Answer:
D.) ±0.01 cm?
Explanation:
Since 32.23 cm has two decimal places, the uncertainty is taken as one-half the last decimal pace.
The last decimal place is 0.03. Half of this is 0.03 cm/2 = 0.015 cm.
Since we cannot go below two decimal places, we ignore the 5 in 0.015 cm.
So, we have our uncertainty as 0.01 cm.
So, the best expression of the uncertainty in the measurement 32.23 cm is ± 0.01 cm.
So, the answer is D. which is ± 0.01 cm.
Help me please , I got 0.003 for a I need help with b and c
Answer:
(a) The moles of CuSO₄ is 3.125 × 10⁻³ moles.
(b) The moles of Cu is 3.125 × 10⁻³ moles.
(c) The mass of Cu is 0.2 g.
Explanation:
Given:
Mass of CuSO₄ = 0.5 g
Molar mass of CuSO₄ = 160 g/mol
The given balanced chemical equation is:
[tex]2Al+3CuSO_4\rightarrow 3Cu+Al_2(SO_4)_3[/tex]
Part (a):
Calculating the moles of CuSO₄.
[tex]\text{Moles of } CuSO_4=\frac{\text{Mass of }CuSO_4}{\text{Molar mass of }CuSO_4}\\\\\text{Moles of } CuSO_4=\frac{0.5g}{160g/mol}\\\\\text{Moles of } CuSO_4=3.125\times 10^{-3}mol[/tex]
Thus, the moles of CuSO₄ is 3.125 × 10⁻³ moles.
Part (b):
Calculating the moles of Cu.
From the balanced chemical equation, we conclude that:
As, 3 moles of CuSO₄ reacts to give 3 moles of Cu
So, 3.125 × 10⁻³ moles of CuSO₄ reacts to give 3.125 × 10⁻³ moles of Cu
Thus, the moles of Cu is 3.125 × 10⁻³ moles.
Part (c):
Calculating the mass of Cu.
Mass of Cu = Moles of Cu × Molar mass of Cu
Molar mass of Cu = 64 g/mol
Mass of Cu = (3.125 × 10⁻³ mole) × (64 g/mol)
Mass of Cu = 0.2 g
Thus, the mass of Cu is 0.2 g.
4. After reaching the final titration endpoint the solution will be cloudy white. As time goes on the solution will turn back to a cloudy dark purple color. Why does this occur if you have already reached the endpoint
Answer: hello some part of your question is missing below is the missing part
In an experiment to determine the % of ascorbic acid in Vitamin C Tablets by Titration with Potassium Bromate,
answer:
Oxidation half reaction of Vitamin C
Explanation:
The solution will turn cloudy dark purple even after reaching endpoint when allowed to settle with time. because of the Oxidation half reaction of Vitamin C. also during the Titration process few drops of starch solution will be added to help determine the endpoint of the experiment .
Predict the products from theses reaction, and balance the equations. Include phase symbols.
Reaction : K(s)+Cl2(g)⟶
Reaction :Cu(s)+O2(g)⟶
Answer:
2 K(s) + Cl₂(g) ⟶ 2 KCl(s)
2 Cu(s) + O₂(g) ⟶ 2 CuO(s)
Explanation:
Both reactions are synthesis reactions (two substances combine to form another).
Reaction: K(s) + Cl₂(g) ⟶
The product is the binary salt KCl. The balanced chemical equation is:
2 K(s) + Cl₂(g) ⟶ 2 KCl(s)
Reaction: Cu(s) + O₂(g) ⟶
The most likely product is the metal oxide CuO. The balanced chemical equation is:
2 Cu(s) + O₂(g) ⟶ 2 CuO(s)
Organic compounds undergo a variety of different reactions, including substitution, addition, elimination, and rearrangement. An atom or a group of atoms in a molecule is replaced by another atom or a group of atoms in a substitution reaction. In an addition reaction, two molecules combine to yield a single molecule. Addition reactions occur at double or triple bonds. An elimination reaction can be thought of as the reverse of an addition reaction. It involves the removal of two atoms or groups from a molecule. A rearrangement reaction occurs when bonds in the molecule are broken and new bonds are formed, converting it to its isomer. Classify the following characteristics of the organic reactions according to the type of organic reaction.
a. Reactions involving the replacement of one atom or group of atoms.
b. Reactions involving removal of two atoms or groups from a molecule.
c. Products show increased bond order between two adjacent atoms.
d. Reactant requires presence of a π bond.
e. Product is the structural isomer of the reactant.
