Answer:
184.113 g/mol
Explanation: The atomic mass of Mg is 24.3 amu. The atomic mass of bromine is 79.9. Therefore, the formula weight of MgBr2 equals 24.3 amu + (2 × 79.9 amu), or 184.1 amu. Because a substance's molar mass has the same numerical value as its formula weight, the molar mass of MgBr2 equals 184.1 g/mol.
Primitive cells, for example, yeasts and microscopic organisms utilize all sugars a similar way, gave that they can transport the starch into the cell. Huge atoms that can't be transported are separated remotely by compounds that the cells discharge.
All dietary starch must be processed into little atoms previously assimilation. Those little particles are chiral, pivoting spellbound light to one side or left.
A) Nitrogen has an ATOMIC mass number of 14, but nitrogen gas consists of N₂ molecules, so the mass to use in this problem is 28 g/mol. Rates of effusion ∝ 1/√(mass), so
<span>√(mass unknown) /√28 = (rate N₂ effusion)/(rate unknown effusion) = 1.59 </span>
<span>∴ mass unknown = (1.59)²(28) = 70.78 g/mol </span>
<span>B) One possible gas that comes close for this mass is NF₃.</span>
A) Mass Number
B) Protons and Electrons
C) (I think group but I'm not 100% sure on this...)
D) 12 (To work out the number of neutrons you do MASS number - ATOMIC number aka the number of protons so in this case you'd do 23-11=12)
Hope this helps
Answer:

Explanation:
Given that:
Half life =
days
Where, k is rate constant
So,
The rate constant, k = 0.00007 days⁻¹
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Given:
5 % is lost which means that 0.05 of
is decomposed. So,
= 1 - 0.05 = 0.95
t = ?
t = 732.76 days = 