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djyliett [7]
4 years ago
8

How many moles of magnesium oxide are formed when 4 moles of magnesium react with oxygen? this is the formula for the reaction:?

Chemistry
1 answer:
Alisiya [41]4 years ago
5 0
Mg + O2(o is a diatomic element)⇒ Mg O(UNBALANCED)
Mg is +2, o is -2, so MgO is the product.
BALANCED EQUATION :  2Mg + O2 ⇒ 2MgO
There are 4 moles of magnesium, so there should be 4/2*2=4 moles of oxygen, and the product will have 4+4=8 moles.
there will be 8 moles of magnesium oxide.


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The two main products of the combustion of gasoline in an automobile engine are
kozerog [31]
There is an excess of oxygen, the two main products should be carbon dioxide gas and water vapor. If the oxygen supply is limited, the fuel will undergo incomplete combustion and produce carbon monoxide, water, and sometimes carbon.

Hope this helped! :)
4 0
3 years ago
Read 2 more answers
Plasma is often in stars, which are likely to contain?
elena-s [515]
Below are the choices that can be found from other sources:

<span>A) elements 
B) ions 
C) molecules 
D) neutrons

the answer is ions. 

</span>A Star<span> and hence our Sun, is an almost entirely ionized ball of </span>plasma<span>, consisting of electrons and ions, in which there is hardly any gas (neutral atoms). The movement of the </span>plasma<span> produces strong magnetic fields and corresponding electric currents.</span>
6 0
3 years ago
In a typical analysis, 15 ml of an aqueous solution containing an unknown amount of acetylcholine had a ph of 7.65. When incubat
melomori [17]

pH of the acetyl choline solution before incubation = 7.65

[H_{3}O^{+}]=10^{-7.65}=2.24*10^{-8}M

pH of the solution after incubation = 6.87

[H_{3}O^{+}]=10^{-6.87}=1.35*10^{-7}M

The difference in concentration of hydronium ion before and after incubation

=1.35*10^{-7}M-2.24*10^{-8}M=1.126*10^{-7}M

This difference in hydronium ion concentration can be attributed to the increase in the concentration of acetic acid, which is formed when acetylcholine is hydrolyzed by acetycholinesterase. The mole ratio of acetylcholine to acetic acid is 1:1.

Therefore the moles of acetylcholine = 15 mL * \frac{1L}{1000mL}*\frac{1.126*10^{-7}mol }{L}=1.689*10^{-9}mol


7 0
3 years ago
One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc
Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

\text{The new partial pressure for }N_2 \ gas}

P_1V_1=P_2V_2

P_2=\dfrac{P_1V_1}{V_2} \\ \\  P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar

\text{The new partial pressure for }Ar \ gas}

P_2=\dfrac{P_1V_1}{V_2} \\ \\  P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar

Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \  \ \  \ \  \ \ \ \   \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \  \ \  \ \  \ \ \ \   \ \ \ \ \ \ \ \ = 2.225 \ Bar

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

2.1  \ bar = 2.07  \ atm \\ \\3.4 \  bar = 3.36 \  atm

For moles N₂:

PV = nRT \\ \\  n = \dfrac{PV}{RT}

n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}

n = 0.08297 \ mol  \ N_2

For moles of Ar:

PV = nRT \\ \\  n = \dfrac{PV}{RT}

n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}

n = 0.2172 \ mol  \ Ar

\mathtt{total \  moles = moles \ of \  N_2 + moles  \ of \ Ar}

=0.08297 mol + 0.2037 mol \\                   = 0.2867 mol gases

Finally;

The final pressure of the mixture is:

PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}

P = 2.217 atm

P ≅ 2.24 bar

7 0
3 years ago
Use this balanced equation to help you solve:
valentina_108 [34]

1. 842g of NaOH will form 547.3 g of Al(OH)₃

2. The yield is 93.55%

<u>Explanation:</u>

3NaOH + Al → Al(OH)₃ + 3Na

1.

Molar mass of NaOH = 40 g/mol

Molar mass of Al = 27 g/mol

Molar mass of Al(OH)₃ = 78 g/mol

According to the balanced equation:

3 moles of NaOH requires 1 mole of Al to form 1 mole of Al(OH)₃

The ratio of NaOH : Al : Al(OH)₃ = 3 : 1 : 1

3 X 40 g of NaOH reacts with 27 g of Al to form 78 g of Al(OH)₃

120 g of NaOH + 27g of Al → 78 g of Al(OH)₃

120g of NaOH form 78g of Al(OH)₃

1g of NaOH will form \frac{78}{120} g of Al(OH)₃

842g of NaOH will form \frac{78}{120} X 842 g of Al(OH)₃

                                   = 547.3 g of Al(OH)₃

Therefore, 842g of NaOH will form 547.3 g of Al(OH)₃

2. Only 512 g of Al(OH)₃ is formed

Yield % = ?

Yield = \frac{512}{547.3} X 100\\\\Yield = 93.55

Therefore, the yield is 93.55%

5 0
3 years ago
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