Answer:
14 ml of water
Explanation:
To find the volume you need to dilute the concentration of a solution, you should use the formula C1 x V1 = C2 x V2 in which:
C1 = initial concentration ( in this case 60 %)
V1 = initial volume ( in this case 70 ml)
C2 = Final concentration ( you want to dilute until 50 %)
V2 = final volume ( the variable you want to search)
So you need to:
1.- Isolate the variable you want to find: V2 = (C1 x V1) / C2
2.- Substitute data: V2 = (60% x 70 ml) /50 %
3.- You do the math, in this case is 84 ml.
4.- Remember that you have a initial volume of 70 ml, so the difference (84 ml - 70 ml = 14 ml) is the volume you need to add to dilute your solution.
Answer:
Six electrons are transferred in the formation of Al₂O₃.
Explanation:
Aluminium metal and Oxygen react to form Al₂O₃ as,
2 Al + 3/2 O₂ → Al₂O₃
Oxidation number of Al on left hand side is zero, while than on right hand side in Al₂O₃ is +3. Means it has lost 3 electrons per one atom and six electrons per two atoms. Also, the oxidation number of O at left hand side in O₂ is zero, while that in Al₂O₃ it is -2 per atom and -6 per 3 atoms.
So, two Al atoms have lost 6 electrons and 3 O atoms have gained six electrons.
Answer:
with ions of calcium in a cyclotron. This produced four atoms of moscovium
Explanation:
Data analysis is a process of inspecting, cleansing, transforming and modeling data with the goal of discovering useful information, informing conclusions and supporting decision-making
Answer:
The flow rate of a tube is the volume of fluid flowing through the tube per unit time. The flowrate is proportional to the product of the velocity of the fluid through the tube, and the cross-sectional area of the tube.
That is
Q = AV
where
A is the area of the tube
V is the velocity of the tube
The cross-sectional area of the tube is proportional to the radius of the tube. From the above equation, we can deduce that if the velocity of the fluid flowing through the tube is held constant, the flowrate of the fluid through the tube will increase with an increase in the radius of the tube, and it will decrease with a decrease in the radius of the tube.
