Answer:
Explanation:
Hello,
In this case, given that 1 inch equals 2.54 cm and 1 lb equals 453.6 g we apply the following conversion factor in order to compute the required density in lb/in³:
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Answer:
There's No Answer Im Sorry
Answer:
ane, al, keto
ol, al, keto
ol, al, one
ol, ane, one.
Explanation:
The suffix –ol is used in organic chemistry principally to form names of organic compounds containing the hydroxyl (–OH) group, mainly alcohols (also phenol). The suffix was extracted from the word alcohol. The suffix also appears in some trivial names with reference to oils (from Latin oleum, oil).
Functional group is a ketone, therefore suffix = -one
Hydrocarbon structure is an alkane therefore -ane
The longest continuous chain is C5 therefore root = pent
The first point of difference rule requires numbering from the left as drawn to make the ketone group locant 2-
pentan-2-one or 2-pentanone
CH3CH2CH2C(=O)CH3
Answer:
The molecules or ions in a solution can freely intermingle.
Explanation:
The reactant molecules and ions can move freely in solution, so thy are more likely to collide with each other and form products.
A is wrong. Solutions are homogeneous.
B and C are wrong. The statements are correct, but they are not the reason why reactions are carried out in solution.
D is wrong. Chemical reactions do not always produce a colour change.
F is wrong. Some chemicals react with water, so other (often flammable) solvents must be used.
The first step is to make a balanced chemical equation.
2AgNO3 + CaCl2 ---> 2AgCl + Ca(NO3)2
Molecular Weights:
CaCl2 = 110.98 g/mol
AgNO3 =170.01
AgCl: 143.45 g/mol
Volume:
CaCl2: 30.0mL=0.03L
AgNO3: 15.0mL=0.015 L
Solving for the limiting reactant one needs to get the mols CaCl2 and mols AgNO3:
CaCl2: 0.150M(mol/L) * 0.03L = 0.0045 moles
AgNO3: 0.100M*0.015L = 0.0015 moles
Since the stoichiometric ratio of AgNO3 to CaCl2 is 2:1
0.0015 mols AgNO3 *(1 mol CaCl2/ 2 mols AgNO3) = 0.00075 mols CaCl2
Since the answer is lesser than CaCl2 then the limiting reactant is AgNO3.
To get the mass of AgCl one will do a stoichiometric calculation with respect to the limiting reactant, AgNO3.
0.0015 moles AgNO3 *