2. Calculate the boiling point of a 1.7 m ethylene glycol solution.

What volume will 12.0 g of oxygen gas occupy at 25 c and a pressure of 52.7 kpa?

Stolb23 [73]

We can use the ideal gas law equation to find the volume occupied by oxygen gas
PV = nRT
where ;
P - pressure - 52.7 kPa
V - volume
n - number of oxygen moles - 12.0 g / 32 g/mol = 0.375 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values in the equation
52 700 Pa x V = 0.375 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
V = 17.6 L
volume of the gas is 17.6 L

8 0

3 years ago

Which statement describes an eletron

Vitek1552 [10]

An electron carries a negative power to the atom. in each atom, the amount of protons and electrons are usually evened out so the atom is stable. the electrons of an atom are also found on the outside ring of an atom, unlike protons and neutrons, which are inside the atom. glad to help! :)

4 0

4 years ago

You’re sitting at a car race and the cars sound differently when they approach as opposed to when they are moving away. Why migh

bulgar [2K]

Answer:

I have no clue this qestion is kinda stumping but it a really good question it got me thinking but i don't know if there is a right answer

Explanation:

I don't know

5 0

2 years ago

When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot

viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the $'n^{th}'$ orbit of hydrogen atom.

Formula used :

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV$

Energy of n = 2 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

$E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV$

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

3 0

3 years ago

if the spin of one electron in an orbital is clockwise , what is the spin of the other electron in that orbital

nata0808 [166]

Here we have to get the spin of the other electron present in a orbital which already have an electron which has clockwise spin.

The electron will have anti-clockwise notation.

We know from the Pauli exclusion principle, no two electrons in an atom can have all the four quantum numbers i.e. principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s) same. The importance of the principle also restrict the possible number of electrons may be present in a particular orbital.

Let assume for an 1s orbital the possible values of four quantum numbers are n = 1, l = 0, m = 0 and s = $\frac{+}{-}$ $\frac{1}{2}$ .

The exclusion principle at once tells us that there may be only two unique sets of these quantum numbers:

1, 0, 0, + $\frac{1}{2}$ and 1, 0, 0, - $\frac{1}{2}$ .

Thus if one electron in an orbital has clockwise spin the other electron will must be have anti-clockwise spin.

6 0

3 years ago