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3 years ago

WILL MARK BRAINLIEST IF ACCURATE AND DETAILED!!!!!!!!!!!

Chemistry

1 answer:

Mazyrski [523]3 years ago

6 0

No, because conflicting evidence and arguments were not considered. there is your answer.

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Which of the following techniques is most appropriate for the recovery of solid KNO3 from an aqueous solution of KNO3?A. Paper c

r-ruslan [8.4K]

Answer:

E. Evaporation to dryness

Explanation:

E. - Evaporation to dryness is the best method for the recovery of solid KNO3 from an aqueous solution of KNO3.

(KNO3 is very soluble, and will violently decompose if overheated.)

Paper chromatography is for separation of different weight molecules in solution.

B. Filtration won't work on a solution

C.Titration would contaminate the salt with something else and is used

to determine concentrations

D. Electrolysis would destroy the salt

7 0

4 years ago

How many moles of chlorine gas and 120°C and 33.3 ATM would occupy a vessel of 12 L?

Mashutka [201]

<h3>Answer:</h3>

12.387 moles

<h3>Explanation:</h3>

We are given;

Temperature of chlorine, T = 120°C

But, K = °C + 273.15

Therefore, T = 393.15 K

Pressure, P = 33.3 Atm

Volume, V = 12 L

We are required to calculate the number of moles of chlorine gas,

To find the number of moles we are going to use the ideal gas equation;

PV = nRT

R is the ideal gas constant, 0.082057 L.atm/mol.K

Therefore, rearranging the formula;

n = PV÷RT

Hence;

n = (33.3 atm × 12 L) ÷ (0.082057 × 393.15 K)

= 12.387 moles

Therefore, the number of moles of chlorine are 12.387 moles

8 0

4 years ago

A compound is formed when 12.2 g Mg combines completely with 5.16 g N. What is the percent composition of this compound? Please

Viefleur [7K]

3Mg + N₂ → Mg₃N₂
n(Mg) = 12,2g÷24,4g/mol = 0,5mol - limiting reagente.
n(N₂) = 5,16g÷28g/mol = 0,18mol
n(Mg₃N₂):n(Mg) = 1:3, n(Mg₃N₂) = 0,166mol
m(Mg₃N₂) = 0,166mol·101,2g/mol = 16,8g.
%(N)= 2·Ar(N)÷Mr(Mg₃N₂) = 2·14÷101,2 = 27,66% = 0,2766
%(Mg) = 3·Ar(Mg)÷Mr(Mg₃N₂) = 3·24,4÷101,2 = 72,34%
or 100% - 27,66% = 72,34%.

7 0

4 years ago

Breathing air that contains 4.0% by volume CO2 over time causes rapid breathing, throbbing headache, and nausea, among other sym

7nadin3 [17]

Answer:

1 mole of any ideal gas occupies the same volume as one mole of any other ideal gas under the same conditions of temperature and pressure.So 1 L CO2 has same number of moles as 1 L O2 and 1 L N2 etc.This means that, assuming all the gases in air are ideal gases, if CO2 is 4.0 % by volume then it is also 4.0 % by moles, because a volume of each gas has the same number of moles.

Explanation:

3 0

3 years ago

50.0kg of nitrogen and 10.0kg 0f hydrogen are mixed to produce ammonia . calculate the ammonia formed, identify the limiting rea

Nana76 [90]

Answer:

Nitrogen is limiting reactant and 30.4kg of ammonia can be formed

Explanation:

Nitrogen, N₂, reacts with hydrogen, H₂ to produce ammonia, NH₃, as follows:

2N₂ + 3H₂ → 2NH₃

<em>Where 2 moles nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia</em>

<em />

To solve this question we must find the moles of each reactant in order to find limiting reactant. With limiting reactant we can find the moles and the mass of ammonia formed as follows:

<em>Moles N2 -Molar mass: 28g/mol-</em>

50000g * (1mol / 28g) = 1786 moles N2

<em>Moles H2 -Molar mass: 2g/mol-</em>

10000g * (1mol / 2g) = 5000 moles H2

For a complete reaction of 5000 moles H2 are needed:

5000 mol H2 * (2mol N2 / 3mol H2) = 3333 moles N2. As there are just 1786 moles, Nitrogen is limiting reactant

The moles of ammonia that can be produced are 1786 moles because 2mol N2 = 2moles NH3.

The mass of ammonia -Molar mass NH3: 17g/mol- is:

1786 moles NH3 * (17g / mol) = 30362g =

30.4kg of ammonia can be formed

3 0

3 years ago