" There will be a net movement of oxygen from outside the cell to inside the cell " Statement is True.
Explanation:
The partial pressure for oxygen in alveoli is greater under normal circumstances, and oxygen moves neatly into the blood. In addition, the partial carbon dioxide pressure throughout the blood usually is higher, such that carbon dioxide migrate clearly into the alveoli.
The few common molecules which can traverse the cell membrane by absorption (or diffusion of a sort recognized as osmosis) are water, carbon dioxide and oxygen. Metabolism is typically oxygen-needed, which is lowest in the cell within the animal and plant, so that net oxygen flows to the cell.
for it to be balanced in this case would be " <em>4</em> C6H6 + <em>6</em> CI2 = <em>3</em> C6H5CI + <em>9</em> HCI" therefore it's be a <u>Double Replacement</u>
Calcium forms an ion with a positive 2 charge and chlorine forms an ion with a negative one charg, so the formula is <span>CaC<span>l2</span></span>
Group 1 metals and group 2 metals form positive ions by losing 1 and 2 electrons respectively. Non-metals in group 17 gain 1, group 16 gain 2 and group 15 gain 3. Elements which lose electrons form positive ions while elements that gain electrons form negative ions.
To write a formula, you must balance charges so the overall charge is zero. A simple way to do this is to swap the # of the ion's charge and make it the subscript of the other ion. However, leave off the number 1 and reduce to lowest whole number ratio.
Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or 
)
Given : kf = 1.86 
m = 0.907 )
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86 * 0.907 )
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C
