The correct answer is 2.53 g of precipitate, BaCrO4.
Answer:
5.758 is the density of the metal ingot in grams per cubic centimeter.
Explanation:
1) Mass of pycnometer = M = 27.60 g
Mass of pycnometer with water ,m= 45.65 g
Density of water at 20 °C = d =
1 kg = 1000 g


Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g
Volume of pycnometer = Volume of water present in it = V


2) Mass of metal , water and pycnometer = 56.83 g
Mass of metal,M' = 9.5 g
Mass of water when metal and water are together ,m''= 56.83 g - M'- M
56.83 g - 9.5 g - 27.60 g = 19.7 g
Volume of water when metal and water are together = v

Density of metal = d'
Volume of metal = v' =
Difference in volume will give volume of metal ingot.
v' = v - V


Since volume cannot be in negative .
Density of the metal =d'
=
Hey there!
Solid Sodium and water will react spontaneously and release energy. This is based on the reactivity series. Sodium is a highly reactive metal and hence, it is placed at the top of the reactivity series. This is because it loses its outermost electron very readily. When it comes in contact with water, it reacts with it violently to form sodium hydroxide and hydrogen gas. This reaction is exothermic and hence, accompanied with a release of energy. Gold lies at the bottom of the reactivity series as it is very stable and does not give away its outermost electrons easily. Therefore, when it comes in contact with water, there is no reaction and no release of energy.
Answer:
Osmosis
Explanation:
The osmosis is a movement from somewhere with a lower concentration of a substance to somewhere with a higher concentration of the same substance. Against the concentration gradient. It occurs through a semi permeable membrane which prevents a non desired substance cross it. Also, it's very important to biological process.
Answer:
0.559 L
Explanation:
Step 1: Given data
- Moles of sodium iodide (n): 0.405 mol
- Molar concentration of sodium iodide (M): 0.724 M (0.724 mol/L)
Step 2: Calculate the volume of solution (V)
The molarity is equal to the moles of solute divided by the liters of solution.
M = n/V
V = n/M
V = 0.405 mol/(0.724 mol/L) = 0.559 L