<span> Pb(NO3)2 + 2 NaBr ------> PbBr2 + 2 NaNO3
2 mol 2 mol
M(NaBr)= 102 g/mol
M(NaNO3) = 85 g/mol
1) 244g NaNO3 *1 mol NaNO3/85 g NaNO3 = 244/85 mol NaNO3
2)</span> Pb(NO3)2 + 2 NaBr ------> PbBr2 + 2 NaNO3
2 mol 2 mol<span>
x mol </span>244/85 mol
<span>
x=(2 mol*</span> 244/85 mol )/2 mol = 244/85 mol NaBr
<span>
3) </span> 244/85 mol NaBr*102g NaBr/1 mol = (244*102/85) g NaBr =292.8 g NaBr<span>
</span>
The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment
Answer:
Compound 2 (2,5-dimethylhexane) will not have the peaks at 29 and 85 m/z
Explanation:
The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.
The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.
Compound 2 (2,5-dimethylhexane) structure shows that the cleavage of C-C bond will not yield a stable ethyl and hexyl carbocation. Hence, no peaks will be observed at 29 and 85 m/z. The absence of these two peaks can be used to distinguish one isomer from the other.
We need to make sure that the pressure is in the same units.
let's convert the atm to mmHg. remember that 1 atm= 760 mmHg
1.03 atm (760 mmHg/ 1 atm)= 783 mmHg
to find the pressure of the gas, you just need to subtract the values
pressure of gas= 783 - 305 = 478 mmHg