Answer:
Temperature of the water
Explanation:
In every study, there must be independent and dependent variables. An independent variable is the variable that is changed in order to obtain a response. In this case, the temperature of the water is being changed, the response in this experiment is the respiration rate of the goldfish.
Thus the respiration rate of the goldfish is the dependent variable because it is controlled by the temperature of the water and changes accordingly.
Summarily, the independent variable is the temperature of the water while the dependent variable is the respiration rate of the goldfish.
The reaction between mercury (Hg) and sulfur (S) to form HgS is:
Hg + S ------------- HgS
Therefore: 1 mole of Hg reacts with 1 mole of S to form 1 mole of HgS
The given mass of Hg = 246 g
Atomic mass of Hg = 200.59 g/mol
# moles of Hg = 246 g/ 200.59 gmol-1 = 1.226 moles
Based on the reaction stoichiometry,
# moles of S that would react = 1.226 moles
Atomic mass of S = 32 g/mol
Therefore, mass of S = 1.226 moles*32 g/mole = 39.23 g
39.2 g of sulfur would be needed to react completely with 246 g of Hg to produce HgS
Answer:
A) 0.95 mol
Explanation:
We will assume the gas given off in the fermentation is an ideal gas because that allows us to use the ideal gas equation.
PV = nRT
First let's convert all measurements to units that we can use
P = 702 mmHg * 1 atm/760 mmHg = 0.92368 atm
V = 25.0 L
R = 0.08206 L-atm/mol-K
T = 22.5 °C +273.15 = 295.65 K
PV = nRT
0.92368 atm * 25.0 L = n * 0.08206 L-atm/mol-K * 295.65 K
n = 0.9518 mol
Answer:
25.42 atm
Explanation:
Data Given:
Volume of a gas ( V )= 2.00 L
temperature of a gas ( T ) = 310 K
number of moles (n) = 2 mol
Pressure of a gas ( P ) = to be find
Solution:
Formula to be used
PV= nRT
Rearrange the above formula
P = nRT / V . . . . . . . . . . (1)
Where R is ideal gas constant
R = 0.08205 L atm mol⁻¹ K⁻¹
Put values in equation 1
P = nRT / V
P = 2 mol x 0.08205 L atm mol⁻¹ K⁻¹ x 310 k / 2 L
P = 50.84 L atm / 2 L
P = 25.42 atm
P ressure of gas (P) will be = 25.42 atm
Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g
