Which of these facts best illustrates why regulation of alcohol consumption is necessary

Calculate the number of grams of xenon in 4.658 g of the compound xenon tetrafluoride.

andrezito [222]

Answer:

The mass of xenon in the compound is 2.950 grams

Explanation:

Step 1: Data given

Mass of XeF4 = 4.658 grams

Molar mass of XeF4 = 207.28 g/mol

Step 2: Calculate moles of XeF4

Moles XeF4 = mass XeF4 / molar mass XeF4

Moles XeF4 = 4.658 grams / 207.28 g/mol

Moles XeF4 = 0.02247 moles

Step 3: Calculate moles of xenon

XeF4 → Xe + 4F-

For 1 mol xenon tetrafluoride, we have 1 mol of xenon

For 0.02247 moles XeF4 we have 0.02247 moles Xe

Step 4: Calculate mass of xenon

Mass xenon = moles xenon * molar mass xenon

Mass xenon = 0.02247 moles * 131.29 g/mol

Mass xenon = 2.950 grams

The mass of xenon in the compound is 2.950 grams

5 0

2 years ago

How many sigma and pi bonds in propionic bond

Ierofanga [76]

Propanoic acid formula is ch ch 2 so it has 8 bonds

7 0

2 years ago

A group of students measured the average monthly temperature of 1000 cities around the world and plotted the cumulative frequenc

mr Goodwill [35]

Answer:

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

Explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

The deviations will be added then.

So the mean absolute deviation for city 1 is 24 ..

For city 2:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

The deviations will be added then.

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

8 0

1 year ago

A student uses visible spectrophotometry to determine the concentration of CoCl2(aq) in a sample solution. First the student pre

k0ka [10]

Answer:

$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

Explanation:

Energy of the photon can be calculated by

$E=h\nu=\frac{h\times c}{\lambda}$ (Planck's equation)

where,

E = energy of photon

h = Planck's constant = $6.63\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of light =

$\nu$ = frequency of the light

we have , $\lambda =510 nm=510\times 10^{-9}m$

Now put all the given values in the above formula, we get the energy of the photons.

$E=\frac{(6.63\times 10^{-34}Js)\times (3\times 10^8m/s)}{510\times 10^{-9}m}$

$E=3.9\times 10^{-19}J\approx 4\times 10^{-9} J$

$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

5 0

2 years ago

A vessel of capacity 400 cc filled with chlorine under 80 cm is connected by a narrow tube and stopcock with another vessel of c

guapka [62]

Answer:

The pressure when the stopcock is opened is opened is 87.783 cm

Explanation:

The given parameters of the question are;

The volume of the vessel of chlorine = 400 cc

The pressure of the vessel of chlorine = 80 cm

The volume of the vessel of chlorine = 250 cc

The pressure of the vessel of chlorine = 100 cm

Daltons law of Partial Pressures states that the total pressure exerted by a volume of a mixture of gases is equal to the partial pressures exerted by the individual gases in the mixture with respect to the given volume

Therefore;

The total volume of the mixture = 400 cc + 250 cc = 650 cc

The partial pressure exerted by the chlorine gas in the total volume is given by Boyles law as follows;

P₁·V₁ = P₂·V₂

P₂ = P₁·V₁/V₂

Where;

P₁ = 80 cm = The pressure in volume V₁ = 400 cc

P₂ₓ = The partial pressure of chlorine in volume V₂ = 650 cc

Substituting, we have;

P₂ₓ = 80 × 400/650 ≈ 49.321 cm

Similarly, the partial pressure exerted by the nitrogen gas in the total volume is given by Boyles Law as follows;

P₂ₐ = P₁·V₁/V₂

Where;

P₁ = 100 cm = The pressure in volume V₁ = 250 cc

P₂ₐ = The partial pressure of nitrogen in volume V₂ = 650 cc

Substituting, we have;

P₂ₐ = 100× 250/650 ≈ 38.462 cm

The pressure of the combined gas, P, when the stopcock is opened is opened is given by Dalstons Law of partial pressure as P = P₂ₐ + P₂ₓ

Therefore, the pressure, P when the stopcock is opened is opened = 49.321 cm + 38.462 cm = 87.783 cm

3 0

1 year ago