Consider two equal-volume flasks of gas at the same temperature and pressure. One gas, oxygen, has a molecular mass of 32. The other gas, nitrogen, has a molecular mass of 28. What is the ratio of the number of oxygen molecules to the number of nitrogen molecules in these flasks

Answers

Answer 1

Answer:

Answer: The ratio of the number of oxygen molecules to the number of nitrogen molecules in these flasks is 1: 1

Explanation:

According to avogadro's law, equal volumes of all gases at same temperature and pressure have equal number of moles.

According to avogadro's law, 1 mole of every substance contains avogadro's number $6.023* 10^(23)$ of particles.

Thus as oxygen and nitrogen are at same temperature and pressure and are in equal volume flasks , they have same number of moles and thus have same number of molecules.

The ratio of the number of oxygen molecules to the number of nitrogen molecules in these flasks is 1: 1

Answer 2

Answer:

Final answer:

The ratio of the number of oxygen molecules to the number of nitrogen molecules in the two flasks is 8/7.

Explanation:

The ratio of the number of oxygen molecules to the number of nitrogen molecules in the two flasks can be determined using Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Since both flasks have the same volume, temperature, and pressure, the ratio of oxygen molecules to nitrogen molecules will be equal to the ratio of their molecular masses.

The molecular mass of oxygen is 32, while the molecular mass of nitrogen is 28. Therefore, the ratio of the number of oxygen molecules to the number of nitrogen molecules will be:

32/28, or 8/7.

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Who Organized elements into four groups based on properties:

Which element has the following Electron configuration? *1s22s22p63s2
O Boron
O Phosphorus
O Manganese
O Magnesium

Answers

Magnesium. You can count the electrons in each level and because the number of electrons is the same with protons you have the atomic number based of which you can get the element in the periodic table

Quartzite is a coarse-grained rock derived from sandstone.Which type of rock is quartzite?

metamorphic

extrusive

sedimentary

igneous

Answers

its a metamorphic rock

Answer:

granoblastic metamorphic rock

Explanation:

Which of the following is a biogeochemical cycle?a. Atmospheric cycle
b. Iconic cycle
c. Neonic cycle
d. Water cycle

Answers

Biogeochemical cycles are cycles that involve biological, geological and chemical processes. It is a pathway by which a chemical substance moves through the biotic and abiotic compartments of the Earth. Among the options, the one which is a biogeochemical cycle is letter "D. Water cycle"

The monoprotic acid from among the following isa. H2CO3.
b. H2SO4.
c. H3PO4.
d. HCl.

Answers

Answer: Option (d) is the correct answer.

Explanation:

An acid which gives only one hydrogen ion or $H^(+)$ ions upon dissociation in an aqueous solution is known as a monoprotic acid.

Whereas when an acid gives two hydrogen or $H^(+)$ ions upon dissociation in an aqueous solution is known as a diprotoc acid.

And when an acid gives three hydrogen or $H^(+)$ ions upon dissociation in an aqueous solution is known as a triprotoc acid.

Therefore dissociation of the given acids in an aqueous solution will be as follows.

$H_(2)CO_(3) \rightarrow 2H^(+) + CO^(2-)_(3)$

$H_(2)SO_(4) \rightarrow 2H^(+) + SO^(2-)_(4)$

$H_(3)PO_(4) \rightarrow 3H^(+) + PO^(2-)_(4)$

$HCl \rightarrow H^(+) + Cl^(-)$

Hence, we can conclude that out of the given options, HCl is the monoprotic acid.

Monoprotic acid are acids having only one hydrogen atoms after dissociation into ions from its compound. The monoprotic acid from among the following is HCl. The answer is letter D. HCl → H+ + Cl-. Note that there is only one H+ ion upon dissociation.

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): 2H2O2(aq) ---> 2H2O(l) + O2(g)What volume of pure O2(g), collected at 27C and 746 torr, would be generated by decomposition of 125 g of a 50.0% by mass hydrogen peroxide solution?

Answers

Answer:

23.0733 L

Explanation:

The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:

$Mass=\frac {50}{100}* 125\ g$

Mass = 62.5 g

Molar mass of $H_2O_2$ = 34 g/mol

The formula for the calculation of moles is shown below:

$moles = (Mass\ taken)/(Molar\ mass)$

Thus, moles are:

$moles= (62.5\ g)/(34\ g/mol)$

$moles= 1.8382\ mol$

Consider the given reaction as:

$2H_2O_2_((aq))\rightarrow2H_2O_((l))+O_2_((g))$

2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.

Also,

1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.

So,

1.8382 moles of hydrogen peroxide decomposes to give $\frac {1}{2}* 1.8382 mole of oxygen gas. Moles of oxygen gas produced = 0.9191 molGiven: Pressure = 746 torrThe conversion of P(torr) to P(atm) is shown below:[tex]P(torr)=\frac {1}{760}* P(atm)$

So,

Pressure = 746 / 760 atm = 0.9816 atm

Temperature = 27 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (27 + 273.15) K = 300.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K

⇒V = 23.0733 L

A gas contains a mixture of NH3(g) and N2H4(g), both of which react with O2(g) to form NO2(g) and H2O(g). The gaseous mixture (with an initial mass of 61.00 g) is reacted with 10.00 moles O2, and after the reaction is complete, 4.062 moles O2 remains. Calculate themass percent of N2H4(g) in the original gaseous mixture.

Answers

Answer:

Mass percent of N2H4 in original gaseous mixture = 31.13 %

Explanation:

Given:

Initial mass of gaseous mixture = 61.00 g

Initial mole of oxygen = 10.0 mol

Moles of oxygen remaining after the reaction = 4.062 mol

Moles of oxygen used = 10.0 - 4.062 = 5.938 mol

$4NH_3 + 7O_2\rightarrow 4NO_2 + 6H_2O$

$N_2H_4 + 3O_2\rightarrow 2NO_2 + 2H_2O$

Total oxygen used in both the reactions = 10.0 parts

out of 10 parts, 3 part react with N2H4.

$No.\;of\;moles\;of \;oxygen \;used = 5.398*(3)/(10) =1.78\; moles$

Now, consider the reaction of N2H4

$N_2H_4 + 3O_2\rightarrow 2NO_2 + 2H_2O$

3 moles of O2 react with 1 mole of N2H4

1.78 moles of oxygen will react with 1.78/3 = 0.5933 mol of N2H4

$Mass = Moles* Molecular mass$

Molecular mass of N2H4 = 32 g/mol

$Mass\;of\;N_2H_4= 0.5933* 32 = 18.99 g$

$Mass\;percent = (Mass\;of\;N_2H_4)/(Total\;mass)* 100$

Total mass = 61.0 g

$Mass\;percent = (18.99)/(61.0)* 100=31.13 \%$

Final answer:

The mass percent of N2H4 in the gaseous mixture can be determined through stoichiometric calculations and determining the limiting reactant. The initial and remaining amounts of O2 are used to calculate the reacted amount of O2, which then allows for the calculation of the amount of N2H4. This information is used in the mass percent formula.

Explanation:

The balanced reaction states that for one mole of NH3, one mole of O2 is required, while for one mole of N2H4, 3 moles of O2 are required. Thus, the initial moles of O2 were 10 moles and after reaction 4.062 moles O2 remained. Thus, the reacted amount of O2 is 10 - 4.062 = 5.938 moles. From calculating the limiting reactant and applying stoichiometry, the amount of N2H4 can be determined. We know the molar mass of N2H4 is 32 g/mole. By calculating the molar ratio, we can then calculate the mass percent of N2H4 in the mixture using the formula: (mass of N2H4 / total mass) * 100%.

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