42.4 ml is the volume in milliliters of the lead ball if a lead ball is added to a graduated cylinder containing 50.6 ml of water.
<h3>What is a graduated cylinder?</h3>
A tall narrow container with a volume scale is used especially for measuring liquids.
The graduated cylinder contains water
mL is a volume unit.
Water volume = 50.6 ml
The lead ball caused an increase in volume from 50.6 ml to 93.0 mL.
The new volume is the lead ball volume plus the original water volume :
Final volume = Vlead ball+ Water original volume
Hence, 42.4 ml is the volume in milliliters of the lead ball.
Learn more about the graduated cylinder here:
brainly.com/question/13386106
#SPJ1
Answer:
V = 552 mL or 0.552 L
Explanation:
First, we need to calculate the number of moles of H2 using the ideal gas equation which is:
PV = nRT
Solving for n:
n = PV / RT
Where:
P = Pressure
V = Volume
R = Gas constant (0.082 L atm / K mol)
T = Temperature in K
Let's convert first both pressure in atm, remember that 1 atm = 760 mmHg
P = 735 / 760 = 0.967 atm
Pwater = 21 / 760 = 0.028 atm
Finally temperature to Kelvin:
T = 23 + 273.15 = 296.15 K
Now, at first the hydrogen was collected by water displacement so pressure is:
P = 0.967 - 0.028 = 0.939 atm
Now the moles of hydrogen:
n = 0.939 * 0.568 / 0.082 * 296.15
n = 0.022 moles
Now that we have the moles, let's calculate the volume when the pressure is 735 mmHg
V = nRT/P
V = 0.022 * 0.082 * 296.15 / 0.967
V = 0.552 L or 552 mL
This is the volume that hydrogen occupies.
Taking into account the reaction stoichiometry, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
4 Al + 3 O₂ → 2 Al₂O₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al: 4 moles
- O₂: 3 moles
- Al₂O₃: 2 moles
The molar mass of the compounds is:
- Al: 27 g/mole
- O₂: 32 g/mole
- Al₂O₃: 102 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al: 4 moles ×27 g/mole= 108 grams
- O₂: 3 moles ×32 g/mole= 96 grams
- Al₂O₃: 2 moles ×102 g/mole= 204 grams
<h3>Mass of Al₂O₃ formed</h3>
The following rule of three can be applied: if by reaction stoichiometry 96 grams of O₂ form 204 grams of Al₂O₃, 48 grams of O₂ form how much mass of Al₂O₃?
<u><em>mass of Al₂O₃= 102 grams</em></u>
Finally, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.
Learn more about the reaction stoichiometry:
brainly.com/question/24741074
brainly.com/question/24653699
#SPJ1
B.
<span>genotype; phenotype
</span><span>Genetic variation determines inherited differences between individuals . Our height or eye color are inherited from our parents, but our phenotype is also affected by environment such as the food we eat (diet), drugs we take, toxins surrounding us, climate, location, culture, physical accidents and lifestyle.</span>
KOH is a strong base, so [OH-] = 0.10 M = 1.0 x 10^-1 M
[H+][OH-] = Kw
Kw = 1.0 x 10^-14
[H+] = 1.0 x 10^-14 / 1.0 x 10^-1 = 1.0 x 10^-13 M
