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2 years ago

7

If 165.0g of nitrogen is reacted with 210.0g of hydrogen, what is the theortical yield of the reaction? What is the excess react

ant? What is the limiting reactant? Show your work.

1 answer:

Alborosie2 years ago

6 0

Answer:

this is your answer 133.0g

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In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide dissolved in molten cryolit

natka813 [3]

The given question incomplete, the complete question is:

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na, Alts).re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

Answer:

The correct answer is 6.2114 grams.

Explanation:

Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.

Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,

w = 27*1800*37 / 3*96500

w = 1798200 / 289500

w = 6.2114 grams

Hence, pure aluminum produced in the process is 6.2114 grams.

7 0

3 years ago

List 10 uses of metals

KatRina [158]

Answer:

1.In electronics

2 In medicine

3.In the construction

4.In jewelry

5.In machinery and automobiles

6.In agriculture

7.In the kitchen

8.Furniture

9. Security system

10 Bullion market

7 0

2 years ago

Using the table of standard formation enthalpies that you'll find under the ALEKS Data tab, calculate the reaction enthalpy of t

Natasha2012 [34]

Hard question !! Is this math .. if not then oh okay cause I’m bad at math but good at everything else so let me think about it !

7 0

3 years ago

How many valence electrons are in oxygen?

Sunny_sXe [5.5K]

i think it is 6 valence electrons

5 0

3 years ago

A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure

Alex

Answer:

B.3/5p

Explanation:

For this question, we have to remember <u>"Dalton's Law of Partial Pressures"</u>. This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.

Additionally, we have a <em>proportional relationship between moles and pressure</em>. In other words, more moles indicate more pressure and vice-versa.

$P_i=P_t_o_t_a_l*X_i$

Where:

$P_i$ =Partial pressure

$P_t_o_t_a_l$ =Total pressure

$X_i$ =mole fraction

With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:

<u>moles of hydrogen gas</u>

The molar mass of hydrogen gas ( $H_2$ ) is 2 g/mol, so:

$6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2$

<u>moles of oxygen gas</u>

The molar mass of oxygen gas ( $O_2$ ) is 32 g/mol, so:

$64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2$

Now, total moles are:

Total moles = 2 + 3 = 5

With this value, we can write the partial pressure expression for each gas:

$P_H_2=\frac{3}{5}*P_t_o_t_a_l$

$P_O_2=\frac{2}{5}*P_t_o_t_a_l$

So, the answer would be <u>3/5P</u>.

I hope it helps!

5 0

3 years ago