Which one of these elements was suggested as transition element by Mendeleev?

2. If 300.0 mL of a 2.5 M solution of nitric acid is added to 500.0 mL of water, what is the molarity of the

Afina-wow [57]

Answer:To calculate the Molarity of a 70 wt. % Nitric Acid the number of moles of HNO 3 present in 1 liter of acid needs to be calculated. Knowing the density of the acid to be 1.413 g/mL, we can calculate the weight of 1 L of 70% HNO 3 to be 1413

Explanation:

8 0

3 years ago

URGENT!??! Please hurry ?!!!

Montano1993 [528]

Answer:

Table which was completed correctly is :

B) Answer choice G

Explanation:

Metals : Most of the elements in the periodic comes in the category of metal. They are found in left and middle of periodic table.

List of metals

Alkali : Li , Na , K ,Rb , Cs ,Fr

Alkaline : Be , Mg , Ca , Sr ,Ba ,Ra

Transition : Sc ,Ti ,V ,Cr ,Mn , Fe , Co, Ni Cu , Zn ,Y ,Zr ,Nb ,Mo ,Tc ,Ru , Rh , Pd , Ag , Cd , In , Sn, La ,Hf ,Ta ,W ,Re ,Os ,Ir ,Pt ,Au ,Hg

Non -Metals: Most of the p-block element are non metal. They are present in upper - right corner of the periodic table.

Non - metals : H , He , C ,N , O , F ,P ,S , Cl, Ar , Br , Xe ,At ,Rn,

Metalloid : They show properties of both the metals and the non - metals.

Also called as semimetals

Metalloid : B ,Si ,Ge ,As ,Sb ,Te ,Po

In table F :

B = metalloid , N = non metal ( wrong-match) , Ra = metal ( wrong-match) , He = Non - metal ( wrong-match)

In table G : Correctly matched

I = Non - metal , Ca = metal , Rb = metal , As = metalloid

In table H:

Ni = metal ( wrong-match) , C = metalloid ( wrong-match) , Mg = metal , Pb = Metal

In table J:

Cu = metal , Sb = Metalloid( wrong-match) , S = non metal( wrong-match) ,Ne = non metal

6 0

3 years ago

Science is an attempt to explain _____________

krek1111 [17]

Answer:

A for sure

Explanation:

6 0

3 years ago

Read 2 more answers

A sample weighing 3.110 g is a mixture of Fe 2 O 3 (molar mass = 159.69 g/mol) and Al 2 O 3 (molar mass = 101.96 g/mol). When he

grandymaker [24]

Answer:

The mass fraction of ferric oxide in the original sample : $\frac{723}{3110}$

Explanation:

Mass of the mixture = 3.110 g

Mass of $Fe_2O_3=x$

Mass of $Al_2O_3=y$

After heating the mixture it allowed to react with hydrogen gas in which all the ferric oxide reacted to form metallic iron and water vapors where as aluminum oxide did not react.

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$