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3 years ago

Calculate the grams of sodium hydroxide formed when 2.24 moles of sodium oxide react with water

1 answer:

6 0

The balanced reaction is
Na2O + H2O --> 2NaOH

If 2.24 moles of sodium oxide react, that means 4.48 moles of NaOH is formed as it is a 1 to 2 stoichiometric relationship.

Now we multiply by the molar mass to get grams.

4.48 moles NaOH * (39.997 grams/1 mole) = 179.2 grams

Your answer is 179. grams.

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The three beakers shown below contain solutions of [cof6]3–, [co(nh3)6]3+, and [co(cn)6]3–. based on the colors of the three sol

vitfil [10]

[Co(CN)₆]³⁻ → Yellow
[Co(NH₃)₆]³⁺ → Orange
[CoF₆]³⁻ → Blue
Explanation:
- All the given compounds have octahedral geometry but the ligand in each are different with the same metal ion.

- Ligands strength order: CN⁻ > NH₃ > F⁻

- The ligand CN will act as a strong field ligand so that the splitting is maximum when compared to NH₃ and F⁻

- If the splitting is more, the energy required for transition is more, and the wavelength is inversely proportional to energy.

- So CN complex will absorb at lower wavelength (yellow color)

3 0

3 years ago

When an electron in an atom spontaneously jumps from a higher energy state to a lower energy state, the atom?

vlabodo [156]

If an atom suffers from a collision, that causes an electron to jump from a lower to higher state, it is called collisional excitation

6 0

3 years ago

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

3 years ago

The daily production of carbon dioxide from an 780.0 mw coal-fired power plant is estimated to be 3.3480 x 104 tons (not metric)

valkas [14]

The production of $CO_{2}$ is $3.3480\times 10^{4} tons/day$ . Converting mass into kg,

1 ton=907.185 kg, thus,

$3.3480\times 10^{4} tons=3.037\times 10^{7} kg$

Thus, production of $CO_{2}$ will be $3.037\times 10^{7} kg/ day$ .

The specific volume of $CO_{2}$ is $0.0120 m^{3}/kg$ .

Volume of $CO_{2}$ produced per day can be calculated as:

$V=Specific volume\times mass$

Putting the values,

$V=0.0120 m^{3}/kg\times 3.037\times 10^{7} kg=364440 m^{3}/day$

Thus, volume of $CO_{2}$ produced per year will be:

$V=\frac{365 days}{1 year}(364440 m^{3}/day)=1.33\times 10^{8}m^{3}/year$

Thus, in 4 year volume of $CO_{2}$ produced will be:

$V=1.33\times 10^{8}m^{3}/year\times 4 years=5.32\times 10^{8}m^{3}$

8 0

2 years ago

Which property of isotopes makes them suitable for determining the dates of ancient artifacts? atomic number of the isotope rate

siniylev [52]

Each isotope has a unique rate of decay, making them suitable for determining the dates of ancient artifacts. The answer is "rate of decay of the isotope."

8 0

3 years ago