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

or

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.
</span>
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


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

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