Answer:
protons+neutrons = atomic mass
11+12 = 23
so the element which has an atomic mass of 23 is
<h2>Sodium (Na)</h2>
Answer:
The process of dissolving can be endothermic (temperature goes down) or exothermic (temperature goes up).
When water dissolves a substance, the water molecules attract and “bond” to the particles (molecules or ions) of the substance causing the particles to separate from each other.
The “bond” that a water molecule makes is not a covalent or ionic bond. It is a strong attraction caused by water’s polarity.
It takes energy to break the bonds between the molecules or ions of the solute.
Energy is released when water molecules bond to the solute molecules or ions.
If it takes more energy to separate the particles of the solute than is released when the water molecules bond to the particles, then the temperature goes down (endothermic).
If it takes less energy to separate the particles of the solute than is released when the water molecules bond to the particles, then the temperature goes up (exothermic).
Explanation:
1 mol of Carbon = 12 grams.
x mol of Carbon = 55 grams
12*x = 1 * 55
x = 55/12
x = 4.583333 mols of carbon
1 mol of anything is 6.02 * 10^23 atoms
4.58333333 mol = x
1/4.5833333 = 6.02 * 10^23/x
x = 4.58333* 6.02*10^23
x = 2.7591 * 10^23 Carbon atoms
Answer:
B
Explanation:
u do the math and you will get the answer
Answer: Thus 0.724 mol of
are needed to obtain 18.6 g of ![NH_3](https://tex.z-dn.net/?f=NH_3)
Explanation:
To calculate the moles :
![\text{Moles of} NH_3=\frac{18.6g}{17g/mol}=1.09moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%7D%20NH_3%3D%5Cfrac%7B18.6g%7D%7B17g%2Fmol%7D%3D1.09moles)
According to stoichiometry :
2 moles of
are produced by = 1 mole of ![CaCN_2](https://tex.z-dn.net/?f=CaCN_2)
Thus 1.09 moles of
will be produced by =
of ![CaCN_2](https://tex.z-dn.net/?f=CaCN_2)
But as yield of reaction is 75.6 %, the amount of
needed is =![\frac{0.545}{75.6}\times 100=0.724](https://tex.z-dn.net/?f=%5Cfrac%7B0.545%7D%7B75.6%7D%5Ctimes%20100%3D0.724)
Thus 0.724 mol of
are needed to obtain 18.6 g of ![NH_3](https://tex.z-dn.net/?f=NH_3)