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BARSIC [14]
3 years ago
11

Chemistry screen shot below plzzzzzzz help i've been stuck forever

Chemistry
2 answers:
Oxana [17]3 years ago
8 0

Answer:

True

Explanation: Imagine the Electrons is by the nucleus which give more energy.

slamgirl [31]3 years ago
4 0
The answer would be true :)
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J.J. Thomson's model of the atom includes all BUT ONE of these features. That is
kakasveta [241]

Answer:

D

Explanation:

usa testprep!!!!!

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3 years ago
What is the different between a graph representing data that is directly proportional and a graph that is inversely proportional
sesenic [268]
A graph depicting a direct relation is a straight line and usually has positive slope. An inverse relation is a curve, typically concave up with negative slope.
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4 years ago
What is relative isotopic mass
lidiya [134]

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Relative atomic mass or atomic weight is a dimensionless physical quantity defined as the ratio of the average mass of atoms of a chemical element in a given sample to the atomic mass constant. The atomic mass constant is defined as being 1/12 of the mass of a carbon-12 atom.

8 0
3 years ago
12. How much mass is in a 3.25-mole sample of NH 4 OH? A. 10.8 g B. 34.0 g C. 35.1 g D. 114 g
Galina-37 [17]

Answer:

D. 114 g

Explanation:

  • NH4OH molecular weight. Molar mass of NH4OH = 35.0458 g/mol This compound is also known as Ammonium Hydroxide.
  • Convert grams NH4OH to moles or moles NH4OH to grams. Molecular weight calculation: 14.0067 + 1.00794*4 + 15.9994 + 1.00794.
8 0
3 years ago
Lead−206 is the end product of 238u decay. one 206pb atom has a mass of 205.974440 amu. (a) calculate the binding energy per nuc
Dovator [93]
The atomic number for Pb is 82
∴ Pb has 82 protons and 206-82 = 14 protons
The actual mass of Pb nuclei is
=(82 × mass of the proton) + (124 × mass of neutron)
=(82× 1.00728) + (124 × 1.008664) amu
= 207.6713 amu
The mass of lead which is given is 205.9744 amu
∴mass defect is
m = 207.6713 - 205.9744 = 1.6969 amu
=1.6969 × 1.66054 × 10⁻²⁷kg
=2.818 × 10⁻²⁷kg
The binding energy is E = mc²
C is the speed of light in vacuum = 2.9979 × 10⁸m/s
∴ E = 2.532 × 10×⁻¹⁰ J/mol
= 2.532 × 10⁻¹⁰ × 6.023 × 10²³ J/mol 
= 1.53811 × 10¹⁴ J/mol

8 0
3 years ago
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