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2 years ago

Select the correct answer.

Chemistry

1 answer:

hichkok12 [17]2 years ago

6 0

Answer:

A. The model could not explain how alpha particles could be deflected at large angles.

Explanation:

The plum pudding model of the atom was proposed by J. J Thomson. He suggested that that an atom is made up of dense particles of electrons surrounded by positive charges.

From the Gold foil experiment carried out by Ernest Rutherford, he noticed that the bulk of the alpha particles targeted at the foil passed through and a little fraction was heavily deflected back.

Rutherford then presented his nuclear model from here. He suggested a massive, dense and tiny nucleus where the protons and neutrons are located. The space outside the mass is dominated by orbiting electrons.

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Given the balanced equation representing a reaction:

notsponge [240]

Answer:

it is a base because they accept H+ ions

Explanation:

4 0

3 years ago

A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

Answer: The mass percent of potassium bromide in the mixture is 9.996%

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For lead (II) bromide:

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0

2 years ago

Through the complete electrolysis of a sample of pure water, a student collects 14.0 grams of hydrogen gas and 112.0 grams of ox

Vadim26 [7]

Answer:

126.0g of water were initially present

Explanation:

The electrolysis of water occurs as follows:

2H₂O(l) ⇄ 2H₂(g) + O₂(g)

Where 2 moles of water produce 2 moles of hydrogen and 1 mole of oxygen.

To find the mass of water we need to determine moles of oxygen and hydrogen, thus:

Moles Hydrogen:

14.0g H₂ ₓ (1mol / 2g H₂) = 7 moles H₂

Moles Oxygen:

112.0g O₂ ₓ (1mol / 32g) = 3.5 moles O₂

Based on the chemical equation, the moles of water initially present were 7 moles (That produce 7 moles H₂ and 3.5 moles O₂). The mass of 7 moles of H₂O is:

7 moles H₂O * (18g / mol) =

<h3>126.0g of water were initially present</h3>

6 0

3 years ago

What are atoms?

liq [111]

Answer:

4. particles of matter

8 0

2 years ago

Read 2 more answers

WILL GIVE BRAINLIEST HURRY ASAP!!!!!