Answer:
They produce ions when dissolved in water.
Explanation:
Acids and bases have the characteristic in common to each other. Both of them have the property of reacting and dissolving in the water. Both acids and bases lead to the production of the ions when they are placed in a water solution. Acids produce Hydrogen ions when they are dissolved in water. Bases produce hydroxide ion when they are dissolved in water.
would be produced from the
complete reaction of 83.7 g carbon
monoxide?
Fe2O3 + 3CO → 2Fe + 3CO2
131.6 grams of carbon dioxide would be produced from the complete reaction of 83.7 g carbon monoxide.
The balanced chemical equation is given below.
Fe2O3 + 3CO → 2Fe + 3CO2
Calculation,
Since, 28g of carbon dioxide produces 44g of carbon monoxide.
So, 83.7 g of carbon dioxide produces 44×83.7/28 grams
83.7 g of carbon dioxide produces 131.6 grams
The symbolic representation of chemical reaction in which reactant represents in left side and product represents in right side is called chemical equation.
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Answer:131.6 g
Explanation:
Answer:
p orbitals only
Explanation:
Carbon has an atomic number of 6 so its electron configuration will be 1s² 2s² 2p². It has two orbitals as indicated with the 2 as its period number with the outer orbital have 4 valence electrons. So carbon is in the p-orbital, period 2 and in group 4.
Carbon's valence electrons reside in the 2s and 2p orbitals. These orbitals hybridize during bond formation to create equivalent sp3 hybrid orbitals, as evidenced in the methane molecule. Carbon's valence electrons are not placed in d orbitals.
Carbon (atomic number 6) has a total of six electrons. Two of these fill the 1s orbital. The next two fill the 2s orbital, and the final two are in the 2p subshell. According to Hund's rule, the most stable configuration for an atom is one with the maximum number of unpaired electrons. Therefore, carbon has two electrons in the 2s subshell and two unpaired electrons in two separate 2p orbitals. When discussing valence electrons, the electrons in the outermost shell are the ones considered, which for carbon are the electrons in the second shell namely 2s and 2p.
The geometry of the methane molecule (CH4) illustrates that in the bonding process, the s and p orbitals hybridize to allow the formation of four equivalent bonds with hydrogen atoms. Without hybridization, we would expect three bonds at right angles (from the p orbitals) and one at a different angle (from the s orbital). Nonetheless, through orbital hybridization, all four bonds in methane are identical, which is explained by the concept of sp3 hybridized orbitals.
Therefore, the valence electrons for carbon would be placed in the s orbital and p orbitals, not in the d orbitals, because carbon does not have electrons in the d subshell in its ground state. Additionally, the s and p orbitals are the only ones involved in bonding for carbon in most of its compounds, such as methane.
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Answer:
Explanation:
check below for explanation.
The mole fraction of HNO3 is 0.225
Explanation:
1.Given data
Density = 1.429 /ml
Mass% = 63.01 g HNO3 / 100g of solution
The mass of 63.01 g is in 100 / 1.142 /ml of solution
Or 63.01 g in 55.7 mL
Molarity = 15.39 moles / L
Mass of water in 100g = 100 - 63.01=36.99 g
So 63.01 grams in 36.99 grams of water
So mass of HNO3 in 1000grams of water = 63.01* x 1000 / 36.99 = 1703
Moles of HNO3 in 1000g = 1703 / 63.01 = 27.03 moles
Molality = 27.03 molal (mole / Kg)
Mole fraction = Mole of HN03 / Moles of water + mole of HNO3
Mole of water = 62/ 18 = 3.44
Moles of HNO3 = 63.01 / 63.01 = 1.000
Mole fraction = 1.000 / 3.44 + 1.000 = 0.225
The mole fraction of HNO3 is 0.225
When a student is warming a chemical in a container using a special burner, it is very important to focus on safety by using the right safety tools.
First, the student needs to wear the right safety clothes like a lab coat, gloves, and goggles to protect themselves from getting splashed or hurt by chemicals. A lab coat stops chemicals from touching the skin, gloves keep the hands safe, and safety goggles protect the eyes from chemicals
and hot things.
Furthermore, using a fume hood is necessary to make sure there is enough fresh air circulating and to remove any dangerous fumes or gases that might be released while heating things up.
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Answer:The student should be wearing a lab coat or maybe an apron to prevent chemicals from spilling or exploding onto their clothes, I do recommend a lab coat better though because it can protect your skin better. Next, make sure while messing with chemicals you are always wearing goggles, if you are not wearing them there is a chance that after touching chemicals you could touch your eyes. And that brings me to washing your hands straight away after messing with chemicals. You could also wear gloves and just take them off when you're done but if you don't have clean hands afterward you could always put the chemicals all over your skin. But in case you do touch your eyes there is always an emergency eyewash station somewhere in the lab room. And if you are to get Chemicals on your skin, in your hair, on your clothes, or to be on fire, there shall be a shower somewhere to get rid of that. But if you read the instructions or listen closely to the teacher you shall have no problem.
Explanation:
I kinda got off topic
Answer:
Structure in attachment.
Explanation:
The oxymercuration-demercuration of an asymmetric alkene usually produces the Markovnikov orientation of an addition. The electrophile ⁺Hg(OAc), formed by the electrophile attack of the mercury ion, remains attached to least substituted group at the end of the double bond. This electrophile has a considerable amount of positive charge on its two carbon atoms, but there is more positive charge on the more substituted carbon atom, where it is more stable. The water attack occurs on this more electrophilic carbon, and the Markovnikov orientation occurs.
In hydroboration, borane adds to the double bond in one step. Boron is added to the less hindered and less substituted carbon, and hydrogen is added to the more substituted carbon. The electrophilic boron atom adds to the less substituted end of the double bond, positioning the positive charge (and the hydrogen atom) at the more substituted end. The result is a product with the anti-Markovnikov orientation.