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alukav5142 [94]
3 years ago
11

The table shows the concentration of a reactant in the reaction mixture over a period of time.

Chemistry
1 answer:
zvonat [6]3 years ago
6 0

Answer:

C. 2.0 x 10⁻³

Explanation:

The average rate of a reaction is the rate of change of the concentration of the reactants (or products) with time.

The average rate of the reaction = - (ΔC)/(Δt).

∵ The concentration of the reactants changes from 1.8 M to 0.6M for 580 s.

∴ The average rate of the reaction over the first 580 seconds

   = - (ΔC)/(Δt) = - [(0.6M) - (1.8 M)]/[(580.0 s) - (0.0 s)] = 2.0 x 10⁻³ M.s⁻¹.

So C. 2.0 x 10⁻³ is the right answer

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How do muscles work in pairs to make parts of the body move?
schepotkina [342]

Answer:

Muscles work in pairs to move a bone.

Skeletal muscles only PULL in one direction. For this reason they always come in pairs. When one muscle in a pair contracts, to bend a joint for example, its counterpart then contracts and pulls in the opposite direction to straighten the joint out again.

4 0
3 years ago
Read 2 more answers
A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to
grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

<u>Step 2:</u> Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0
3 years ago
In a redox reaction, how does the total number of electrons lost by the oxidized substance compare to the total number of electr
snow_lady [41]
The answer is (2) equal to. In redox reactions, you can't just lose electrons somewhere. If an electrons is lost by one, it must be gained by another. Hence, the importance of balancing redox reactions.
3 0
3 years ago
The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th
MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = 25^oC=273+25=298 K

Now put all the given values in this formula, we get

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

0.4284 is the ratio of the rate constants.

7 0
3 years ago
The factor that affects how easily an electron can be removed from an atom is the
Zolol [24]

Answer: D

Explanation:

The answer is D

4 0
3 years ago
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