoms of the element gold? (1 point) The nucleus is missing in both. Each has a
klio [65]

i have the same question idk the anwser tho

5 0
3 years ago
How much of a 0.250 M lithium hydroxide is required to neutralize 20.0 mL of 0.345M chlorous acid?
Bumek [7]

Answer:

27.6mL of LiOH 0.250M

Explanation:

The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:

LiOH + HClO₂ → LiClO₂ + H₂O

<em>That means, 1 mole of hydroxide reacts per mole of acid</em>

Moles of  20.0 mL = 0.0200L of 0.345M chlorous acid are:

0.0200L ₓ (0.345mol / L) = <em>6.90x10⁻³ moles of HClO₂</em>

To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:

6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =

<h3>27.6mL of LiOH 0.250M</h3>
6 0
3 years ago
A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of
Bad White [126]

Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

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

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}

The number of moles of the ideal gas is:

n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

n = 5.541\,mol

The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

T_{2} = 360.955\,K

The final pressure is:

P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

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

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

4 0
4 years ago
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