Given:
No of atoms present= 8.022 x 10^23 atoms
Now we know that 1 mole= 6.022 x 10^23 atoms
Hence number of moles present in 8.022 x 10^23 atoms is calculated as below.
Number of moles
= 8.022 x 10^23/6.022x 10^23
=1.3 moles.
Hence we have 1.3 moles present.
Answer:
0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidised. ... Thus, 90g of lower oxide contains as much metal as 100g of higher oxide, i.e., 80g (given). Hence, 80g of metal combines with 10g of oxygen in the lower oxide and 20g of oxygen in the higher oxide.
If the density is higher than water than the object will sink
<span>Ernest Rutherford was the scientist that developed a model of the atom that looked like a nucleus surrounded by electrons.
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The heat from the hotter water will go into the colder water untl equilibrium is reached. Equilibrium is same temperature!
Now, the heat is proportional to the mass, the specific heat and the temperature difference. The specific heat does not matter since all is water, it will cancel out:
m_1 * c_H20 * ( T_final - T_1 ) = -m_2 * c_H20 * ( T_final - T_2)
Notice the minus, because one wins the heat of the one who loses it. In this way both sides have the same sign:
m_1*(T_final - T_1)=-m_2*(T_final-T_2), or after some simple algebra:
T_final = (m_1 * T_1 + m_2 * T_2 )/(m_1+m_2),
which looks like an arithmetic mean, and one could have gone for this, but the above shows all the work. Notice that if T_1=T_2, T_final=T_1 always, which makes sense.
Now you can convert volume to mass with the density, but since mass = density*volume and it is all water, the density will cancel out and you can work with volumes. If you prefer just say: 120 ml->120 g , etc ...
T_final = (120*95+320*25)/(320+120)=44.0909 degrees Celsius, or ~ 44.09 degrees with two decimal precision as your statement (beware of precision always!).
