Hello!
First you need to calculate q
<span>delta U is change in internal energy </span>
<span>delta U = q + w </span>
<span>q is heat and w work done </span>
<span>here work was done by the system means energy leaving the system so w is negative </span>
<span>delta U = q + w </span>
<span>q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ </span>
<span>q = m x c x delta T </span>
<span>7211 J = 80.0 g x c x (225-25) °C </span>
<span>c = 0.451 J /g °C
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Hope this Helps! Have A Wonderful Day! :)
You can search that up online it’s not that hard but good luck !!
Answer : Option 1) nuclei of 
and nuclei of 
only.
Explanation : Radiation is spontaneously emitted from nuclei of 
because this isotope of hydrogen is highly radioactive as compared to other isotopes of hydrogen namely; nuclei of and nuclei of .
They have much stable nucleus as compared to nuclei of .
The more it is unstable the more radiations will be emitted from its nucleus.
Molar mass :
NaBr = 103 g/mol
Pb(NO3)2 = 331.20 g/mol
<span><span /><span>Balanced chemical equation :
</span></span>2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr<span>2
</span><span>
2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
g NaBr -------------------> 311 g Pb(NO3)2
331.20 g = 2*103*311
331.20 g = 64066
mass ( NaBr ) = 64066 / 331.20
mass ( naBr) = 193,43 g of NaBr
hope this helps!.
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