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2 years ago

How many grams of nano3 would you add to 500g of h2o in order to prepare a solution that is 0.500 molal in nano3?

1 answer:

4 0

When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.

0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3

Since the molar mass of NaNO3 is 85 g/mol, the mass is

0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed

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At higher elevations, the boiling point of water decreases, due to the decrease in atmospheric pressure. As a result, what could

amid [387]

When we say decrease in boiling point, that means, we achieve boiling at a more lower temperature (lower than 100deg C). This is due to the lower atmospheric pressure. Boiling happens when the vapor pressure is equal the atmospheric pressure. Lower atmospheric pressure takes lower temperature for vapor pressure to equate with the atmospheric pressure. The answer here is letter B.
At higher elevations, it would take longer to hard boil an egg, because there is a lower boiling point, so the egg is boiling in water at a lower temperature.

8 0

1 year ago

Arranges the following molecules in order of increasing dipole moment: <br> H2O, H2S, H2Te, H2Se.

erastova [34]

Explanation:

Dipole moment is defined as the measurement of the separation of two opposite electrical charges.

$H_{2}O$ is a bent shaped molecule with a dipole moment of 1.87.

$H_{2}S$ is also a bent shaped molecule with a dipole moment of 1.10.

$H_{2}Te$ is a also a bent shaped molecule and has a negligible dipole moment.

$H_{2}Se$ has a dipole moment of 0.29.

Therefore, given molecules are arranged according to their increasing dipole moment as follows.

$H_{2}Te$ < $H_{2}Se$ < $H_{2}S$ < $H_{2}O$

7 0

2 years ago

Identify the Lewis acids and Lewis bases in the following reactions:

postnew [5]

Answer: 1. $H^++OH^-\rightarrow H_2O$ Lewis acid : $H^+$ , Lewis base : $OH^-$

2. $Cl^-+BCl_3\rightarrow BCl_4^-$ Lewis acid : $BCl_3$ , Lewis base : $Cl^-$

3. $K^++6H_2O\rightarrow K(H_2O)_6$ Lewis acid : $K^+$ , Lewis base : $H_2O$

Explanation:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1. $H^++OH^-\rightarrow H_2O$

As $H^+$ gained electrons to complete its octet. Thus it acts as lewis acid. $OH^-$ acts as lewis base as it donates lone pair of electrons to electron deficient specie $H^+$ .

2. $Cl^-+BCl_3\rightarrow BCl_4^-$

As $BCl_3$ is short of two electrons to complete its octet. Thus it acts as lewis acid. $Cl^-$ acts as lewis base as it donates lone pair of electrons to electron deficient specie $BCl_3$ .

3. $K^++6H_2O\rightarrow K(H_2O)_6$

As $K^+$ is short of electrons to complete its octet. Thus it acts as lewis acid. $H_2O$ acts as lewis base as it donates lone pair of electrons to electron deficient specie $K^+$ .

8 0

2 years ago

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

8 0

1 year ago

According to the Bohr model, the energy of the hydrogen atom is given by the equation: E = (-21.7 x 10 -19 J)/ n 2 Calculate the

Anit [1.1K]

Answer:

91.6 nm

Explanation:

The energy of the hydrogen atom can be calculated by the emission of a photon. When an electron is excited it goes from to the next energetic level, and when it returns to its ground state, it emits a photon. Hydrogen has only one electron, which is at the level n = 1. So, the equation is given:

E = (-21.7x10⁻¹⁹J)/1²

E = -21.7x10⁻¹⁹J

The energy of the photon is the energy absorbed, and because of that is positive (the opposite of the energy released by the electron). This energy can be calculated by:

E = h*c/λ

Where h is the Planck's constant (6.626x10⁻³⁴ J.s), c is the speed of the light (3.00x10⁸ m/s), and λ is the wavelength of the photon.

21.7x10⁻¹⁹ = 6.626x10⁻³⁴ * 3.00x10⁸/λ

λ = 9.16x10⁻⁸ m

λ = 91.6 nm

7 0

2 years ago