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ki77a [65]
3 years ago
5

How many molecules of manganese (IV) sulfate are there if you are given 542.32 g?

Chemistry
1 answer:
Tomtit [17]3 years ago
5 0

1.3 x 10²⁴molecules

Explanation:

Given parameters:

Mass of Mn(SO₄)₂ = 542.32g

Unknown:

Number of molecules in the compound = ?

Solution:

The number of moles is the unit used to express the number of particles in a substance.

A mole contains 6.02 x 10²³ molecules

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

Molar mass Mn(SO₄)₂ = 55 + 2(32 + 4(16)) = 247g/mol

Number of moles = \frac{542.32}{247} = 2.2moles

  1 mole  =  6.02 x 10²³ molecules

2.2 moles =  2.2 x  6.02 x 10²³ molecules = 1.3 x 10²⁴molecules

learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

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Galvanized nails are iron nails that have been plated with zinc to prevent rusting. The relevant reaction is
fomenos

Answer:

1.18 × 10⁷ c

Iron is the anode and zinc is the cathode.

Explanation:

Let's consider the reduction of Zn²⁺.

Zn²⁺(aq) + 2 e⁻ ⇒ Zn(s)

<em>How many coulombs of charge are needed to produce 61.2 mol of solid zinc?</em>

<em />

We can establish the following relations:

  • When 2 moles of electrons circulate, 1 mol of Zn is produced.
  • 1 mole of electrons have a charge of 96468 c (Faraday's constant).

Then, for 61.2 mol of Zn:

61.2molZn.\frac{2mole^{-} }{1molZn} .\frac{96468c}{1mole^{-}} =1.18 \times 10^{7} c

<em>Identify the anode and cathode when plating an iron nail with zinc.</em>

The anode is where the oxidation takes place and the cathode is where the reduction takes place.

Anode (oxidation): Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻

Cathode (reduction): Zn²⁺(aq) + 2 e⁻ ⇒ Zn(s)

3 0
3 years ago
A guy swam a 50 meter freestyle in 21.93 seconds. How fast is this in miles per hour?
lara [203]

Answer:

<h2>Speed = 5.107 mile/hr</h2>

Explanation:

<h3>Given:</h3>

<u>distance</u> =50m = 0.0311 mile

<u>Time</u>= 21.93 seconds/3600 =0.00609 hour

<u>Speed</u><u>=</u>?

<h3>Speed= Distance / Time </h3><h3> = 0.0311 / 0.00609</h3><h3> = 5.107 mile/hr</h3>

5 0
3 years ago
Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t
Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

Q_1=m\times \Delta H_{fusion}

where,

Q_1 = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

\Delta H_{fusion} = enthalpy change for fusion = 3.35\times 10^5J/Kg

Now put all the given values in Q_1, we get:

Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J

<u>For process 2 :</u>

Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})

where,

Q_2 = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

c_{p,l} = specific heat of liquid water = 4186J/Kg^oC

T_1 = initial temperature = 0^oC

T_2 = final temperature = 100^oC

Now put all the given values in Q_2, we get:

Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC

Q_2=4.186\times 10^5J

From this we conclude that, Q_1 that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0
3 years ago
Consider the reaction NOBr(g) =&gt; NO(g) + 1/2 Br2(g). A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M-1
timama [110]

Answer:

Thus, the order of the reaction is 2.

The rate constant of the graph which is :- 2.00 M⁻¹s⁻¹

Explanation:

The kinetics of a reaction can be known graphically by plotting the concentration vs time experimental data on a sheet of graph.

The concentration vs time graph of zero order reactions is linear with negative slope.

The concentration vs time graph for a first order reactions is a exponential curve.  For first order kinetics the graph between the natural logarithm of the concentration vs time comes out to be a straight graph with negative slope.

The concentration vs time graph for a second order reaction is a hyberbolic curve. Also, for second order kinetics the graph between the reciprocal of the concentration vs time comes out to be a straight graph with positive slope.

Considering the question,

A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M⁻¹s⁻¹.

<u>Thus, the order of the reaction is 2.</u>

<u>Also, slope is the rate constant of the graph which is :- 2.00 M⁻¹s⁻¹</u>

5 0
4 years ago
Which type of relationship exists between Alaskan brown bears and salmon? A. predation B. mutualism C. competition D. commensali
aksik [14]

Answer:

A. predation.

Explanation:

Predation is a type of ecological interaction in which one organism which is known as predator feeds upon another organism thereby killing it. The organism which is killed by predator is known as prey. In such interaction one organism is benefited while another one is harmed.

Brown bears of Alaska feed upon salmon. In order to so they have developed many ways which include waiting at the bottom of falls from where salmons pass by or standing at the bottom of the falls.

3 0
3 years ago
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