Answer:
2gcm⁻³
Explanation:
Given parameters:
Mass of the object = 20g
Volume = 10mL
Unknown:
Density of the object = ?
Solution:
Density of a body is its mass per unit volume;
Density = 
We need to convert mL to cm³ for the volume
1mL = 1cm³;
10mL is therefore, 10cm³
So;
Density = 
= 2gcm⁻³
Answer:
D
Explanation:
Because the silk took away the electrons to allow the rod to become positive. They cant be destroyed because of the law of conservation of charges. Also protons cannot move only electrons can.
Answer:
When nitric acid combine with sodium hydroxide the salt formed is called sodium nitrate. option B
Explanation:
It is the strong acid strong base reaction. When acid and base react with each other salt and water are formed.
In given reaction nitric acid combine with sodium hydroxide base and form sodium nitrate salt and water.
Chemical equation:
HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H₂O(l)
Ionic equation:
H⁺NO₃⁻(aq) + Na⁺OH⁻(aq) → Na⁺NO₃⁻(aq) + H₂O(l)
Net ionic equation:
H⁺(aq) + OH⁻(aq) → H₂O(l)
The Na⁺(aq) and NO₃⁻(aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
16.4 grams is the mass of solute in a 500 mL solution of 0.200 M
.
sodium phosphate
Explanation:
Given data about sodium phosphate
atomic mass of Na3PO4 = 164 grams/mole
volume of the solution = 500 ml or 0.5 litres
molarity of sodium phosphate solution = 0.200 M
The formula for molarity will be used here to know the mass dissolved in the given volume of the solution:
The formula is
molarity = 
putting the values in the equation, we get
molarity x volume = number of moles
0.200 X 0.5= number of moles
number of moles = 0.1 moles
Atomic mass x number of moles = mass
putting the values in the above equation
164 x 0.1 = 16.4 grams
16.4 grams of sodium phosphate is present in 0.5 L of the solution to make a 0.2 M solution.
