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Andreyy89
3 years ago
9

Question 2 (1 point)

Chemistry
1 answer:
gregori [183]3 years ago
5 0

Answer:

John Dalton

Explanation:

Both John Dalton and Democritus thought that the atom was an indivisible sphere until J.J. Thompson came out with the plum pudding model. Hope I helped!

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If you have 547.3 grams of Ni2O, how many molecules would be present?
jasenka [17]

Answer: 8.830418848725065

Explanation:

8.830418848725065

5 0
3 years ago
Find the total number of atoms in a sample of cocaine hydrochloride, c17h22clno4, of mass 23.0 mg .
Nataliya [291]
From the periodic table:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
mass of nitrogen = 14 grams
mass of oxygen = 16 grams
mass of chlorine = 35.5 grams
Therefore,
molar mass of <span>c17h22clno4 = 17(12) + 22(1) + 35.5 + 14 + 4(16) = 339.5 grams

number of moles = mass / molar mass
number of moles = (23*10^-3) / (339.5)
number of moles = 6.77 * 10^-5 moles

number of atoms = number of moles * Avogadro's number
number of atoms = 6.77*10^-5 * 6.022*10^-23
number of atoms = 4.079 * 10^-27 atoms</span>
3 0
4 years ago
3.5 grams is equivalent to how many kilograms? 3500 kg 35,000 kg 0.35 kg 0.0035 kg
bagirrra123 [75]
There are 1000 grams in a kg.
To convert g to kg, dovide by 1000.

3.5/1000= 0.0035 kg

Final answer: D
4 0
3 years ago
En una cucharadita de glucosa ( c6 h12 o6) hay en 3'4x10^22 moléculas, ¿Cuántos gramos de glucosa hay?
Alika [10]

Answer:

1 \: mole = 6.02 \times  {10}^{23}  \: molecules \\ ( \frac{3.4 \times  {10}^{22} }{6.02 \times  {10}^{23} } ) \: moles \:  =  \: 3.4 \times  {10}^{22}  \\  = 0.056 \: moles

\small{ \star{ \underline{ \blue{becker}}}}

6 0
3 years ago
gThe atomic radii of a divalent cation and a monovalent anion are 0.085 nm and 0.125 nm, respectively. (a) Calculate the force o
frozen [14]

Answer:

1.05 × 10⁻⁸ N

Explanation:

Using Coulomb's law

F force of attraction between them = k \frac{q^{2} }{r^{2} } = k \frac{z1z2q^{2} }{r^{2} }

K coulomb constant = 8.99 × 10⁹ N m²C⁻²

Z₁ = + 2 ( valency electron)

Z₂ = -1 ( valency electron)

r is the distance between then since they just touch one another = 0.085 nm + 0.125 nm = 0.21 nm = 0.21 × 10⁻⁹ m

q charge = 1.602 × 10⁻¹⁹ C of an electron

F = ( 2 × 1 × 8.99 × 10⁹ N m²C⁻² × (1.602 × 10⁻¹⁹ C)² / ( 0.21 × 10⁻⁹ m )² = 1046.34 × 10 ⁻¹¹ = 1.05 × 10⁻⁸ N

6 0
3 years ago
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