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9966 [12]
3 years ago
11

A student placed 3.0 grams of Mg into some HCl in two different experiments. In each case, it reacted according to the following

equation:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g).

In the first experiment, it took 32 minutes for all of the Mg to react. In the second experiment, it took 54 minutes for all of the Mg to react. Which of the following could account for the change in reaction rate of the second experiment?



a catalyst was added



the Mg was powdered into H2



the H2 was decreased



the temperature was decreased
Chemistry
1 answer:
lana [24]3 years ago
6 0

Answer: -

A catalyst is a substance that speeds up the chemical reaction. In the first experiment it took 32 minutes for all the magnesium to react.

In the next it took 54 minutes for all the magnesium to react. So the reaction was slower the second time.

Thus a catalyst was not used.

Powdering up the Magnesium would mean increasing the surface area. More the surface area faster the reaction. Since the reaction is slowing down,

Magnesium was not powdered

If the temperature was decreased then kinetic energy would decrease of the reactant molecules and the reaction would slow down due to less number of collisions between reactant moleules.

So the temperature was decreased is a correct option.

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3 years ago
V2 = (P1 × V1 × T2) (P2 × T1) A gas with a beginning pressure of 2 atm at a temperature of 300 K has a volume of 20 ml. What wil
rusak2 [61]

6.6ml will be the new volume if the pressure increases to 4 atm and the temperature are lowered to 200 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given data:

V_1=20 ml

T_1=3

P_1=2 atm

T_2=200 K

P_2=4 atm

V_2=?

Using equation:

V_2 = \frac{P_1 X V_1 XT_2}{P_2 X T_1}

V_2 =\frac{2 atm  X 20 ml X200 K}{4 atm  X 300 K}

V_2 = 6.6 ml

Hence, 6.6ml will be the new volume if the pressure increases to 4 atm and the temperature are lowered to 200 K.

Learn more about the ideal gas equation here:

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6 0
2 years ago
A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,
inna [77]

Answer:

3.676 L.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R  is the general gas constant,

T is the temperature of the gas in K.

  • If n and P are constant, and have different values of V and T:

(V₁T₂) = (V₂T₁)

  • Knowing that:

V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,

V₂ = ??? L, T₂ = 40°C + 273 = 313 K,

  • Applying in the above equation

(V₁T₂) = (V₂T₁)

∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.

7 0
3 years ago
Read 2 more answers
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