Identify the following parts of the chemical equation

What type of electrons are involved in chemical reactions?

sweet-ann [11.9K]

The outermost energy shell of an atom

because they are involved in forming bonds

7 0

3 years ago

A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co

DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>

43058 mol air×29g/mol <em>1249 kg air</em>

Percent of oxygen is: $\frac{289kg}{1249 kg}$ =<em>0,231 kg O₂/ kg air</em>

I hope it helps!

4 0

2 years ago

A 11.97g sample of NaBr contains 22.34 % Na by mass. Considering the law of constant composition (definite proportions), how man

kap26 [50]

<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

Mass of NaBr sample is 11.97 g

% composition by mass of Na in the sample is 22.34%

We are required to determine the mass of 9.51 g of a NaBr sample.

Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.

In this case,

A sample of 11.97 g of NaBr contains 22.34% of Na by mass

Therefore;

A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

Therefore;

Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

Mass of sodium = (22.34% × 9.51 g) ÷ 100

= 2.125 g

Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

3 0

2 years ago

3. A girl<br> ran 105 metres in 15 seconds. What was her speed

sweet [91]

Answer:

7 meters per second was her speed

5 0

2 years ago

Ammonia, NH 3 , may react with oxygen to form nitrogen gas and water. 4 NH 3 ( aq ) + 3 O 2 ( g ) ⟶ 2 N 2 ( g ) + 6 H 2 O ( l )

Alex Ar [27]

Answer:

The limiting reactant is NH₃

0.0186moles of N₂ are the one produced by the limiting reactant

0.020 moles of N₂ are the one produced by the reactant in excess

Explanation:

This is the reaction

4NH₃ + 3O₂ → 2N₂ + 6H₂O

We should calculate the moles of each reactant

Mass / Molar mass = Moles

3.55 g / 17g/m = 0.208 moles NH₃

5.33 g / 32g/m = 0.166 moles O₂

4 moles of ammonia react with 3 moles of oxygen

0.208 moles of ammonia react with (0.208 .3)/4 = 0.156 moles O₂

We have 0.166 moles of O₂ and we need 0.156 moles, so O₂ is the reactant in excess.

3 moles of O₂ react with 4 moles of NH₃

0.166 moles of O₂ react with (0.166 . 4)/ 3 = 0.221 moles

We have 0.208 moles NH₃ and we need 0.221, so NH₃ is the limiting reactant.

To know the moles of N₂, let's apply the Ideal Gas Law

P.V =n.R.T

1atm . 0.450L = n . 0.082 . 295K

0.450 / (0.082 .295) = 0.0186 moles

If we have 100 % yield reaction:

4 moles NH₃ make 2 moles N₂

0.208 moles NH₃ make (0.208 .2)/4 = 0.104 moles

So the % yield reaction is.

0.104 moles ___ 100%

0.0186 moles ___ 17.9%

0.0186moles of N₂ are the one produced by the limiting reactant.

3 moles of O₂ produce 2 moles N₂

0.166 moles O₂ produce (0.166 .2)/3 = 0.111 moles

Now, we apply the yield.

100% ____ 0.111 moles

17.9% = 0.020 moles

8 0

2 years ago

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