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vladimir1956 [14]
3 years ago
9

How many significant figures are in 11.005 g​

Chemistry
1 answer:
Annette [7]3 years ago
8 0

Answer:

Number of Significant Figures: 5

The Significant Figures are 1 1 0 0 5

Explanation:

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__________ ___________ can release different forms of energy such as electricity or light, in addition to the release of heat en
Stells [14]

Answer:

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Explanation:

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8 0
2 years ago
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A hypothetical covalent molecule, x–y, has a dipole moment of 1.44 d and a bond length of 163 pm. calculate the partial charge o
Aneli [31]
As we know,
                                     1 D  =  3.34 × 10⁻³⁰ C.m
So,
                                     1.44 D  =  ?
Solving for 1.44 D,
                                     =  (3.34 × 10⁻³⁰ C.m × 1.44 D) ÷ 1 D
                    
                         1.44 D  =  4.80 × 10⁻³⁰ C.m

Dipole Moment 
is given as,
 
                         Dipole Moment  =  q  ×  r    
Solving for q,
                         q  =  Dipole Moment / r    ------ (1)
Where,
                         Dipole Moment  =  4.80 × 10⁻³⁰ C.m

                         r  =  163 pm  =  1.63 × 10⁻¹⁰ m

Putting values in eq. 1,

                            q  =  4.80 × 10⁻³⁰ C.m / 1.63 × 10⁻¹⁰ m

                            q  =  2.94 × 10⁻²⁰ C

As,
                            1.602 × 10⁻¹⁹ C  =  1 e⁻
So,
                             2.94 × 10⁻²⁰ C  =  X e⁻

Solving for X,

                            X  =  (2.94 × 10⁻²⁰ C × 1 e⁻) ÷ 1.602 × 10⁻¹⁹ C

                                = 0.183 e⁻

Result:
           
So one element is containing + 0.183 e⁻ while the other element is containing - 0.183 e⁻.
4 0
3 years ago
IS THISS RIGHTTTTT???
shtirl [24]

Answer: No

Explanation: For it to be a divergent boundary, the arrows would have to be pointing in opposite directions. (one points left, one points right).

3 0
2 years ago
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How many molecules does Ba3(PO4)2 contain
Natalija [7]

Answer:

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Explanation:

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6 0
2 years ago
The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl
nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0
2 years ago
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