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Aleks04 [339]
2 years ago
13

Element 'X' forms a chloride with the formula XCl2, which is a solid with high melting point. X would most likely be in the same

group of the periodic table as:
I) Si
2) Al
3) Mg
4) Na



pls pls answer fast ​
Chemistry
1 answer:
notka56 [123]2 years ago
8 0

Answer:

Mg

Explanation:

Mg forms 2+ ions and mgcl2 has a melting point over 700°C I believe.

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7 0
3 years ago
What is the volume of 14.0g of nitrogen gas at STP?
lozanna [386]

Answer:

  • <em>The volume of 14.0 g of nitrogen gas at STP is </em><u><em>11.2 liter.</em></u>

Explanation:

STP stands for standard pressure and temperature.

The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:

  •   Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).

  •    After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).

Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1)  of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.

With the later definition, the volume of a mol of gas at STP is 22.7 liter.

I will use the traditional measure of 22.4 liter per mole of gas.

<u>1) Convert 14.0 g of nitrogen gas to number of moles:</u>

  • n = mass in grams / molar mass
  • Atomic mass of nitrogen: 14.0 g/mol
  • Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol
  • n = 14.0 g / 28.0 g/mol = 0.500 mol

<u>2) Set a proportion to calculate the volume of nitrogen gas:</u>

  • 22.4 liter / mol = x  / 0.500 mol
  • Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.

<u>Conclusion:</u> the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.

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The generalized rate expression may be written as:
r = k[A]ᵃ[B]ᵇ

We may determine the order with respect to B by observing the change in rate when the concentration of B is changed. This can be done by comparing the first two runs of the experiment, where the concentration of A is constant but the concentration of B is doubled. Upon doubling the concentration of B, we see that the rate also doubles. Therefore, the order with respect to concentration of B is 1.
The same can be done to determine the concentration with respect to A. The rate increases 4 times between the second and third trial in which the concentration of B is constant, but that of A is doubled. We find that the order with respect to is 2. The rate expression is:

r = k[A]²[B]
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