Answer: Limestone is a sedimentary rock.
Explanation:
1. Limestone is a carbonate sedimentary rock made up of calcite(CaCO₃).This rock type forms calcareous shells and tests of organisms that was deposited in a sedimentary basin.
Limestone is used in building and construction works. It also finds application in chemical industries.
2. Over a long period of time, we would take a look at the rock "limestone" through the rock cycle.
Limestone being a sedimentary rock would be converted to marble, a metamorphic rock if subjected to metamorphic conditions over an extensive period of time. With series of metamorphic transformation, marble can grade to higher metamorphic facies of rocks as it combines with other minerals in the crust. The minerals would eventually change and as the changed rock approaches its melting temperature, melt would result.
From the other spectrum, limestone can be weathered if subjected extensively to denudation forces such as wind, water and glaciation. Water is more potent for the chemical weathering of limestone. Limestone would easily and readily dissolve in it over a long period of time.
<span>significant figures: 7, decimals: 3</span>
C. If the banks of the river are eroding, then the river will be wider.
Moles of Zn present= 2.36/65.4= 0.0361 moles
Therefore maximum moles of ZnO= 0.0722
Mass of one mole of ZnO= 81.4
Mass of ZnO produced= 0.0722 x 81.4= 5.87g
Answer:
We need 7.5 mL of the 1M stock of NaCl
Explanation:
Data given:
Stock = 1M this means 1 mol/ L
A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL
Step 2: Calculate the volume of stock we need
The moles of solute will be constant
and n = M*V
M1*V1 = M2*V2
⇒ with M1 = the initial molair concentration = 1M
⇒ with V1 = the volume we need of the stock
⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L
⇒ with M2 = the concentration of the new solution = 0.15 M
1*V1 = 0.15*(50)
V1 = 7.5 mL
Since 0.0075 L of 1M solution contains 0.0075 moles
50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M
We need 7.5 mL of the 1M stock of NaCl
