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stiks02 [169]
3 years ago
15

I need help ASAP!!! I need to know the model and the person who discovered it please!

Chemistry
2 answers:
QveST [7]3 years ago
6 0

Answer:

its actually the first modern atomic theory and it was created by john dalton in 1808.

Explanation:

brainliest?

Crank3 years ago
4 0

Answer:

i think electron cloud model

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What should go in the blank?
dem82 [27]

Answer:

³⁸₂₀Ca.

Explanation:

³⁸₁₉K –> __ + ⁰₋₁β

Let ʸₓA represent the unknown.

Thus the equation above can be written as:

³⁸₁₉K –> ʸₓA + ⁰₋₁β

Thus, we can obtain the value of y an x as follow:

38 = y + 0

y = 38

19 = x + (–1)

19 = x – 1

Collect like terms

19 + 1 = x

x = 20

Thus,

ʸₓA => ³⁸₂₀A => ³⁸₂₀Ca

Therefore, the equation is:

³⁸₁₉K –> ³⁸₂₀Ca + ⁰₋₁β

6 0
3 years ago
How many moles of sulfur dioxide are produced with 6.0 mols of carbon dioxide?
kodGreya [7K]

Answer:

228.38g of CO2

Explanation:

6 0
3 years ago
11.23 mm = ____ m 1,123 m 0.01123 m 0.1123 m 11,230 m
kaheart [24]
For every meter, the equivalent measurements is 1000 millimeters. Hence in the problem where the number of millimeters is given, we divide the number by 1000 to get the number of meters. The answer here is 0.01123 m.
3 0
3 years ago
Read 2 more answers
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
3 years ago
PLEASE HELP WITH ALL
Gala2k [10]

Answer:

Question 1: D

Question 2: B C E F

Question 3: A

Question 4: A

Question 5: A

5 0
4 years ago
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