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Paul [167]
3 years ago
11

Which of the following observations indicates that there is a small, dense, positively charged part in the center of an atom?

Chemistry
1 answer:
ser-zykov [4K]3 years ago
7 0

Answer:

Option D

Explanation:

Rutherford deduced that the atomic nucleus was positively charged because the alpha particles that he fired at the metal foils were positively charged, and like charges repel. Alpha particles consist of two protons and two neutrons, so they are positively charged. In Rutherford's experiments most of the alpha particles passed straight through the foil without being deflected. However, occasionally the alpha particles were deflected in their paths, and rarely the alpha particles were deflected backward at a 180 degree angle.

Since like charges repel, Rutherford concluded that the cause of the deflections of the positively charged alpha particles had to be something within the atom that was also positively charged. Rutherford concluded from his metal foil experiments that most of an atom is empty space with a tiny, dense, positively charged nucleus at the center that contains most of the mass of the atom.

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A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

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4 0
3 years ago
Brainliest for correct answer please show all work
Korvikt [17]

Answer:

1) Na₃PO₄ + 3 KOH ➙ 3 NaOH + K₃PO₄

2) MgF₂ + Li₂CO₃➙ MgCO₃ + 2 LiF

3) P₄ + 3 O₂➙ 2 P₂O₃

Explanation:

To balance an equation, ensure that the number of atoms of each element is equal on both sides.

Reactants would be those on the left of the arrow while products are on the right of the arrow.

Balance O and H atoms last.

<u>Question 1:</u>

__Na₃PO₄ + __KOH ➙ __NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 1Na, 1P, 3K, 5O, 1H

<em>Balance the number of Na:</em>

__Na₃PO₄ + __KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 3Na, 1P, 3K, 7O, 3H

<em>Balance the number of K:</em>

__Na₃PO₄ + 3 KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 3K, 7O, 3H

Products: 3Na, 1P, 3K, 7O, 3H

<em>The equation is now balanced.</em>

<u>Question 2:</u>

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + __LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 1F, 1Li, 1C, 3O

<em>Balance</em><em> </em><em>n</em><em>u</em><em>m</em><em>b</em><em>e</em><em>r</em><em> </em><em>of</em><em> </em><em>L</em><em>i</em><em> </em><em>and</em><em> </em><em>F</em><em> </em><em>atoms</em><em>:</em>

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + 2 LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 2F, 2Li, 1C, 3O

<em>The</em><em> </em><em>equation</em><em> </em><em>is</em><em> </em><em>now</em><em> </em><em>balanced</em><em>.</em>

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Reactants: 4P, 2O

Products: 2P, 3O

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__P₄ + __O₂➙ 2 P₂O₃

Reactants: 4P, 2O

Products: 4P, 6O

<em>Balance</em><em> </em><em>the</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>O</em><em> </em><em>atoms</em><em>:</em>

__P₄ + 3 O₂➙ 2 P₂O₃

Reactants: 4P, 6O

Products: 4P, 6O

<em>The</em><em> </em><em>equation</em><em> </em><em>is</em><em> </em><em>now</em><em> </em><em>balanced</em><em>.</em>

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Answer: 960 degrees celsius

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