<h3>
Answer:</h3>
19.3 g/cm³
<h3>
Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
Answer:
H₃PO₄/H₂PO₄⁻ and HCO₃⁻/CO₃²⁻
Explanation:
An acid is a proton donor; a base is a proton acceptor.
Thus, H₃PO₄ is the acid, because it donates a proton to the carbonate ion.
CO₃²⁻ is the base, because it accepts a proton from the phosphoric acid.
The conjugate base is what's left after the acid has given up its proton.
The conjugate acid is what's formed when the base has accepted a proton.
H₃PO₄/H₂PO₄⁻ make one conjugate acid/base pair, and HCO₃⁻/CO₃²⁻ are the other conjugate acid/base pair.
H₃PO₄ + CO₃²⁻ ⇌ H₂PO₄⁻ + HCO₃⁻
acid base conj. conj.
base acid
Answer:
The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³
(Option C)
Explanation:
Given;
concentration of HA, 
= 6.0mol/dm³
volume of HA, 
= 25.0cm³, = 0.025dm³
Concentration of HB, 
= 3.0mol/dm³
volume of HB, 
= 45.0cm³ = 0.045dm³
To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;
where;
is initial concentration
is initial volume
is final concentration of the solution
is final volume of the solution
Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³
The molar masses of H2S and NH3 are 34 and 17 g/mol, respectively. The equation that would best represent the given is,
Rate A/Rate B = √(molar mass B/molar mass A)
Substituting,
x/77 = √(17 /34 )
x = 54.4
Thus, it will take 54.4 seconds for NH3 to travel through the container.
Answer:
The air pressure in the ears increases
The volume of air in the ears increases
The change in volume causes discomfort
It takes time for the ears to dispell excess air past the ear drum.
Explanation:
As the plane engages in a steep incline into the atmosphere, the outside atmospheric pressure decreases with altitude. The air pressure in the ear, therefore, become greater than atmospheric pressure. The air volume in the ear therefore grows and pushes on the ear causing discomfort. As the air in the cabin pressurizes the discomfort eases away as pressure equalization is restored relative to the ear.
