Is a electron found in a subtonic particles in a atomic nucleus

PLEASE HELP ME!!!

ohaa [14]

Answer:

1045

Answer: 1045 J of energy was released on cooling the down the water from 20 °C to 10 °C.

4 0

3 years ago

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In what stage of wastewater treatment are large objects such as rags, sticks and garbage removed from influent?

Reptile [31]

Primary treatment !(:

8 0

3 years ago

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5. How much heat (in calories) is absorbed by a reaction when

Wewaii [24]

Answer:

1.2 × 10⁴ cal

Explanation:

Given data

Mass of water (m): 300 g

Initial temperature: 80 °C

Final temperature: 40°C

We can calculate the heat released by the water ( $Q_w$ ) when it cools using the following expression.

$Q_w = c \times m \times (T_f - T_i)$

where

c is the specific heat capacity of water (1 cal/g.°C)

$Q_w = \frac{1cal}{g.\°C} \times 300g \times (40\°C - 80\°C) = -1.2 \times 10^{4} cal$

According to the law of conservation of energy, the sum of the heat released by the water ( $Q_w$ ) and the heat absorbed by the reaction ( $Q_r$ ) is zero.

$Q_w + Q_r = 0\\Q_r = -Q_w = 1.2 \times 10^{4} cal$

7 0

3 years ago

Explain what is meant by the term "excited state" as it applies to an electron. Is an electron in an excited state higher or low

nikdorinn [45]

Answer:

Excited state of an electron is the state attained by an electron after it has absorbed energy and it moves further from the nucleus.

an electron is at higher energy when excited and at lower energy when at ground state.

an excited electron is less stable due to the decrease in the nuclear force of attraction and the grounded electron is more stable due to it's close distance to the nucleus.

6 0

3 years ago

Answers to all of these

Paul [167]

Answer:

1. Percent composition of Al = 13.423 %

2.

Percent composition of Zn = 28.02 %

Percent composition of Cl = 30.6 %

Percent composition of O = 41.3 %

3. The empirical formula is C₅O₁₆

4. Molecular Formula= P₄O₆

Explanation:

Part first :

Data Given

Formula of the Molecule = Al₂ (CrO₄)₃

% of Al₂ = ?

> First of all find the atomic masses of each component in a molecule

For Al₂ (CrO₄)₃ atomic masses are given below

Al = 27 g/mol

Cr = 52 g/mol

O = 16 g/mol

> Then find the total masses of each component

2 atoms of Al = 27 g/mol x 2

= 54 g/mol

3 atoms of Cr = 52 g/mol x 3

= 156 g/mol

12 atoms of O = 16 g/mol x 12

= 192 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Al₂ (CrO₄)₃ = [27x2 + 52x3 + 16x12]

Molar Mass of Al₂ (CrO₄)₃ = 402

Now to find the mass percent of Al

Formula used to find the Mass percent of a component

Percent composition of Al = mass of Al in Molecula / molar mass of Al₂(CrO₄)₃ x 100%

Put the values

Percent composition of Al = 54 (g/mol) / 402 (g/mol) x 100%

Percent composition of Al = 13.423 %

_______________________________________

Part 2

Data Given

Formula of the Molecule = Zn(ClO₃)₂

% Zn = ?

% Cl = ?

% O = ?

> First of all find the atomic masses of each component in a molecule

For Zn(ClO₃)₂ atomic masses are given below

Zn = 65 g/mol

Cl = 35.5 g/mol

O = 16 g/mol

> Then find the total masses of each component

1 atoms of Zn= 65 g/mol x 1

= 65 g/mol

2 atoms of Cl = 35.5 g/mol x

= 71 g/mol

6 atoms of O = 16 g/mol x 6

= 96 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Zn(ClO₃)₂ = [65x1 + 35.5x2 + 16x6]

Molar Mass of Zn(ClO₃)₂ = 232g/mol

Now to find the mass percent of of each component one by one

1. Formula used to find the mass percent of Zn

Percent composition of Zn= mass of Zn in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Zn = 65(g/mol) / 232 (g/mol) x 100%

Percent composition of Zn = 28.02 %

-------------------

2. Formula used to find the mass percent of Cl

Percent composition of Cl = mass of Cl in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Cl = 71 (g/mol) / 232 (g/mol) x 100%

Percent composition of Cl = 30.6 %

---------------------

3. Formula used to find the mass percent of O

Percent composition of O = mass of O in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of O = 96 (g/mol) / 232 (g/mol) x 100%

Percent composition of O = 41.3 %

________________________________________

Part 3:

Data Given

Percentage of C = 27.3 %

Percentage of O = 72.7 %

Emperical Formula of the compound = ?

Solution:

So the compound has 27.3 % C and 72% O

First, find the mass of each of the elements in 100 g of the Compound.

C = 27.3 g

O = 72 g

Now find how many moles are there for each element in 100 g of compound

For this molar mass are required

That is

C = 12 g/mol

O = 16 g/mol

Formula Used

mole of C = mass of C / Molar mass of C

mole of C = 27.3 / 12 g/mol

mole of C = 2.275

Formula Used

mole of O = mass of O / Molar mass of O

mole of O = 72g / 16 g/mol

mole of O = 7.2

Divide each one by the smallest number of moles

C = 2.275 / 2.275

C = 1

O = 7.2 / 2.275

O = 3.2

Multiply the mole fraction to a number to get the whole number.

C = 1 x 5 = 5

O = 3.2 x 5 = 16

So, the empirical formula is C₅O₁₆

______________________________________

Part 4

Data Given

Percentage of P= 56.38 %

Percentage of O = 43.62%

Molar Mass = 219.9g

Molecular Formula of the compound = ?

Solution:

First, find the mass of each of the elements in 100 g of the Compound.

Mass of P= 56.38g

Mass of O = 43.62g

Now find how many moles are there for each element in 100 g of compound

find the moles in total compounds

Formula Used

mole of P = mass of / Molar mass of P

mole of P = 56.38 g / 31 g/mol

mole of P = 1.818

Formula Used

mole of O = mass of O / Molar mass of O

mole of O = 43. 62 / 16 g/mol

mole of O = 2.7262

Now

first find the Emperical formula

Divide each one by the smallest number of moles

P = 1.818 /1.818

P= 1

for oxygen

O = 2.7262 / 1.818

O = 1.5

Multiply the mole fraction to a number to get the whole number.

P = 1 x 2 = 2

O = 1.5 x 2 = 3

So, the empirical formula is P₂O₃

Now

Find molar mass of the empirical formula P₂O₃

2 (31) + 3 (16) = 62 + 48 = 110

Now find that how many empirical units are in a molecular unit.

(219.9 g/mol) / ( 110 g/mol) = empirical units per molecular unit

empirical units per molecular unit = 1.999 =2

A here we get two empirical units in a molecular unit,

So the molecular formula is:

2 (P₂O₃) = P₄O₆

7 0

3 years ago