Elements in the same group tend to have very similar properties (D). This is due to the number of valence electrons each group has.
Equation is as follow,
<span> 4 Na (s) + O</span>₂ <span>(g) → 2Na</span>₂<span>O (s)
According to equation,
91.92 g (4 moles) of Na produces = 123.92 g (2 moles) of Na</span>₂O
So,
17.4 g of Na will produce = X g of Na₂O
Solving for X,
X = (17.4 g × 123.92 g) ÷ 91.92 g
X = 23.45 g of Na₂O
<h3>
Answer: 386.67 g/mol </h3>
Explanation:
Molar Mass = Mass ÷ Mole
= 0.406 g ÷ 0.00105 mol
= 386.67 g/mol
∴ molar mass of cholesterol = 386.67 g/mol
For stainless steel different kinds of compositions are used. Based on that different series of stainless steel has been coined.
1. Series 200 - Iron alloyed with <span>chromium, nickel and manganese.
2. Series 300 - It has
a. Stainless Steel 304 - it has composition of 18% chromium and 8% Nickel
b. </span>Stainless Steel 316 - This has 18% chromium and 10% Nickel
Each kind of stainless steel is of different cost and has different applications.
Answer:
a) Aqueous LiBr = Hydrogen Gas
b) Aqueous AgBr = solid Ag
c) Molten LiBr = solid Li
c) Molten AgBr = Solid Ag
Explanation:
a) Aqueous LiBr
This sample produces Hydrogen gas, because the H+ (conteined in the water) has a reduction potential higher than the Li+ from the salt. Therefore the hydrogen cation will reduce instead of the lithium one and form the gas.
b) Aqueous AgBr
This sample produces Solid Ag, because the Ag+ has a reduction potential higher than the H+ from the water. Therefore the silver cation will reduce instead of the hydrogen one and form the solid.
c) Molten LiBr
In a molten binary salt like LiBr there is only one cation present in the cathod. In this case the Li+, so it will reduce and form solid Li.
c) Molten AgBr
The same as the item above: there is only one cation present in the cathod. In this case the Ag+, so it will reduce and form solid Ag.
