Mass of the shells will remain constant.
Explanation:
Sedimentary rocks are formed by the rocks that have been weathered. The microscopic shells, small pieces of rocks and mineral particles get deposited on this weathered sedimentary rocks.
The texture or chemical properties of the microscopic shell get changed. but the amount of microscopic shell deposited to make sedimentary rocks remains the same.
Chalk and coral reefs are the example of sedimentary rocks made up with microscopic shells.
The skeletal part of the shells takes part in rock formation whose mass remains unchanged although physical and chemical properties go changes.
Answer:
60
Explanation:
The formula of the compound is given as:
HC₂H₃O₂
To find the formula mass we sum up the atomic masses of the component elements;
Atomic mass:
H = 1
C = 12
O = 16
HC₂H₃O₂ = 1 + 2(12) + 3(1) + 2(16)
= 1 + 24 + 3 + 32
= 60
Answer:
C. An atom consists of positively charged matter that contains negatively charged particles.
Explanation:
Answer:fH = - 3,255.7 kJ/mol
Explanation:Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
Answer:
Amount of mercury is 1.0*10⁻⁵ g
Explanation:
<u>Given:</u>
Mercury content of stream = 0.68 ppb
volume of water = 15.0 L
Density of water = 0.998 g/L
<u>To determine:</u>
Amount of mercury in 15.0 L of water
<u>Calculation:</u>
where 1 μg (micro gram) = 10⁻⁶ g
0.68 ppm implies that there is 0.68 *10⁻⁶ g mercury per Liter of water
Therefore, the amount of mercury in 15.0 L water would be:
