Answer:
1) CO₂
2) 0.2551 g
Explanation:
The balanced reactions are:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂
1) The gas produced is CO₂.
2) Calculate mass of CaCO₃:
(0.5236 g) (0.4230) = 0.2215 g CaCO₃
Convert to moles:
(0.2215 g CaCO₃) (1 mol / 100.1 g) = 0.002213 mol CaCO₃
Find moles of CaCO₃:
(0.002213 mol CaCO₃) (1 mol CO₂ / mol CaCO₃) = 0.002213 mol CO₂
Convert to mass:
(0.002213 mol CO₂) (44.01 g / mol) = 0.09738 g CO₂
Calculate mass of MgCO₃:
(0.5236 g) (0.5770) = 0.3021 g MgCO₃
Convert to moles:
(0.3021 g MgCO₃) (1 mol / 84.31 g) = 0.003583 mol MgCO₃
Find moles of MgCO₃:
(0.003583 mol MgCO₃) (1 mol CO₂ / mol MgCO₃) = 0.003583 mol CO₂
Convert to mass:
(0.003583 mol CO₂) (44.01 g / mol) = 0.1577 g CO₂
Total mass of CO₂:
0.09738 g CO₂ + 0.1577 g CO₂ = 0.2551 g CO₂
Answer:
The correct answer is C. differentiation.
Explanation:
Cell differentiation is a process via which generic embryonic cells are transformed into specialized cells through a process of gene expression. Certain signals from inside and outside our body initiates gene expression.
Cell differentiation occurs at various stages of development, which causes the size and the shape of the cell to change dramatically.
Importance of cell differentiation:
- Plays vital role in embryonic development
- Helps in developing organisms throughout their life
Octopus are invertebrates
Answer: c. alluvial fan.
A wide sloping deposit of sediment formed where a stream leaves a mountain range is called an alluvial fan. An alluvial fan is the deposition of sediment on a landform like mountain. It forms as an open fan or cone of sediment. The sediments are left by natural drainage system like river on the landform.
There are two types of alluvial fans.
1. Debris dominated: These includes viscous mixture of water, mud, gravel along with woody debris. This transfer large boulders of landform soil.
2. Floodwater dominated: Water will spill in the alluvial fan in the form of thin sheets. This can transfer fine particles of landform soil.
Answer:The ratio of the concentrations of 
and 
when the buffer has a pH of 7.02 is 0.69
Explanation:
The dissociation constant for formic acid = 
Concentration of HA= 0.5 mM
Concentration of = 0.1 mM
pH = 6.16
First we have to calculate the value of 
.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[A^-}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BA%5E-%7D%7B%5BHA%5D%7D)
Now put all the given values in this expression, we get:
![6.16=pK_a+\log (\frac{[0.1]}{0.5})](https://tex.z-dn.net/?f=6.16%3DpK_a%2B%5Clog%20%28%5Cfrac%7B%5B0.1%5D%7D%7B0.5%7D%29)
To calculate the ratio of the concentrations of and when the buffer has a pH of 7.02.
Using Henderson Hesselbach equation :
![7.02=6.86+\log \frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=7.02%3D6.86%2B%5Clog%20%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
![\frac{[A^-]}{[HA]}=1.44](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%3D1.44)
![\frac{[HA]}{[A^-]}=0.69](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHA%5D%7D%7B%5BA%5E-%5D%7D%3D0.69)
Thus the ratio of the concentrations of and when the buffer has a pH of 7.02 is 0.69
