First situation:
IV: soda, gatorade, orange juice, and water
DV: state of IV listed above
Control: freezer, and ice tray
Second Situation:
IV: laundry detergent, water
DV: result of the squares after being washed
Control: chocolate, type of cloth, squares of cloth
Third Situation:
IV: Water used, pea plant
DV: growth of pea plant
Control: pots and amount of water plant gets each day
Answer:
A polysaccharide (n) can be formed by linking several monosaccharides through glycosidic linkages.
Explanation:
Polysaccharides are carbohydrates or complex carbohydrates, where monosaccharides join with glucosidic bonds to form a more complex structure that would be the polysaccharide.
An example of a polysaccharide is starch, or glycogen.
Starch is found in many foods such as potatoes or rice, and glycogen is a form of energy reserve of our organism housed in muscles and liver to fulfill locomotion, physical activity, and other activities that consist of glycolysis.
Polysaccharides are degraded in our body by different stages, and several enzymes unlike monosoccharides or disaccharides, since they have more unions and a more complex structure to disarm in our body and thus assimilate it.
Polysaccharides are also part of animal structures, such as insect shells or nutritional sources, among others.
<u>Answer:</u> The number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively
<u>Explanation:</u>
In a chemical equation, the chemical species are termed as reactants or products.
Reactants are defined as the species which react in the reaction and are written on the left side of the reaction arrow.
Products are defined as the species which are produced in the reaction and are written on the right side of the reaction arrow.
For the given chemical equation:
On the reactant side:
Number of carbon atoms = (6 × 2) = 12
Number of hydrogen atoms = (14 × 2) = 28
Number of oxygen atoms = (2 × 19) = 38
Hence, the number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively
Answer:
A Reaction
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
B Non Reaction
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
Y > Q > W > Z > X
Explanation:
The first question is whether a reaction will occur base on the chemical equation below.
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
Firstly, some element are more reactive than others , base on this criteria element can be arranged base on it reactivity .
1. Mn(s) + Ca(NO₃)₂(aq)
This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.
2. KOH(aq) + Fe(s)
The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.
3. Pt(NO₃)₂(aq) + Cu(s)
Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.
4. Cr(s) + H₂SO₄(aq)
Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.
Base on the reaction
Q + W+ Reaction occurs
Since the reaction occurred element Q is more reactive as it displace element w from it compound.
X + Z+ No reaction
No reaction occurred because element x is less reactive than z therefore, it cannot displace z from it compound.
W + Z+ Reaction occurs
Element w is more reactive than z as it displaces z form it compound.
Q+ + Y Reaction occurs
Element Y is more reactive than element Q as it displaces Q from it compound.
Therefore, the order of reactivity from the most reactive to the least reactive will be Y > Q > W > Z > X
0.17 M is the is the molal concentration of this solution
Explanation:
Data given:
freezing point of glucose solution = -0.325 degree celsius
molal concentration of the solution =?
solution is of glucose=?
atomic mass of glucose = 180.01 grams/mole
freezing point of glucose = 146 degrees
freezing point of water = 0 degrees
Kf of glucose = 1.86 °C
ΔT = (freezing point of solvent) - (freezing point of solution)
ΔT = 0.325 degree celsius
molality =?
ΔT = Kfm
rearranging the equation:
m = 
m= 0.17 M
molal concentration of the glucose solution is 0.17 M
