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stich3 [128]
3 years ago
14

Ammonia (NH3) burns in air to form nitrogen dioxide and water. Calculate grams of nitrogen dioxide produced if 4.0 grams of ammo

nia are consumed
Chemistry
1 answer:
kenny6666 [7]3 years ago
8 0
The balanced equation for the above reaction is;
4NH₃ + 7O₂  --> 4NO₂  + 6H₂O
stoichiometry of NH₃ to NO₂ is 4:4
the number of NH₃ moles consumed are - 4.0 g / 17 g/mol = 0.24 mol
number of NH₃ moles reacted are equivalent to number of NO₂ moles formed 
therefore number of NO₂ moles formed - 0.24 mol x 46 g/mol = 11.04 g
mass of NO₂ formed is 11.04 g
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A mineral sample is obtained from a region of the country that has high arsenic contamination. An elemental analysis yields the
Gnesinka [82]

Answer:

<em><u>CaAsHO₄</u></em>

Explanation:

The data has a mistake in one of the values there. I believe the mistake is on the hydrogen. So, I'm going to assume the value of Hydrogen is 0.6%, so the total percent composition would be 100.1% (Something better). All you have to do is replace the correct value of H (or the value with the mistaken option) and do the same procedure.

Now, to calculate the empirical formula, we can do this in three steps.

<u>Step 1. Calculate the amount in moles of each element.</u>

In these case, we just divide the percent composition with the molar mass of each one of them:

Ca: 22.3 / 40.078 = 0.5564

As: 41.6 / 74.9216 = 0.5552

O: 35.6 / 15.9994 = 2.2251

H: 0.6 / 1.00794 = 0.5953

Now that we have done this, let's calculate the ratio of mole of each of them. This is doing dividing the smallest number of mole between each of the moles there. In this case, the moles of As are the smallest so:

Ca: 0.5564/0.5552 = 1.0022

As: 0.5552/0.5552 = 1

O: 2.2251/0.5552 = 4.0077

H: 0.5953/0.5552 = 1.0722

Now, we round those numbers, and that will give us the number of atoms of each element in the empirical formula

<u>Step 3. Write the empirical formula with the rounded numbers obtained</u>

In this case we will have:

Ca: 1

As: 1

O: 4

H: 1

The empirical formula would have to be:

<em><u>CaAsHO₄</u></em>

3 0
3 years ago
What is the enthalpy in KJ of reaction for the initial combustion if the delta H for vaporization is 75.75 KJ and the net delta
Gnesinka [82]

Answer:

-775.75

Explanation:

4 0
3 years ago
Why can't all mixtures be classified as solutions?
shusha [124]
D. Not all mixtures are heterogeneous
7 0
3 years ago
Solid potassium chlorate decomposes upon heating to form
olga55 [171]

Answer:

32.6%

Explanation:

Equation of reaction

2KClO₃ (s) → 2KCl (s) + 3O₂ (g)

Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)

Molar volume of Oxygen at s.t.p = 22.4L / mol

since the gas was collected over water,

total pressure = pressure of water vapor + pressure of  oxygen gas

0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C

pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1

P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p

Using ideal gas equation

\frac{P1V1}{T1} = \frac{P2V2}{T2}

V2 = \frac{P1V1T2}{T1P2}

V2 = 664.1052 ml

245.2   yielded 67.2 molar volume of oxygen

0.66411 will yield = \frac{245.2 * 0.66411}{67.2}  = 2.4232 g

percentage of potassium chlorate in the original mixture = \frac{2.4232 * 100}{7.44} = 32.6%

3 0
3 years ago
HELP MEE please with this question this question is 100pts and i will give branliest i only have 5 min left guys thanx:) Which s
LenaWriter [7]

Answer:

The cell grows into its full size

The cell copies it’s dna

Explanation:

hope this helps pls mark brainliest

8 0
3 years ago
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