1. Substitution reaction
2. Addition reaction
3. Elimination reaction
4. Rearrangement reaction
Answer:
Reactions involving the replacement of one atom or group of atoms. - Substitution reaction
Reactions involving removal of two atoms or groups from a molecule - Elimination reaction
Products show increased bond order between two adjacent atoms - Elimination reaction
Reactant requires presence of a π bond - Addition reaction
Product is the structural isomer of the reactant - Rearrangement reaction
Explanation:
When an atom or a group of atoms is replaced by another in a reaction, then such is a substitution reaction. A typical example is the halogenation of alkanes.
A reaction involving the removal of two atoms or groups from a molecule resulting in increased bond order of products is called an elimination reaction. A typical example of such is dehydrohalogenation of alkyl halides.
Any reaction that involves a pi bond is an addition reaction because a molecule is added across the pi bond. A typical example is hydrogenation of alkenes.
Rearrangement reactions yield isomers of a molecule. Rearrangement may involve alkyl or hydride shifts in molecules.
Reactions involving the replacement of one atom or group of atoms is substitution reaction, reactions involving removal of two atoms or groups from a molecule and products show increased bond order between two adjacent atoms is elimination reaction, reactant requires presence of a π bond in addition reaction and product is the structural isomer of the reactant is rearrangement reaction.
What is chemical reaction?Chemical reactions are those reactions in which reactants undergoes through a variety of changes for the formation of new product.
Substitution reaction: In this reaction any atom or molecule of reactant is replaced by any outside atom or molecule.Addition reaction: In this reaction addition of any reagent takes place across the double or triple bond of any reactant for the formation of product.Elimination reaction: In this reaction any molecule or two atoms will eliminate from the reactant as a result of which we get a bond order increased product.Rearrangement reaction: In this reaction atoms or bonds of a reactant get rearranged for the formation of new product.Hence, classification of above points are done according to their characteristics.
To know more about chemical reactions, visit the below link:
https://brainly.com/question/26018275
What is the observation of heating of iodine crystals
Answer:
On heating, the van der Waals dispersion forces existing then will easily break as it has a low boiling point and sublimates into gas. On heating iodine in the test tube, iodine evolves as violet fuming gas.
Explanation:
A student named a particular compound 2-ethyl-3-methyl-2-butene. Assuming that the student's choice actually corresponded to the correct distribution of the double bond and the substituents, what is the correct IUPAC name for this compound
Answer:
2-ethyl-3-methylbut-2-ene
Explanation:
The whole idea of IUPAC nomenclature is to devise a universally accepted system of writing the name of a compound from its structure.
According to IUPAC nomenclature, the root of the compound is the longest carbon chain. The substituents are named in alphabetical order and in such a way as to give each one the lowest number. The position of the functional group is indicated accordingly.
For the compound in question, its correct IUPAC name is 2-ethyl-3-methylbut-2-ene.
A 14.570 g sample of CaCl2 was added to 12.285 g of K2CO3 and mixed in water. A 3.494 g yield of CaCO3 was obtained.
What is the limiting reagent?
-CaCO3
-K2CO3
-CaCl2
Calculate the percent yield of CaCO3.
yield of CaCO3= %
Answer:
Limiting reagent is the potassium carbonate.
Percent yield of calcium carbonate is: 39.3 %
Explanation:
The reaction is:
CaCl₂ + K₂CO₃ → CaCO₃ + 2KCl
Formula for percent yield is:
(Produced yield / Thoeretical yield) . 100
Firstly we determine the moles of each reactant, in order to say what is the limiting reagent: ratio is 1:1.
1 mol of chloride need 1 mol of carbonate.
14.570 g . 1 mol /110.98 g = 0.131 moles of CaCl₂
12.285 g . 1 mol / 138.2g = 0.0889 moles of carbonate.
Limiting reagent is carbonate. For 0.131 moles of CaCl₂ we need the same amount of carbonate and we have less moles.
Ratio is also 1:1, with calcium carbonate.
