Answer:
The area around the nucleus must be of low mass.
Explanation:
Rutherford`s experiment showed that there are some positive charges in the center of the atoms, and because they are all together, they will give a great mass to the atom.
It was quite different from Thomson`s experiment, in which it was thought that the negative charges were mixed with the positive charges, around the atom (like a Pudding Model). In Rutherford`s experiment, because the direction of beta particles, it was the prediction of the positive nucleus.
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Glucose and a plants and ur welcome
Explanation:
Z = atomic mass of the element and , A = atomic mass of the element .
a) Z = 11, A = 23
Element = Sodium
symbol: ²³₁₁Na .
b) Z = 28, A = 64
Element = Nickel
symbol: ⁶⁴₂₈Ni .
c) Z = 50, A = 115
Element = tin
symbol: ¹¹⁵₅₀Sn .
d) Z = 20, A = 42
Element = Calcium
symbol: ⁴²₂₀Ca .
Answer:
The molarity of the solution increases.
Explanation:
Molarity is the measure of the concentration of the solute in the solution. In this case, the solvent is the sugar solution and the solute is the sugar.
If sugar is ADDED to the already sugary solution, then there would be more sugar. Therefore, the sugar (solute) would increase in number.
This means that the answer is the third choice: The molarity of the solution increases.
The answer would not be the first or second choice because there isn't anything in the question that implies water. It just says sugar solution.
The answer is not the last choice because the sugar concentration does not decrease after you have added more sugar to it. It increases.
Answer:
a) 90 kg
b) 68.4 kg
c) 0 kg/L
Explanation:
Mass balance:
w is the mass flow
m is the mass of salt
v is the volume flow
C is the concentration
![-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)](https://tex.z-dn.net/?f=-%5Bln%282000L%2B3%2AL%2Fmin%2At%29-ln%282000L%29%5D%3Dln%28m%29-ln%2890kg%29)
![-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)](https://tex.z-dn.net/?f=-ln%5B%282000L%2B3%2AL%2Fmin%2At%29%2F2000L%5D%3Dln%28m%2F90kg%29)
![m=90kg*[2000L/(2000L+3*L/min*t)]](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2At%29%5D)
a) Initially: t=0
![m=90kg*[2000L/(2000L+3*L/min*0)]=90kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A0%29%5D%3D90kg)
b) t=210 min (3.5 hr)
![m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A210min%29%5D%3D68.4kg)
c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.
