When an electron in an atom moves from a higher, less stable level, down to a lower, more stable level, a photon is

An experiment shows that a 250 ?ml gas sample has a mass of 0.430 g at a pressure of 736 mmhg and a temperature of 28 ?c.

dmitriy555 [2]

What we're looking for here is the gas sample's molar mass given its mass, pressure, volume, and temperature. Recalling the gas law, we have

$PV = nRT$ or
$n = \frac{PV}{RT}$

where R is <span>0.08206 L atm / mol K, P is the given pressure, T is the temperature, and V is the volume.

Before applying the values given, it is important to make sure that they are to be converted to have consistent units with that of R.
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Thus, we have

P = 736/ 729 = 0.968 atm
T = 28 + 273.15 = 301.15 K
V = 250/1000 = 0.250 L

Now, applying these converted values into the gas law, we have

$n = \frac{(0.968 atm)(0.250 L)}{(0.08206 L.atm/mol.K)(301.15 K)}$
$n = 0.00979 moles$

Given that the mass of the sample is 0.430 g, we have

$molar mass = \frac{mass}{number of moles}$
$molar mass = \frac{0.430}{0.00979} = 43.9$

Thus, the gas sample has a molar mass of 43.9 g/mol.

4 0

3 years ago

Which group of elements will readily lose one electron to meet the octet rule? Alkaline Earth Metals c. Halogens Alkali Metals d

vova2212 [387]

Answer:

Alkali metals

Explanation:

The alkali metals are a group of metallic elements which are present in the first group of the periodic table. In other words, they are present in group 1 of the periodic table. These elements have one electron in their valence shell, the reason why they are placed in group 1.

They ionize by losing one electron to achieve the configuration of the nearest Noble gas or inert gas. Because they need to offset only one electron in their outermost shell, they are very chemically reactive and hence rarely occur in the free state.

Examples of elements in this group include lithium, potassium and sodium. They each have one electron only in their outermost shells.

4 0

3 years ago

A six-carbon organic compound containing oxygen is suspected of being either a secondary alcohol or a ketone. which chemical or

White raven [17]

Following are the possible isomers of secondary alcohol and ketones for six carbon molecules. In order to distinguish between sec. alcohol and ketone we can simply treat the unknown compound with acidified Potassium Dichromate (VI) in the presence of acid. If with treatment with unknown compound the colour of K2Cr2O7 (potassium dichromate VI) changes from orange to green then it is confirmed that the unknown compound is sec. alcohol, or if no change in colour is detected then ketone is confirmed. This is because ketone can not be further oxidized while, sec. alcohol can be oxidized to ketones as shown below,

6 0

3 years ago

The standard enthalpy of formation of BrCl(g) is 14.7 kJmol-1 . The standard enthalpies for the atomization of Br2(l) and Cl2(g)

natita [175]

Explanation:

Equation of the reaction:

Br2(l) + Cl2(g) --> 2BrCl(g)

The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.

The standard enthalpy change of formation for a compound,

ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.

This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction

1/2Br2(g) + 1/2Cl2(g) → BrCl(g)

Here, ΔH°rxn = ΔH°f

This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl

Using Hess' law,

ΔH°f = total energy of reactant - total energy of product

= (1/2 * (+112) + 1/2 * (+121)) - 14.7

= 101.8 kJ/mol

ΔH°rxn = 101.8 kJ/mol.

8 0

3 years ago

How to do an equation

pav-90 [236]

What do you mean by that Oh wait do you mean like 2×2=4 or 2+2=4 something like that maby?? I am still confused?? XD

7 0

3 years ago