In the pot of tea the molecules would be moving faster than in the cooled cup of tea. As liquid is heated the atoms vibrate faster which increases the distance between them. When heat leaves a substance, the molecules vibrate slower and get closer.
Answer:
The volume of water to be added is 0.175 liters of water
Explanation:
The given concentration of the nitric acid = 55% (M/M)
The mass of the nitric acid solution = 100 gm
The concentration solution is to diluted to = 20% (M/M)
The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution
Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get
Let "x" represent the volume of the resulting solution, we have;
20% of x = 55 g of nitric acid
∴ 20/100 × x = 55 g
x = 55 g × 100/20 = 275 g
The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid
The mass of extra water to be added = 275 g - 100 g = 175 g
Volume = Mass/Density
The density of water ≈ 1 g/ml
∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l
The volume of water to be added = 0.175 liters of water.
Answer:
<u>2</u> Fe + <u>3</u> H2SO4 —> Fe2(SO4)3 + <u>3</u> H2
I hope I helped you^_^
Answer:
Concentration of the barium ions = ![[Ba^{2+}] = 0.4654 M](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%20%3D%200.4654%20M)
Concentration of the chloride ions = ![[Cl^{-}]=0.9308 M](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%5D%3D0.9308%20M%20)
Explanation:
Moles of hydrogen chloride = n
Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L
Molarity of the hydrogen chloride = 0.1355 M
According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.
Then 0.05947 moles of HCl will react with:
barium hydroxide
Moles of barium hydroxide = 0.029735 mol
1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:
Volume of solution after neutralization reaction :
= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L
Concentration of the barium ions =![[Ba^{2+}]](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D)
![[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%3D%5Cfrac%7B0.029735%20mol%7D%7B0.06389%20L%7D%3D0.4654%20M)
1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.
Then concentration of chloride ions will be:
![[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D2%5Ctimes%20%5BBa%5E%7B2%2B%7D%5D%3D2%5Ctimes%200.4654%20M%3D0.9308%20M)
They are examples of elements.
