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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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Elena-2011 [213]

Answer:2.86x10^-7m

Explanation:E=hc/^

E=6.94x10^-19J

c = 2.9979x10^8m/s

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^ =( 6.626x10^-34)x( 2.9979x 10^8)/ 6.94x10^-19

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5 0
2 years ago
Which statement about the electron-cloud model is true? It is the currently accepted atomic model. It can easily be replaced by
yan [13]

Answer:

3 choice

Explanation:

7 0
1 year ago
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How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation? 2HBr(aq)+
Mkey [24]

Full Question:

A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?

How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?

2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)

Answer:

13.1 g K2CO3 required to neutralize spill

Explanation:

2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)

Number of moles = Volume * Molar Concentration

moles HBr= 0.42L x .45 M= 0.189 moles HBr

From the stoichiometry of the reaction;

1 mole of K2CO3 reacts  with 2 moles of HBr

1 mole = 2 mole

x mole = 0.189

x = 0.189 / 2 = 0.0945 moles

Mass = Number of moles * Molar mass

Mass = 0.0945 * 138.21  = 13.1 g

3 0
2 years ago
6. From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 25ºC: Reaction A: ΔH = 10.5 kJ
qwelly [4]

Answer:

Both reaction A and reaction B are non spontaneous.

Explanation:

For a spontaneous reaction, change in gibbs free energy (\Delta G) should be negative.

We know, \Delta G=\Delta H-T\Delta S, where T is temperature in Kelvin scale.

Reaction A: \Delta G=(10.5\times 10^{3})-(298\times 30)J/mol=1560J/mol

As \Delta G is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

\Delta H-T\Delta S< 0

or, T> \frac{\Delta H}{\Delta S}

or, T> \frac{10.5\times 10^{3}}{30}

or, T> 350

So at a temperature greater than 350 K, the reaction is spontaneous.

Reaction B: \Delta G=(1.8\times 10^{3})-(-113\times 298)J/mol=35474J/mol

As \Delta G is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

\Delta H-T\Delta S< 0

or, T> \frac{\Delta H}{\Delta S}

or, T> \frac{1.8\times 10^{3}}{-113}

or, T> -16

So at a temperature greater than -16 K, the reaction is spontaneous.

3 0
2 years ago
A driver with a nearly empty fuel tank may say she is "running on fumes."
tensa zangetsu [6.8K]

Answer:

The farthest the vehicle could travel (if it gets 20.0 miles per gallon on liquid gasoline) is 1.62 miles.

Explanation:

The automobile gas tank has a volume capacity of 15 gallons which can be converted to liters: 15 × 3.7854 = 56.781 liters

We can find the moles of gasoline by using the ideal gas equation: PV = nRT.

Make n (number of moles) the subject of the formula: n = PV/RT, where:

P = 747 mmHg

V = 56.781 liters

R (universal gas constant) = 0.0821 liter·atm/mol·K

T = 25 ∘C = (273 + 25) K = 298 K

1 atm (in the unit of R) = 760 mmHg

Therefore n = 747 × 56.781/(0.0821 × 760 × 298) = 2.281 mol.

Given that the molar mass of the gasoline = 101 g/mol,

the mass of gasoline = n × molar mass of gasoline = 2.281 mol × 101 g/mol = 230.38 g

the density of the liquid gasoline = 0.75 g/mL

In order to calculate the distance the vehicle can travel, we have to calculate volume of gasoline available = mass of the liquid gasoline ÷ density of liquid gasoline

= 230.38 g ÷ 0.75 g/mL = 307.17 mL = 0.3071 liters = 0.3071 ÷ 3.7854 = 0.0811 gallons

since the vehicle gets 20.0 miles per gallon on liquid gasoline, the distance traveled by the car = gallons available × miles per gallon = 0.0811 × 20 = 1.62 miles.

6 0
2 years ago
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