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mixer [17]
3 years ago
8

How many atoms are in 26.8 g Fe

Chemistry
1 answer:
Tatiana [17]3 years ago
6 0

Answer:

2.881x10^23 atoms

Explanation:

From the studies of Avogadro's hypothesis, we discovered that 1mole of any substance contains 6.02x10^23 atoms.

Therefore 1mole of Fe contains 6.02x10^23 atoms.

Molar Mass of Fe = 56g/mol

56g of Fe contains 6.02x10^23 atoms.

Therefore, 26.8g of Fe will contain = (26.8x6.02x10^23) / 56 = 2.881x10^23 atoms

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Determine if the individual bonds in each compound are polar or nonpolar.
almond37 [142]

Answer:

See explanation

Explanation:

For SCl2

S = 2.8, Cl = 3.16 difference = 0.4 polar

H2S

H = 2.2  S= 2.8  difference = 0.6  polar

CF4

C = 2.55,  F= 3.98 difference = 1.43

This compound is nonpolar irrespective of the difference in electronegativity because the molecule is symmetrical

PCl5

P= 2.19,  Cl= 3.16  difference = 0.97

This compound is nonpolar irrespective of the difference in electronegativity because the molecule is symmetrical

CH4

C = 2.55    H = 2.2    difference = 0.35 the molecule is nonpolar

CaBr2

Ca= 1    Br= 2.96  difference = 1.96

The compound is not just polar but ionic in nature due to large electronegativity difference between the bonding atoms

SF2

S = 2.8   F= 3.98     difference = 1.18

The molecule is polar

CO2

C= 2.55    O= 3.44  difference = 0.89

This compound is nonpolar irrespective of the difference in electronegativity because the molecule is symmetrical

NH3

N= 3.04     H = 2.2    difference =0.84

The molecule is polar

NaS

Na= 0.93   S = 2.8    difference = 1.87

The compound is not just polar but ionic in nature due to large electronegativity difference between the bonding

4 0
3 years ago
An ion with 5 protons, 6 neutrons, and a charge of 3+ has an atomic number of
user100 [1]

Answer:1. 5

Explanation: the element is boron and the atomic number is 5

6 0
3 years ago
How much heat in joules and in calories is required to heat at 28.4g( 1 oz) ice cube from -23C TO 1.0C
Tom [10]

1426.58 J  and 340.90 calories heat in joules and in calories is required to heat at 28.4g( 1 oz) ice cube from -23C TO 1.0C.

Explanation:

Data given:

mass = 28.4 gram

initial temperature = -23 degrees

final temperature = 1degress

change in temperature  ΔT = Tfinal - Tinitial

ΔT =  1 -(-23)

 ΔT = 24 degrees

specific heat capacity of ice cube c = 2.093 J/g C

Formula used:

q = mc ΔT

putting the values in the equation:

q= 28.4 x 2.093 x 24

  = 1426.58 J

ENERGY IN CALORIES:

340.90 calories is the energy is required in the process.

3 0
3 years ago
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