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Novay_Z [31]
3 years ago
8

Consider the reversible reaction, A+B⇌C+D. If the concentration of product D is increased, the rate of the reverse reaction woul

d ________________.
increase

decrease

be unaffected
Chemistry
1 answer:
natali 33 [55]3 years ago
4 0

Answer:

If the concentration of product D is increased, the rate of the reverse reaction would increase.

Explanation:

Chemical reaction:

A + B ⇄ C + D

In given condition the equilibrium is disturb by increasing the concentration of product.

When the concentration of product D is increased the system will proceed in backward direction in order to regain the equilibrium. Because when the product concentration is high it means reaction is not on equilibrium state the reaction will proceed backward direction to regain the equilibrium state.

According to the Le- Chatelier principle,

At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.

The equilibrium can be disturb,

By changing the concentration

By changing the volume

By changing the pressure

By changing the temperature

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Since there is more energy added as heat rises, the particles disperse and have larger movements.
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Calculate how many grams of o2(g) can be produced from heating 87.4 grams of kclo3(s).
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<span>34.2 grams Lookup the atomic weights of the involved elements Atomic weight potassium = 39.0983 Atomic weight Chlorine = 35.453 Atomic weight Oxygen = 15.999 Molar mass KClO3 = 39.0983 + 35.453 + 3 * 15.999 = 122.5483 g/mol Moles KClO3 = 87.4 g / 122.5483 g/mol = 0.713188188 mol The balanced equation for heating KClO3 is 2 KClO3 = 2 KCl + 3 O2 So 2 moles of KClO3 will break down into 3 moles of oxygen molecules. 0.713188188 mol / 2 * 3 = 1.069782282 mols So we're going to get 1.069782282 moles of oxygen molecules. Since each molecule has 2 atoms, the mass will be 1.069782282 * 2 * 15.999 = 34.23089345 grams Rounding the results to 3 significant figures gives 34.2 grams</span>
4 0
3 years ago
Use the equation 2cu+o2=2cuo to find how many miles of copper react to form 3.64 mil cuo
Firlakuza [10]
<h3><u>Answer;</u></h3>

3.64 moles

<h3><u>Explanation;</u></h3>

From the equation;

2Cu + O2 → 2CuO

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Moles of CuO = 3.64 Moles

Therefore;

moles of Cu will be ;

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Moles of Cu = <u>3.64 moles </u>

8 0
3 years ago
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How much heat is needed to melt 10.0 grams of ice at -10°C until it is water at 10°C?
zhannawk [14.2K]

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<h3>What is Sensible heat? </h3><h3 />

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state.

Q= c×m×∆T

<h3>What is Latent heat? </h3><h3 />

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

In this case, the heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q= m×L

Where,

L is the latent heat

<h3>-10°C to 0 °C</h3><h3 />

C= specific heat capacity of ice= 2.108 J/gK

M= 10 g

ΔT= T(final)– T(initial) = 0 °C – (-10 °C)= 10 °C= 10 K

Sensitive heat Q(1) = 2.108×10×10

= 210.8J

<h3>Heat needed to melt ice</h3><h3 />

The specific heat of melting of ice is 334 J/g, the heat needed to melt 10 grams of ice is

Q(2) = 10× 334

= 3340J

<h3>0°C to 10 °C</h3><h3 />

C= specific heat capacity of liquid water is 4.187 J/gK

M= 10 g

ΔT= T(final) – T(initial) = 10 °C – 0 °C= 10 °C= 10 K because being a temperature difference, the difference is the same in °C and K.

Q(3) = 4.187×10×10

= 418.7 J.

Total heat required= Q1 + Q2 + Q3

Total heat required= 210.8 J + 3,340 J + 418.7 J

= 3969.5J

Thus, the heat needed to melt 10 gram of ice from temperature-10°C to 10°C is 3969.5. Therefore, option B is correct option.

learn more about heat :

brainly.com/question/16818736

#SPJ1

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This makes sense because as temperature increases pressure increases, as well as when temperature decreases, pressure decreases. Since it’s a colder day the pressure will be lower.
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3 years ago
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