1 mol of potassium carbonate produces 1 mol of calcium carbonate
then, 0.0889 moles will produce the same amount of CaCO₃
We convert moles to mass: 0.0889 mol . 100.08g /mol = 8.89 g
That's the theoretical yield; to find the percent yield:
(3.494 g / 8.89g) . 100 = 39.3%
list some applications of chemistry in your dail life
Chemistry and chemical reactions are not just limited to the laboratories but also the world around you.
Chemistry in Food Production:
Plants produce food for themselves through photosynthesis; which is a complex chemical reaction in itself. The chemical reaction that takes place in photosynthesis is the most common and vital chemical reaction.
6 CO2 + 6 H2O + light → C 6H12O6 + 6 O2
Chemistry in Hygiene:
Right before you consume your food, you make it a point to wash your hands with soap. Isn’t it? The cleaning action of soap is based on its ability to act as an emulsifying agent. Soaps are fatty acids salts of sodium or potassium; produced by a chemical reaction called saponification. Soaps interact with the grease or oil molecule, which, in turn, results in a cleaner surface.
The Chemistry of an Onion:
Ever wondered why you shed tears while chopping an onion? This also happens because of the underlying chemistry concepts. As soon as you slice an onion, sulfenic acid is formed from amino acid sulfoxides. Sulfenic acid is responsible for the volatile gas, propanethiol S-oxide, that stimulates the production of tears in the eyes.
Chemistry in Baking:
Who does not like to eat fluffy freshly baked bread? Baking soda is an efficient leavening agent. The addition of baking soda to food items before cooking leads to the production of carbon dioxide (CO2); which causes the foods to rise. This whole process of rising of baked good is called chemical leavening.
Chemistry in Food Preservatives:
In case you ever read the ingredients on the bottle of ketchup, jams or pickles, you might be surprised to see a never-ending list of chemicals. What are they? These chemicals are called food preservatives; which delay the growth of microorganisms in foods. The chemical food preservatives not only prohibit the growth of bacteria, virus, fungi but also hinder the oxidation of fats, which is responsible for making the foods rancid. The most common chemical food preservatives are sodium benzoate, sorbic acid, potassium sorbate, calcium sorbate, sodium sorbate, propionic acid, and the salts of nitrous acid.
Chemistry in Digestion
The moment you put food in your mouth, a number of different chemical reactions start in your digestive tract. Saliva contains the enzyme amylase, which is responsible for breaking down carbohydrates, the stomach starts producing hydrochloric acid, the liver releases bile and the list of compounds released during digestion goes on. How do they work? All these enzymes undergo chemical reactions so that proper digestion, as well as assimilation of the food, occurs.
The Working of a Sunscreen
Before going out on a sunny day, you make it a point to wear sunscreen. Even the principle, behind the working of a sunscreen, has a chemistry background. The sunscreen uses a combination of organic and inorganic compounds to act as a filter for incoming ultraviolet rays. Sunblocks, on the other hand, scatter away UV light; so that it is unable to penetrate deep into the skin. Sunblocks contain complex chemical compounds like zinc oxide or titanium oxide, which prevent the UV rays to invade deeper into the skin.
Chemistry in Rust Formation
With time, your iron instruments start developing an orange-brown flaky coating called rust. The rusting of iron is a type of oxidation reaction. The atoms in the metal iron undergo oxidation and reduction; causing rusting. The formation of verdigris on copper and the tarnishing of silver are also the other everyday examples of chemical reactions. The chemical equation underlying rusting is:
Fe + O2 + H2O → Fe2O3. XH2O
Hope it helps.
How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?
Answer:
0.375mL of solution of nalorphine must be injected
Explanation:
A solution of 0.40% (w/v) contains 0.40g of solute (In this case, nalorphine), in 100mL of solution. To obtain 1.5mg of nalorphine = 1.5x10⁻³g of nalorphine are needed:
1.5x10⁻³g * (100mL / 0.40g) =
0.375mL of solution of nalorphine must be injectedTrộn 100ml dung dịch H2SO4 0,03M với 200ml dung dịch HCl 0,03M và 0,001mol Ba(OH)2 0,05M . Hãy tính pH của dung dịch này?
Answer:
pH = 1.92Explanation:
[H+] = 0.1x0.03x2 + 0.2x0.03 = 0.012 mol
[OH-] = 0.001x0.05x2 = 0.0001 mol
=> [H+] dư = 0.012 - 0.0001 =0.0119 mol
pH = -log[H+] = 1.92
