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9966 [12]
4 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]4 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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If 42 grams of carbon and 52 grams of oxygen are used, how many grams of CO2 will be produced (Hint: find the
zavuch27 [327]

Answer:

71.5g

Explanation:

The reaction equation is given as:

               C  +  O₂  →  CO₂

Mass of C = 42g

Mass of O₂  = 52g

Unknown:

Mass of CO₂ produced  = ?

Solution

Now to solve this problem, we have to find limiting reactant which is the one given in short supply in this reaction.

 The extent of the reaction is controlled by this reactant.

Find the number of moles of the given species;

 Number of moles  = \frac{mass}{molar mass}

      Number of moles of C  = \frac{42}{12}   = 3.5mol

     Number of moles of O₂   = \frac{52}{32}   = 1.63mol

Now;

   From the balanced reaction equation;

           1 mole of C reacted with 1 mole of O₂

We see that C is in excess and O₂ is the limiting reactant.

            1 mole of O₂ will produce 1 mole of CO₂

 So;      1.63mole of O₂ will produce 1.63 mole of CO₂

Mass of CO₂ = number of moles x molar mass

       Molar mass of CO₂ = 44g/mol

Mass of CO₂ = 1.63 x 44 = 71.5g

8 0
3 years ago
How do you know if a reaction is endothermic
Karo-lina-s [1.5K]
If the temperature rises in a reaction. Exothermic is if it loses heat. 

Have a nice day, brainliest would be fantastic.
7 0
3 years ago
Particles that are smaller and have ____ generally dissolve faster.(1 point)
saw5 [17]

Answer:

For many solids dissolved in liquid water, the solubility increases with temperature. The increase in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions.

Explanation:

7 0
3 years ago
Read 2 more answers
What's the difference between a mole of propane and a gram of propane? <br>​
kobusy [5.1K]
B. A gram would have a lot more molecules of propane than a mole
5 0
3 years ago
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
cricket20 [7]

Answer:

21.2 gm

Explanation:

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

butane is the hydrocarbon C4H10  

in combustion, we react hydrocarbons with O2 to form CO2 and H2O

so

C4H10  + O2---------------->  CO2 + H2O

BALANCE

2C4H10 + 1302--------> 8CO2 + 10 H2O

the molar mass of CO2 is 12 + 16X2 = 44

64.1 gm of CO2 is

64.1/44 = 1.46 MOLES OF  CO2,

FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE  IT IS A

8:2 OR 4:1 RATIO.  THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2

SO

THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES

THE MOLAR MASS OF BUTANE IS 58.12

0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm

6 0
3 years ago
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