a M is a Group 2 melal that forms the nitrate M(NO_(3))_(2) 0320 g of M(NO_(3))_(2) is heated strongly and decomposes completely 2N(NO_(3))_(2)(s)arrow 2NO(s)+4NO_(2)(g)+O_(2)(g) The mixture of gases formed has a volume of 225cm^3 at 450^circ C and 101000Pa Determine the M, of M(NO_(3))_(2) 450+2+3=158 Identify M The gas constant, R=831JK^-1mol^-1 15 m

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a M is a Group 2 melal that forms the nitrate M(NO_(3))_(2) 0320 g of M(NO_(3))_(2) is heated strongly and decomposes completely 2N(NO_(3))_(2)(s)arrow 2NO(s)+4NO_(2)(g)+O_(2)(g) The mixture of gases formed has a volume of 225cm^3 at 450^circ C and 101000Pa Determine the M, of M(NO_(3))_(2) 450+2+3=158 Identify M The gas constant, R=831JK^-1mol^-1 15 m

Answer 1

Requires calculations using standard values and constants for different elements to determine \( M_{1_{M(NO3)2}} \) and then comparing \( M \) to the masses of Group II elements. High precision required. No absolute answer without calculations.

Answer 2

## Step 1: Identify the premise This is a chemistry problem involving the application of the gas law and basic stoichiometry. The premise is that 0.320g of M(NO3)2, unknown group 2 metal nitrate, is heated and decomposes into a mix of gasses. These gasses have a given volume, temperature, and pressure. We are tasked to find the molar mass \(M_1\) of M(NO3)2 and identify the \(M \) in the compound. ## Step 2: Use the ideal gas law First, we'll start with the application of the Ideal Gas Law \( PV = nRT \) which includes Pressure (P), Volume (V), number of moles of gas (n), ideal gas constant (R), and absolute temperature (T). ### The ideal gas law formula: ### \( PV = nRT \) We use this law to calculate the amount in moles for the mixture of the gases produced when M(NO3)2 is heated. In this case, P is 101000 Pa (pressure stated in pascal), V is 2.25 x 10^-4 m^3 (volume convert from cm^3 to m^3), R is 8.31 J • K−1 • mol −1, T is 723 K (temperature convert from Celsius to Kelvin), n is what we are looking to find. Rearranging: ### \( n = \frac{PV}{RT} \) ## Step 3: Stoichiometry to find moles of M(NO3)2 From the balanced equation in the question, we can tell that the moles of the gas produced is a result from one mole of M(NO3)2 decomposing into 4 mol of NO and 1 mol O2 (i.e., a total of 5 mol per 2 mol M(NO3)2 ). Utilizing stoichiometry we can determine the moles of M(NO3)2 from the moles of gasses. ### \(n_{_{M(NO3)2}} = \frac{n_{_{total gas}}}{2.5}\) Because each 2 mol of M(NO3)2 gives 5 mol of gas. ##Step 4: Calculate for molar mass After obtaining the moles of M(NO3)2 , we can then determined its molar mass by taken the mass and divide it by the moles calculated in the step above. ### \(M_{1_{M(NO3)2}} = \frac{mass_{_{M(NO3)2}}}{mol_{_{M(NO3)2}}}\) ## Step 5: Identify M once we have found the required atomic mass of M, Identify the element by comparing with the given periodic masses. The Element that is found serves as \( M \) in \( M(NO3)2 \).

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