Answer:
Index fossil, any animal or plant preserved in the rock record of the Earth that is characteristic of a particular span of geologic time or environment.
Explanation:
True
Answer:
The volume of carbon dioxide gas generated 468 mL.
Explanation:
The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%
Mass of tablet = 3.45 g
Mass of bicarbonate =
Moles of bicarbonate ion = 

According to reaction, 1 mole of bicarbonate ion gives with 1 mole of carbon dioxide gas , then 0.01840 mole of bicarbonate ion will give:
of carbon dioxide gas
Moles of carbon dioxide gas n = 0.01840 mol
Pressure of the carbon dioxide gas = P = 1.00 atm
Temperature of the carbon dioxide gas = T = 37°C = 37+273 K=310 K
Volume of the carbon dioxide gas = V
(ideal gas equation)

1 L = 1000 mL
0.468 L =0.468 × 1000 mL = 468 mL
The volume of carbon dioxide gas generated 468 mL.
Answer: OB
Explanation:
The empirical formula of an element is the simplest formula of an element which shows the ratios of atoms of the each element present in a molecule of a compound .
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