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alexgriva [62]
2 years ago
6

3. A 0.500 g sample of nitrogen gas combines with 1.140 g of oxygen gas to form NO2. If the atomic mass of oxygen is 16.000, cal

culate the atomic mass of nitrogen from this data.
Chemistry
2 answers:
DedPeter [7]2 years ago
6 0

Answer;

= 18.24

Explanation;

The ratio of N and O in the formula NO2 IS 1:2

Mass of nitrogen gas is 0.500 g

Moles of nitrogen will be;

= 0.500/16 = 0.03125 moles

Therefore;

The moles of Oxygen from the ratio will be;

= 0.03125 × 2 = 0.0625 moles

But; 0.0625 moles is equal to 1.140 g of Oxygen

The atomic number (mass in 1 mole) will be;

= 1.140 /0.0625

= 18.24

Thus the atomic number of Oxygen from the data is 18.24

astra-53 [7]2 years ago
6 0

The balanced chemical equation for the given reaction is:

N₂ + 2O₂ → 2NO₂

According to the given balanced equation, 2 moles of O₂ combines with 1 mole of O₂

Given, Mass of O₂ = 1.140 g

Atomic mass of O = 16 amu

Molar mass of O₂ = 16 x 2 = 32 g/mol

Now to calculate the moles of N₂:

1.14 g of O₂ x (1 mole of O₂/ 32 g of O₂) x (1 mol of N₂/ 2 mol O₂) = 0.0178 mol of N₂

Molar mass = mass/ moles  

Given, Mass of N₂ = 0.500 g

Molar mass of N₂ = 0.500 g / 0.0178 mol = 28 g/mol

Atomic mass of N = 28/2 = 14 amu

Therefore, the atomic mass of N is 14 amu


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If a type of radiation is attracted to the negative electrode, then it must have a positive charge. Beta particles are electrons and are negative. X-rays and gamma rays are essentially the same thing and are electromagnetic energy. They have no charge. The difference between X-rays and gamma rays is how they are produced*. Your answer is alpha radiation. An alpha particle is the nucleus of a helium atom and carries a +2 charge.  

* Gamma rays are a product of the random nuclear decay of unstable nuclei. X-rays are produced when high speed electrons are caused to stop or slow down. The kinetic energy of the electron is converted to electromagnetic energy. X-rays and gamma rays are not particles and do not carry a charge, and cannot be attracted or deflected from charged electrodes.

4 0
1 year ago
A solution of SO2 in water contains 0.00023 g of SO2 per liter of solution. What is the concentration of SO2 in ppm? in ppb?
Mnenie [13.5K]

Answer:

= 230 ppb

Explanation:

Considering that;

1ppm = 1mg/L  

Then;

0.00023g = 0.23mg  

Therefore;

0.00023 g/L = 0.23 mg/L

0.23 mg/L = 0.23 ppm

1 ppm = 100 ppb

Therefore;

0.23 ppm = 0.23 ×1000

                = 230 ppb

5 0
2 years ago
Read 2 more answers
You decide to establish a new temperature scale on which the melting point of ammonia (-77.75 ∘c) is 0∘a, and the boiling point
Hoochie [10]
First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:

-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
 x = 177.75*100/44.4 = 400.33

The boiling point of water in ∘a would be 400.33∘a.

7 0
1 year ago
Read 2 more answers
If excess caso4(s) is mixed with water at 25 ∘c to produce a saturated solution of caso4, what is the equilibrium concentration
Ray Of Light [21]

<em>Answer:</em>

The equlibrium concentration sof Ca+2 ion willl be 4.9×10∧-3 M

<em>Data Given:</em>

              Ksp of CaSO4 = 2.4 × 10∧-5

              CaSO4 ⇔ Ca+2   +  SO4∧-2

<em>Solution:</em>

                Ksp = [Ca+2].[ SO4∧-2]

                 2.4 × 10∧-5 = [x].[x]= x²

                 x =  4.9×10∧-3 M

<em>Result:</em>

  • The conc. of Ca+2 ion is 4.9×10∧-3 M
3 0
2 years ago
A 2.50 g sample of zinc is heated, and then placed in a calorimeter containing 65.0 g of water. Temperature of water increases f
swat32

718.65 degrees is the initial temperature of the zinc metal sample.

Explanation:

Data given:

mass of zinc sample = 2.50 grams

mass of water  = 65 grams

initial temperature of water = 20 degrees

final temperature of water = 22.5 degrees

ΔT  = change in temperature of water is 2.50 degrees

specific heat capacity of zinc cp= 0.390 J/g°C

initial temperature of zinc sample = ?

cp of water = 4.186 J/g°C

heat absorbed = heat released (no heat loss)

formula used is

q = mcΔT

q water = 65 x 4.286 x 2.5

q water = 696.15 J

q zinc = 2.50 x 0.390 x (22.50- Ti)

equating the two equations

696.15 = - 22.50+ Ti

Ti = 718.65 degrees is the initial temperature of zinc.

6 0
2 years ago
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