Density = mass/volume = 80 g/10cm³ = 8 g/cm³
Answer:
D. chlorine, oxygen, nitrogen, hydrogen.
Explanation:
- Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.
<em>ν ∝ 1/√M</em>
where ν is the rate of effusion and M is the atomic or molecular mass of the gas particles.
- The molecular mass for the listed gases are:
O₂: 32.0 g/mol,
Cl₂: 70.906 g/mol,
N₂: 28.0 g/mol,
H₂: 2.0 g/mol.
- Hence, the smallest molecular mass of the gas, the fastest rate of effusion.
So, the order from the slowest to the fastest rate of effusion is:
<em>Chlorine, oxygen, nitrogen, hydrogen.</em>
Explanation:
Two acids we come into contact with in an average week
- Vinegar is an 10% solution of acetic acid
in water. Used in salad dressing and while cooking food. It has a sour taste. - Citric acid present in fruits and vegetables like : lemons, orange, tomatoes etc. It is a weak organic acid with sour taste.
Two bases we come into contact with in an average week.
- Baking soda (
) is used in baking food like: cakes, cookies, breads. Baking soda is one of the ingredient while baking breads and cakes. - Caustic soda (NaOH) is used for preparation of detergents, papers , soaps etc. We use soaps and detergents for washing.
Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.