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Monica [59]
3 years ago
14

What type of energy is stored for use at a later time ?

Chemistry
2 answers:
Alecsey [184]3 years ago
4 0
I guess it’s potential energy.
tankabanditka [31]3 years ago
3 0
l think potential energy
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For 13 points plz. Plz
Ludmilka [50]

Answer:

Individual versus nature

Explanation:

The type of conflict is <em>individual</em><em> </em><em>versus</em><em> </em><em>nature</em><em> </em>because the cat faced many problems in nature such as facing the mouse, grass, people, etc. These are all considered as nature

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3 years ago
g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc
Gala2k [10]

\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}

\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}

\text{and in the liquid form it is easily transported. An industrial chemist studying this}

\text{reaction fills a} \ \mathbf{100 \  L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \  \text{of oxygen gas, }

\text{to be} \  \mathbf{2.6\  mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}

\text{ammonia at the final temperature of the mixture. Round your answer to  2 significant digits.}

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

From the above reaction; the ICE table can be represented as:

                    \mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

I (mol/L)     0.086            0.28                 0              0

C                   -4x                -3x               +2x           +6x

E                 0.086 - 4x     0.28 - 3x      +2x             +6x

At equilibrium;

The water vapor = \dfrac{2.6 \ mol}{100 \ L} = 6x

x = \dfrac{2.6}{100} \times \dfrac{1}{6}

x = 0.00433

\text{equilibrium constant}  ({k_c}) =  \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }

\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\

Replacing the value of x, we have:

K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}

K_c = \mathbf{5.5 \times 10^{-8} \ to  \ 2 \ significant \ figures}

5 0
3 years ago
Does it require more force to stop a bus or a bike moving at the same speed
iVinArrow [24]
Well depends how fast they're going if it's a slow speed a bus but at a fast speed a bike because you wanna be careful while stopping
7 0
3 years ago
The earths magnetic poles are located in different areas close to its geographic poles?
s2008m [1.1K]

Answer:

Explanation:

The answer is true

4 0
2 years ago
Read 2 more answers
Calculate the new volume of 1.23 mL of a gas at 32 C is subjected to drop in temperature of 20 degrees Celsius
kotykmax [81]

Answer:

1,15mL = V₂

Explanation:

Based on Charle's law the volume is directely proportional to the absolute temperature in a gas under constant pressure. The equation is:

V₁T₂ = V₂T₁

<em>Where V is volume and T absolute temperature of a gas where 1 is initial state and 2, final state.</em>

The V₁ is 1.23mL

T₁ = 32°C + 273.15 = 305.15K

T₂ = T₁ - 20°C = 285.15K

Replacing:

1.23mL*285.15K = V₂*305.15K

<h3>1,15mL = V₂</h3>

<em />

8 0
2 years ago
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