round up given mass to a whole number



Hence our answer is Rhodium-103
<span>This can be solved using basic algebra. X equals the number of liters of 2% brine to add. Therefore we use this equation. 3 + 0.02x = 2.4 + 0.08x 0.6 = 0.6x x = 0..6/0.06 x = 10</span>
Answer:
The volume of solution in liters required to make a 0.250 M solution from 3.52 moles of solute is 14.08 liters of solution
Explanation:
The question relates to the definition of the concentration of a solution which is the number of moles per liter (1 liter = 1 dm³) of solution
Therefore we have;
The concentration of the intended solution = 0.250 M
Therefore, the number of moles per liter of the required resolution = 0.250 moles
Therefore, the concentration of the required solution = 0.250 moles/liter
The volume in liters of the required solution that will have 3.52 moles of the solute is given as follows;
The required volume of solution = The number of moles of the solute/(The concentration of the solution)
∴ The required volume of solution = 3.52 moles/(0.250 moles/liter) = 14.08 liters
The required volume of solution to make a 0.250 M solution from 3.52 moles of solute = 14.08 liters.
Therefore the number of liters required to make a 0.250 M solution from 3.52 moles of solute = 14.08 liters.
Answers:
A) 2040 kg/m³; B) 58 600 km
Explanation:
A) Density


<em>B) Radius</em>



![r= \sqrt [3]{ \frac{3V }{4 \pi } }](https://tex.z-dn.net/?f=r%3D%20%5Csqrt%20%5B3%5D%7B%20%5Cfrac%7B3V%20%7D%7B4%20%5Cpi%20%7D%20%7D)
![r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}](https://tex.z-dn.net/?f=r%3D%20%5Csqrt%20%5B3%5D%7B%20%5Cfrac%7B3%5Ctimes%208.268%20%5Ctimes%2010%5E%7B23%7D%20%5Ctext%7B%20m%7D%5E%7B3%7D%7D%7B4%20%5Cpi%20%7D%20%7D%3D%20%5Csqrt%20%5B3%5D%7B%201.974%20%5Ctimes%2010%5E%7B23%7D%20%5Ctext%7B%20m%7D%5E%7B3%7D%7D%3D%205.82%20%5Ctimes%2010%5E%7B7%7D%20%5Ctext%7B%20m%7D%3D%5Ctext%7B58%20200%20km%7D)