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2 years ago

What is the composition of 'Clean dry air' - Idk what subject its just science

1 answer:

7 0

Dry air is a mixture of nitrogen, oxygen, carbon dioxide etc.
air is a mixture of gases 78% nitrogen an 21% oxygen and other components.

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Please help me i keep getting this wrong :(

solmaris [256]

A) A chandelier has been hanging in the kitchen for years
B) A log floats on top of the lake
C) You place your book on the top of a flat table

Those are the answers. In each case, there is always a force that balances the weight of the object and keeps them in a static equilibrium. Tension, Buoyancy and Normal force.

7 0

2 years ago

The molar mass of an unknown gas was measured by an effusion experiment. It was found that it took 60 s for the gas to effuse, w

Marizza181 [45]

Answer:

The molar mass of the gas is 44 g/mol

Explanation:

It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:

$\frac{r_1}{r_2} =\frac{\sqrt{M_2} }{\sqrt{M_1} }$

If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):

$\frac{X/48s}{X/60s} =\frac{\sqrt{M_2} }{\sqrt{28g/mol} }$

6,61 = √M₂

44g/mol = M₂

The molar mass of the gas is 44 g/mol

I hope it helps!

4 0

3 years ago

Element x reacts with oxygen to produce x2o3 in an experiment it is found that 1.0000 g of x produces 1.1xxx g of x2o3 what is t

8_murik_8 [283]

Since X is 1 g, therefore O must be 0.1 g. Therefore:

moles O = 0.1 g / (16 g / mol) = 0.00625 mol

We can see that for every 3 moles of O, there are 2 moles of X, therefore:

moles X = 0.00625 mol O (3 moles X / 2 moles O) = 0.009375 mol

Molar mass X = 1 g / 0.009375 mol

Molar mass X = 106.67 g/mol

4 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

2 years ago

What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?

mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

$C=\frac{n}{V}$ .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is $K_{2}SO_{4}$ thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass

Putting the values,

$n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol$

Thus, number of moles of $K_{2}SO_{4}$ will be $0.1061\times 0.5=0.053 mol$ .

The volume of solution is 225 mL, converting this into L,

$1 mL=10^{-3}L$

Thus,

$225 mL=0.225 L$

Putting the values in equation (1),

$C=\frac{(0.053 mol}{0.225 L}=0.236 M$

Therefore, concentration of potassium sulfate solution is 0.236 M.

4 0

3 years ago