Answer:
19.8% of Nitrogen
Explanation:
In the Al(NO₃)₃ there are:
1 atom of Al
3 atoms of N
And 9 atoms of O
The molar mass of Al(NO₃)₃ is:
1 Al * (26.98g/mol) = 26.98g/mol
3 N * (14g/mol) = 42g/mol
9 O * (16g/mol) = 144g/mol
26.98 + 42 + 144 = 212.98g/mol
We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:
2.57g * (42g/mol / 212.98g/mol) =
0.51g N
Percent composition of nitrogen is:
0.51g N / 2.57g * 100
= 19.8% of Nitrogen
Moles of Zn: 26 / 65 = 0.4
Moles of S: 12.8 / 32 = 0.4
Molar ratio of Zn : S = 1 : 1
Empircal formula: ZnS
The answer is C
5. 25 x 10⁻⁷mg
Explanation:
This is mass conversion from mg to kg;
The kg is a quantity of mass used to measure the amount of matter in a substance.
Given mass = 5.25 x 10⁻¹³kg
The kilo- is a prefix that denotes 10³
therefore;
1000g = 1kilogram
the milli- is a prefix that denotes 10⁻⁻³
1000mg = 1g
Now that we know this, we can convert:
5.25 x 10⁻¹³kg x 
= 5. 25 x 10⁻¹³ x 10⁶mg
= 5. 25 x 10⁻⁷mg
learn more:
Conversion brainly.com/question/1548911
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There are 66 neutrons in a single atom of indium-115. The atomic number of indium-115 is 49, meaning there are 49 protons. Then the atomic mass is 115, so 115-49 = 66.
Answer:
B = (2.953 × 10⁻⁹⁵) N.m⁹
Explanation:
At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.
That is,
Fa + Fr = 0
Fa = - Fr
Fa = (|q₁q₂|)/(4πε₀r²)
Fr = -B/(r^n) but n = 9
Fr = -B/r⁹
(|q₁q₂|)/(4πε₀r²) = (B/r⁹)
|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C
(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²
r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m
(k|q₁q₂|)/(r²) = (B/r⁹)
(k × |q₁q₂|) = (B/r⁷)
B = (k × |q₁q₂| × r⁷)
B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]
B = (2.953 × 10⁻⁹⁵) N.m⁹
