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never [62]
2 years ago
5

Is god real I don’t really no but say yes

Chemistry
1 answer:
vredina [299]2 years ago
7 0

Answer:

Yes

Explanation:

We have been created by him therefore we would not have been created if there was no God.

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Name two control substance?
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Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of
bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

<u>Step 1: </u>Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

<u>Step 2:</u> The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

<u>Step 3:</u> Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

<u>Step 4:</u> Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

<u>Step 5: </u>Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

<u>Step 6:</u> Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

<u>Step 7</u>: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

<u>Step 8:</u> Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0
3 years ago
Stoichiometry<br> Please show how to get this I don't understand any part of it!!
tatiyna

Answer:

The following is a rather lengthy discussion on chemical stoichiometry, but it will lead you through the mole concept and its application to chemical reaction stoichiometry. The problems are different than the one posted, but if you can follow this, then you can work your problem of interest.

Explanation:

Stoichiometry is very easy to master if you understand the ‘mole concept’ and how it is used to define and describe chemical process mathematically. A ‘mole’ – in chemistry – is the mass of substance containing one Avogadro’s Number of particles. That is, N₀ = 6.023 x 10²³ particles / mole. When working with chemical reactions and equations data should be first converted to moles using the following conversions.  

1 mole = 1 formula weight = 6.023 x 10²³ particles = 22.4 liters at STP(0⁰, 1atm).

In this problem you are given the equation 2Na + 2H₂O => 2NaOH + H₂. ‘Reading the equation’ there are 2 moles of Na, 2 moles of water, 2 moles of NaOH and 1 mole of H₂. In another example 3H₂ + N₂ => 2NH₃ there are 3 moles of H₂, 1 mole of N₂ and 2 moles of NH₃. The mole values can be multiples or fractions but if one mole value increases all the remaining mole values increase or decrease proportionally. For example:

Using the equation 2Na + 2H₂O => 2NaOH + H₂, one could multiply the equation by 2 giving 4Na + 4H₂O => 4NaOH + 2H₂. The equation shows balance but is not in standard form. If one multiplies the equation by ½ gives Na + H₂ => NaOH + ½H₂. Again, the equation shows balance but is not in standard form.

Standard form of equation also implies the equation conditions are at 0⁰C & 1atm pressure and 1 mole of any gas phase substance occupies 22.4 Liters volume. Such is the significance of converting given data to moles as all other substance mass (in moles) are proportional.  

1st problem, 1.76 x 10²⁴ formula units of Na will react with water (usually read as an excess) to produce (?) grams of H₂.

1st write the equation followed by listing the givens below the respective formulas… That is…

                         Na                      +            H₂O       =>    NaOH    +    H₂,

Given:      1.76 x 10²⁴ atoms                excess             ---------          ? grams

               = 1.76 x 10²⁴atoms/6.023 x 10²³atoms/mole

               = 2.922moles                                        produces  =>    2.922moles H₂ (b/c coefficients of Na & H₂ are same)

                                                      Convert moles to grams =>    2.922moles H₂  x  2.000 grams H₂/mole H₂

                                                                                                              = 5.8443 grams H₂

2nd problem, 3.5 moles Na will react with H₂O (in excess) to produce (?) moles of NaOH.

Again, write equation and assign values to each formula unit in the equation.

                         Na                      +            H₂O        =>           NaOH    +    H₂,

Given:            3.5moles                       excess                      ? grams       ----

Since coefficients of balanced std equation are equal …

                      3.5 moles Na                          produces =>     3.5 moles NaOH

                                             Convert moles to grams =>     3.5 moles NaOH x 40 g Na/mole Na

                                                                                                  = 140 grams NaOH

3rd problem, 2.75 x 10²⁵ molecules H₂O will react with (?) atoms of Na.

Same procedure, convert to moles, solve problem by ratios then convert to needed dimension at end of problem.

                         Na          +            H₂O                                   =>           NaOH           +    H₂

Given:           excess            2.75 x 10²⁵ molecules H₂O    =>      ? atoms NaOH       ----

Convert to moles =>          2.75 x 10²⁵ molecules H₂O / 6.023 x 10²³ molecules H₂O/mole H₂O  

                                             = 45.658 moles H₂O              =>  45.658 moles NaOH  (coefficients are equal)

Convert moles NaOH to grams NaOH                            =>   45.658 moles NaOH x  40 grams NaOH/mole NaOH

                                                                                                     = 1826.33 grams NaOH  

Master the mole concept and you master a lot of chemistry! Good luck.

8 0
3 years ago
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