Answer:
C). Electronegativity.
Explanation:
'Electronegativity' is described as the tendency or a measure of the ability of an atom or molecule to attract electrons that leads to the formation of chemical bonds. It is denoted by 'χ', a Greek letter. The elements like Fluorine having a very high rate of electronegativity, its atom would attract the electrons more strongly while the elements with lesser electronegativity like Cesium would possess a lesser tendency to attract electrons towards it in order to form a bond. Thus, <u>option C</u> is the correct answer.
Answer:
1.09 grams
Explanation:
According to the following chemical equation:
HF + NaNO₃ -> HNO₃ + NaF
1 mol of hydrogen fluoride (HF) produces 1 mol of sodium fluoride (NaF). Thus, we first convert from mol to grams by using the molar mass (MM) of each compound:
MM(HF)= (1 g/mol x 1 H) + (19 g/mol x 1 F) = 20 g/mol HF
1 mol HF x 19.9 g/mol HF = 20 g
MM(NaF) = (23 g/mol x 1 Na) + (19 g/mol x 1 F) = 42 g/mol NaF
1 mol NaF x 42 g/mol NaF = 42 g
Thus, from 20 g of HF are produced 42 g of NaF ⇒ 20 g HF/42 g NaF. We multiply this stoichiometric ratio by the mass of NaF produced to calculate the required mass of HF:
20 g HF/42 g NaF x 2.3 g NaF = 1.09 g HF
Therefore, 1.09 grams of HF are necessary to produce 2.3 g of NaF.
Sodium chloride has a high melting and boiling point. There are strong electrostatic attractions between the positive and negative ions, and it takes a lot of heat energy to overcome them. Ionic substances all have high melting and boiling points
Answer:
Painting with an oil based paint
Galvinization
Alloying
Explanation:
Answer:
Percentage mass of copper in the sample = 32%
Explanation:
Equation of the reaction producing Cu(NO₃) is given below:
Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)
From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.
Molar mass of copper = 64 g/mol
mass of copper = number of moles * molar mass
mass of copper = 0.01 mol * 64 g/mol = 0.64 g
Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%
Percentage mass of copper in the sample = 32%
