You would get four moles of magnesium nitrate :) you would have to
“ ?molesmg(oh)2 = 8molmg(no3)2 x molmg(oh)2 / 2molhno3 = 4 moles of magnesium nitrate :))) hopefully this helps! <3
For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.
For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.
Answer: The kilograms of water must evaporate from 8kg of a 25% salt solution to produce 40% salt solution is 3 kg.
Explanation:
According to the ratio and proportion:
where,
= concentration of ist solution = 25%
= mass of ist solution = 8 kg
= concentration of second solution = 40%
= mass of second solution = ? kg
Thus the final solution must have a mass of 5 kg , i.e (8-5)= 3 kg of mass must be evaporated.
Therefore, the mass that must be evaporated from 8kg of a 25% salt solution to produce 40% salt solution is 3 kg.
Answer:
190 °C
Step-by-step explanation:
The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:
V₁/T₁ = V₂/T₂ Invert both sides of the equation.
T₁/V₁ = T₂/V₂ Multiply each side by V₂
T₂ = T₁ × V₂/V₁
=====
V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K
V₂ = 5.6 L; T₂ = ?
=====
T₂ = 310.15 × 5.6/3.75
T₂ = 310.15 × 1.49
T₂ = 463 K
t₂ = 463 – 273.15
t₂ = 190 °C
