Answer:
Sedimentary rocks are the product of 1) weathering of preexisting rocks, 2) transport of the weathering products, 3) deposition of the material, followed by 4) compaction, and 5) cementation of the sediment to form a rock. The latter two steps are called lithification.
It is a binary direct bandgap semiconductor commonly used in light-emitting diodes since the 1990s
Answer:
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Explanation:
Mass of ethylene glycol = m = 100 g
Specific heat capacity of ethylene glycol = c = 3.5 J/g°C
Change in temperature of ethylene glycol = ΔT
Heat loss by the ethylene glycol = Q = 350 J


ΔT = 1°C
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Answer:
0.0611M of HNO3
Explanation:
<em>The concentration of the NaOH solution must be 0.1198M</em>
<em />
The reaction of NaOH with HNO3 is:
NaOH + HNO3 → NaNO3 + H2O
<em>1 mole of NaOH reacts per mole of HNO3.</em>
That means the moles of NaOH used in the titration are equal to moles of HNO3.
<em>Moles HNO3:</em>
12.75mL = 0.01275L * (0.1198mol / L) = 0.0015274 moles NaOH = Moles HNO3.
In 25.00mL = 0.025L -The volume of the aliquot-:
0.00153 moles HNO3 / 0.025L =
<h3> 0.0611M of HNO3</h3>
Answer:
Supersaturated
Explanation:
The tea has absorbed and dissolved as much sugar as it could. If there is sugar left at the bottom, it means the solution is supersaturated because it can't absorb any more.