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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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The solubility of copper chloride at 20 °C is 73 g/100 g of water. Kiera adds 100 g of copper chloride to 100 g of water and sti
OLEGan [10]

Answer:

m_{undissolved}=27g

Explanation:

Hello there!

In this case, according to the given information of the solubility of copper chloride, as the maximum amount of this salt one can dissolve without having a precipitate, we infer that since just 73 grams are actually dissolved, the following amount will remain solid as a precipitate:

m_{undissolved}=100g-73g\\\\m_{undissolved}=27g

Best regards!

3 0
2 years ago
Calculate the activity coefficients for the following conditions:
uysha [10]

<u>Answer:</u>

<u>For a:</u> The activity coefficient of copper ions is 0.676

<u>For b:</u> The activity coefficient of potassium ions is 0.851

<u>For c:</u> The activity coefficient of potassium ions is 0.794

<u>Explanation:</u>

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}       ........(1)

where,

\gamma_i = activity coefficient of ion

Z_i = charge of the ion

\mu = ionic strength of solution

\alpha _i = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)        ......(2)

where,

C_i = concentration of i-th ions

Z_i = charge of i-th ions

  • <u>For a:</u>

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M

Now, calculating the activity coefficient of Cu^{2+} ion in the solution by using equation 1:

Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{  (known)}\\\mu=0.01M

Putting values in equation 1, we get:

-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676

Hence, the activity coefficient of copper ions is 0.676

  • <u>For b:</u>

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M

Now, calculating the activity coefficient of K^{+} ion in the solution by using equation 1:

Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.025M

Putting values in equation 1, we get:

-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851

Hence, the activity coefficient of potassium ions is 0.851

  • <u>For c:</u>

We are given:

0.02 M K_2SO_4 solution:

Calculating the ionic strength by using equation 2:

C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M

Now, calculating the activity coefficient of K^{+} ion in the solution by using equation 1:

Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.06M

Putting values in equation 1, we get:

-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794

Hence, the activity coefficient of potassium ions is 0.794

6 0
3 years ago
When dipentyl ether is treated with HI, what type of reaction occurs? both SN1 and SN2 SN2 E1 SN1 E2
masha68 [24]

Answer:

SN2

Explanation:

The first step of ether cleavage is the protonation of the ether since ROH is a better leaving group than RO-.

The second step of the reaction may proceed by either SN1 or SN2 mechanism depending on the structure of the ether. Methyl and primary ethers react with HI by SN2 mechanism while tertiary ethers react with HI by SN1 mechanism. Secondary ethers react with HI by a mixture of both mechanisms.

Dipentyl ether is a primary ether hence when treated with HI, the reaction with HI proceeds by SN2 mechanism as explained above.

7 0
3 years ago
PLEASE HELP ME ASAPPPP
sattari [20]
Boyle’s law gives the relationship between pressure and volume of gases. It states that at constant temperature the pressure of gas is inversely proportional to volume of gas.
PV = k
Where P is pressure V is volume and k is constant
P1V1 = P2V2
Parameters at STP are on the left side and parameters for the second instance are on the right side of the equation
P1 - standard pressure - 1.0 atm
Substituting the values in the equation
1.0 atm x 5.00 L = P x 15.0 L
P = 0.33 atm
New pressure is 0.33 atm
5 0
3 years ago
Anyone help plz <br> And ty
andreyandreev [35.5K]

Answer: do that yourself dip

Explanation:

6 0
3 years ago
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