Answer:
Explanation:
A substance that produces an excess of hydroxide ion (-OH) in aqueous solution.
This is an arrhenius Base
According to the arrhenius theory, a base is a substance that combines with water to produce excess hydroxide ions, OH⁻ in an aqeous solution. Examples are :
- Sodium hydroxide NaOH
- Potassium hydroxide KOH
A substance that produces an excess of hydrogen ion (H+) in aqueous solution
This is an arrhenius Acid
An arrhenius acid is a substance that reacts with water to produce excess hydrogen ions in aqueous solutions.
Examples are;
- Hydrochloric acid HCl
- Hydroiodic acid HI
- Hydrobromic acid HBr
We are given with a compound, Zinc (Zn) having a 1.7 x 10
^23 atoms. We are tasked to solve for it's corresponding mass in g. We need to
find first the molecular weight of Zinc, that is
Zn= 65.38 g/mol
Not that 1 mol=6.022x10^{23} atoms, hence,
1.7 x 10 ^23 atoms x 1 mol/6.022x10^{23} atoms x65.38
g/ 1mol
=18.456 g of Zn
Therefore, the mass of Zinc 18.456 g
Answer:
The new pressure of the pump is 26.05 atm or 2639.4 kPa
Explanation:
Step 1: Data given
Volume of the bicycle tire pump = 252 mL = 0.252 L
Pressure of air = 995 kPa = 9.81989 atm
The volume of the pump is reduced to 95.0 mL = 0.095 L
Step 2: Calculate the new pressure
V1*P1 = V2*P2
⇒with V1 = the initial volume of the bicycle tire pump = 0.252 L
⇒with P1 = the initial pressure of the pump = 9.81989 atm = 995 kPa
⇒with V2 = the reduced volume of the pump = 0.095 L
⇒with P2 = the new pressure = TO BE DETERMINED
0.252 L * 9.81989 atm = 0.095 L * P2
P2 = 26.05 atm
The new pressure is 26.05 atm
OR
0.252 L * 995 = 0.095 L * P2
P2 = 2639.4 kPa
The new pressure of the pump is 26.05 atm or 2639.4 kPa
Answer:
1HydrogenH2HeliumHe3LithiumLi4BerylliumBe5BoronB6CarbonC7NitrogenN8OxygenO9FluorineF10NeonNe11SodiumNa12MagnesiumMg13AluminiumAl14SiliconSi15PhosphorusP16SulfurS17ChlorineCl18ArgonAr19PotassiumK20CalciumCa
70.0 g. The decomposition of 125 g CaCO3 produces 700 g CaO.
MM = 100.09 56.08
CaCO3 → CaO + CO2
Mass 125 g
a) Moles of CaCO3 = 125 g CaCO3 x (1 mol CaCO3/100.09 g CaCO3)
= 1.249 mol CaCO3
b) Moles of CaO = 1.249 mol CaCO3 x (1 mol CaO/1 mol CaCO3)
= 1.249 mol CaO
c) Mass of CaO = 1.249 mol CaO x (56.08 g CaO/1 mol CaO) = 70.0 g
