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Katarina [22]
3 years ago
6

The table below provides some information about an unidentified element. Based on this information, the unidentified element is

best classified as which of the following?
A) a metal in group 1
B) a metal in group 14
C) a nonmetal in period 4
D) a metalloid in period 5

Chemistry
1 answer:
Leona [35]3 years ago
5 0

In the physical properties, it is mentioned that the element has 4 valence electrons. The elements in the periodic table are arranged such that, the number of valence electrons present in the neutral atoms belonging to a particular group  is equal to the group number.

Thus, the unidentified element can be best classified as a nonmetal in period 4

Ans C)

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A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
Read 2 more answers
Acidic solutions contain high concentrations of
finlep [7]

Answer:

hydrogen ions

Explanation:

because acid is the specie that have ability to donate proton or forming bond with electron pair

6 0
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Calculate the amount of HCl in grams required to react with 3.75 g of CaCO3 according to the following reaction: CaCO3(s) + 2 HC
wolverine [178]

Answer:

The correct answer is 2.75 grams of HCl.

Explanation:

The given balanced equation is:  

CaCO₃ (s) + 2HCl (aq) ⇒ CaCl₂ (aq) + H₂O (l) + CO₂ (g)

Based on the given information, one mole of calcium carbonate is reacting with two moles of HCl. The molecular mass of HCl is 36.5 grams, thus, the mass of 2 moles of HCl will be, 36.5 × 2 = 73 grams

The molecular mass of CaCO₃ is 100 gram per mole, that is, the mass of 1 mole of CaCO₃ is 100 grams, therefore, the mass of HCl required for reacting with 3.75 grams of CaCO₃ will be,  

= 3.75 × 2 × 36.5 / 100 = 2.74 grams of HCl.  

8 0
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I think c is correct answer
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